Hi: I have a table called clients and each client has many lists (a has_many and belongs_to has been created in the class. What I need to do is get a count of the rows in each list for each client. I was hoping I could do something like client.lists.count but it does not seem to work. Any suggestions? bruce
client.lists.size ? Does each list object have more than one element? That''d make sense given the name, but I''m not entirely sure. If that''s the case, you could do it with a join. Pat On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote:> Hi: > > I have a table called clients and each client has many lists (a > has_many and belongs_to has been created in the class. > > What I need to do is get a count of the rows in each list for each > client. I was hoping I could do something like > > client.lists.count but it does not seem to work. Any suggestions? > > bruce > > > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails >
bruce balmer wrote: > client.lists.count but it does not seem to work. Any suggestions?> > bruceTry: clients.lists.length jt -- Posted via http://www.ruby-forum.com/.
Pat: Thanks. That''s great. However, (if only life were that simple) I also want to get a subset of all the rows, so I don''t think ".size" will work. I know how to do it with a join, I was just trying to do it by associations because it seemed neater. Anyhow, I appreciate your help so far and will apply it immediately. bruce On 28-Jan-06, at 8:16 PM, Pat Maddox wrote:> client.lists.size ? > > Does each list object have more than one element? That''d make sense > given the name, but I''m not entirely sure. If that''s the case, you > could do it with a join. > > Pat > > On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >> Hi: >> >> I have a table called clients and each client has many lists (a >> has_many and belongs_to has been created in the class. >> >> What I need to do is get a count of the rows in each list for each >> client. I was hoping I could do something like >> >> client.lists.count but it does not seem to work. Any suggestions? >> >> bruce >> >> >> _______________________________________________ >> Rails mailing list >> Rails@lists.rubyonrails.org >> http://lists.rubyonrails.org/mailman/listinfo/rails >> > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails
Actually, it was not the number of lists per client that I was seeking but the number of orgs per list. Hence why I was hoping to do something like client.lists.orgs.count() but it seems that the associations don''t go that deep. I could well be wrong, this is my first day tackling this. Bruce On 28-Jan-06, at 8:16 PM, Pat Maddox wrote:> client.lists.size ? > > Does each list object have more than one element? That''d make sense > given the name, but I''m not entirely sure. If that''s the case, you > could do it with a join. > > Pat > > On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >> Hi: >> >> I have a table called clients and each client has many lists (a >> has_many and belongs_to has been created in the class. >> >> What I need to do is get a count of the rows in each list for each >> client. I was hoping I could do something like >> >> client.lists.count but it does not seem to work. Any suggestions? >> >> bruce >> >> >> _______________________________________________ >> Rails mailing list >> Rails@lists.rubyonrails.org >> http://lists.rubyonrails.org/mailman/listinfo/rails >> > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails
Hey Bruce,
I thought that was what you might want. Not sure what the association
in the list in named, but if you want to try associations, do
something like:
count = 0
client.lists.each { |list| count += list.elements.size }
The problem with that is that I''m sure it will do a bunch of queries
that you don''t need..I''m pretty sure it''s possible to
do what you want
in a single SQL statement with a join, but unfortunately I don''t know
enough SQL to do it :(
Pat
On 1/28/06, Bruce Balmer <brucebalmer@mac.com>
wrote:> Pat:
>
> Thanks. That''s great.
>
> However, (if only life were that simple) I also want to get a subset
> of all the rows, so I don''t think ".size" will work. I
know how to
> do it with a join, I was just trying to do it by associations because
> it seemed neater.
>
> Anyhow, I appreciate your help so far and will apply it immediately.
>
> bruce
>
>
> On 28-Jan-06, at 8:16 PM, Pat Maddox wrote:
>
> > client.lists.size ?
> >
> > Does each list object have more than one element? That''d
make sense
> > given the name, but I''m not entirely sure. If
that''s the case, you
> > could do it with a join.
> >
> > Pat
> >
> > On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote:
> >> Hi:
> >>
> >> I have a table called clients and each client has many lists (a
> >> has_many and belongs_to has been created in the class.
> >>
> >> What I need to do is get a count of the rows in each list for each
> >> client. I was hoping I could do something like
> >>
> >> client.lists.count but it does not seem to work. Any
suggestions?
> >>
> >> bruce
> >>
> >>
> >> _______________________________________________
> >> Rails mailing list
> >> Rails@lists.rubyonrails.org
> >> http://lists.rubyonrails.org/mailman/listinfo/rails
> >>
> > _______________________________________________
> > Rails mailing list
> > Rails@lists.rubyonrails.org
> > http://lists.rubyonrails.org/mailman/listinfo/rails
>
> _______________________________________________
> Rails mailing list
> Rails@lists.rubyonrails.org
> http://lists.rubyonrails.org/mailman/listinfo/rails
>
gah, forgot to mention that you should change ''elements'' to whatever the assocation name is. On 1/28/06, Pat Maddox <pergesu@gmail.com> wrote:> Hey Bruce, > > I thought that was what you might want. Not sure what the association > in the list in named, but if you want to try associations, do > something like: > > count = 0 > client.lists.each { |list| count += list.elements.size } > > The problem with that is that I''m sure it will do a bunch of queries > that you don''t need..I''m pretty sure it''s possible to do what you want > in a single SQL statement with a join, but unfortunately I don''t know > enough SQL to do it :( > > Pat > > > On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: > > Pat: > > > > Thanks. That''s great. > > > > However, (if only life were that simple) I also want to get a subset > > of all the rows, so I don''t think ".size" will work. I know how to > > do it with a join, I was just trying to do it by associations because > > it seemed neater. > > > > Anyhow, I appreciate your help so far and will apply it immediately. > > > > bruce > > > > > > On 28-Jan-06, at 8:16 PM, Pat Maddox wrote: > > > > > client.lists.size ? > > > > > > Does each list object have more than one element? That''d make sense > > > given the name, but I''m not entirely sure. If that''s the case, you > > > could do it with a join. > > > > > > Pat > > > > > > On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: > > >> Hi: > > >> > > >> I have a table called clients and each client has many lists (a > > >> has_many and belongs_to has been created in the class. > > >> > > >> What I need to do is get a count of the rows in each list for each > > >> client. I was hoping I could do something like > > >> > > >> client.lists.count but it does not seem to work. Any suggestions? > > >> > > >> bruce > > >> > > >> > > >> _______________________________________________ > > >> Rails mailing list > > >> Rails@lists.rubyonrails.org > > >> http://lists.rubyonrails.org/mailman/listinfo/rails > > >> > > > _______________________________________________ > > > Rails mailing list > > > Rails@lists.rubyonrails.org > > > http://lists.rubyonrails.org/mailman/listinfo/rails > > > > _______________________________________________ > > Rails mailing list > > Rails@lists.rubyonrails.org > > http://lists.rubyonrails.org/mailman/listinfo/rails > > >
Pat: Worked beautifully. Thanks again. I think I get the concept. Associations only go two deep. From client to list for example and from list to orgs. But you can step through them with the code you wrote. Got it! Thanks. bruce On 28-Jan-06, at 9:45 PM, Pat Maddox wrote:> gah, forgot to mention that you should change ''elements'' to whatever > the assocation name is. > > > On 1/28/06, Pat Maddox <pergesu@gmail.com> wrote: >> Hey Bruce, >> >> I thought that was what you might want. Not sure what the >> association >> in the list in named, but if you want to try associations, do >> something like: >> >> count = 0 >> client.lists.each { |list| count += list.elements.size } >> >> The problem with that is that I''m sure it will do a bunch of queries >> that you don''t need..I''m pretty sure it''s possible to do what you >> want >> in a single SQL statement with a join, but unfortunately I don''t know >> enough SQL to do it :( >> >> Pat >> >> >> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>> Pat: >>> >>> Thanks. That''s great. >>> >>> However, (if only life were that simple) I also want to get a subset >>> of all the rows, so I don''t think ".size" will work. I know how to >>> do it with a join, I was just trying to do it by associations >>> because >>> it seemed neater. >>> >>> Anyhow, I appreciate your help so far and will apply it immediately. >>> >>> bruce >>> >>> >>> On 28-Jan-06, at 8:16 PM, Pat Maddox wrote: >>> >>>> client.lists.size ? >>>> >>>> Does each list object have more than one element? That''d make >>>> sense >>>> given the name, but I''m not entirely sure. If that''s the case, you >>>> could do it with a join. >>>> >>>> Pat >>>> >>>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>>>> Hi: >>>>> >>>>> I have a table called clients and each client has many lists (a >>>>> has_many and belongs_to has been created in the class. >>>>> >>>>> What I need to do is get a count of the rows in each list for each >>>>> client. I was hoping I could do something like >>>>> >>>>> client.lists.count but it does not seem to work. Any >>>>> suggestions? >>>>> >>>>> bruce >>>>> >>>>> >>>>> _______________________________________________ >>>>> Rails mailing list >>>>> Rails@lists.rubyonrails.org >>>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>>> >>>> _______________________________________________ >>>> Rails mailing list >>>> Rails@lists.rubyonrails.org >>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>> >>> _______________________________________________ >>> Rails mailing list >>> Rails@lists.rubyonrails.org >>> http://lists.rubyonrails.org/mailman/listinfo/rails >>> >> > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails
Hey Bruce, Try doing something like this num_orgs = Organization.find_by_sql(["SELECT lists.id, lists.cliend_id, organizations.list_id FROM clients, lists, organizations WHERE lists.client_id = ? AND organizations.list_id lists.id", client.id]) I don''t know if that will work or not, it''s just an idea. However if you''re going to be going through a lot of records you will only want to do one query so it''s a lot faster. I''m sure someone here will be able to come up with the exact SQL, and if not then just hop on #mysql for some help. The point is you want to do this in one statement, because it is possible to do. Pat On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote:> Pat: > > Worked beautifully. Thanks again. I think I get the concept. > Associations only go two deep. From client to list for example and > from list to orgs. But you can step through them with the code you > wrote. Got it! Thanks. > > bruce > > > On 28-Jan-06, at 9:45 PM, Pat Maddox wrote: > > > gah, forgot to mention that you should change ''elements'' to whatever > > the assocation name is. > > > > > > On 1/28/06, Pat Maddox <pergesu@gmail.com> wrote: > >> Hey Bruce, > >> > >> I thought that was what you might want. Not sure what the > >> association > >> in the list in named, but if you want to try associations, do > >> something like: > >> > >> count = 0 > >> client.lists.each { |list| count += list.elements.size } > >> > >> The problem with that is that I''m sure it will do a bunch of queries > >> that you don''t need..I''m pretty sure it''s possible to do what you > >> want > >> in a single SQL statement with a join, but unfortunately I don''t know > >> enough SQL to do it :( > >> > >> Pat > >> > >> > >> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: > >>> Pat: > >>> > >>> Thanks. That''s great. > >>> > >>> However, (if only life were that simple) I also want to get a subset > >>> of all the rows, so I don''t think ".size" will work. I know how to > >>> do it with a join, I was just trying to do it by associations > >>> because > >>> it seemed neater. > >>> > >>> Anyhow, I appreciate your help so far and will apply it immediately. > >>> > >>> bruce > >>> > >>> > >>> On 28-Jan-06, at 8:16 PM, Pat Maddox wrote: > >>> > >>>> client.lists.size ? > >>>> > >>>> Does each list object have more than one element? That''d make > >>>> sense > >>>> given the name, but I''m not entirely sure. If that''s the case, you > >>>> could do it with a join. > >>>> > >>>> Pat > >>>> > >>>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: > >>>>> Hi: > >>>>> > >>>>> I have a table called clients and each client has many lists (a > >>>>> has_many and belongs_to has been created in the class. > >>>>> > >>>>> What I need to do is get a count of the rows in each list for each > >>>>> client. I was hoping I could do something like > >>>>> > >>>>> client.lists.count but it does not seem to work. Any > >>>>> suggestions? > >>>>> > >>>>> bruce > >>>>> > >>>>> > >>>>> _______________________________________________ > >>>>> Rails mailing list > >>>>> Rails@lists.rubyonrails.org > >>>>> http://lists.rubyonrails.org/mailman/listinfo/rails > >>>>> > >>>> _______________________________________________ > >>>> Rails mailing list > >>>> Rails@lists.rubyonrails.org > >>>> http://lists.rubyonrails.org/mailman/listinfo/rails > >>> > >>> _______________________________________________ > >>> Rails mailing list > >>> Rails@lists.rubyonrails.org > >>> http://lists.rubyonrails.org/mailman/listinfo/rails > >>> > >> > > _______________________________________________ > > Rails mailing list > > Rails@lists.rubyonrails.org > > http://lists.rubyonrails.org/mailman/listinfo/rails > > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails >
I need to stop sending so quickly. num_orgs = Organization.find_by_sql(["SELECT lists.id, lists.cliend_id, organizations.list_id FROM clients, lists, organizations WHERE lists.client_id = ? AND organizations.list_id lists.id", client.id]).size On 1/28/06, Pat Maddox <pergesu@gmail.com> wrote:> Hey Bruce, > > Try doing something like this > > num_orgs = Organization.find_by_sql(["SELECT lists.id, > lists.cliend_id, organizations.list_id FROM clients, lists, > organizations WHERE lists.client_id = ? AND organizations.list_id > lists.id", client.id]) > > I don''t know if that will work or not, it''s just an idea. However if > you''re going to be going through a lot of records you will only want > to do one query so it''s a lot faster. I''m sure someone here will be > able to come up with the exact SQL, and if not then just hop on #mysql > for some help. The point is you want to do this in one statement, > because it is possible to do. > > Pat > > > On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: > > Pat: > > > > Worked beautifully. Thanks again. I think I get the concept. > > Associations only go two deep. From client to list for example and > > from list to orgs. But you can step through them with the code you > > wrote. Got it! Thanks. > > > > bruce > > > > > > On 28-Jan-06, at 9:45 PM, Pat Maddox wrote: > > > > > gah, forgot to mention that you should change ''elements'' to whatever > > > the assocation name is. > > > > > > > > > On 1/28/06, Pat Maddox <pergesu@gmail.com> wrote: > > >> Hey Bruce, > > >> > > >> I thought that was what you might want. Not sure what the > > >> association > > >> in the list in named, but if you want to try associations, do > > >> something like: > > >> > > >> count = 0 > > >> client.lists.each { |list| count += list.elements.size } > > >> > > >> The problem with that is that I''m sure it will do a bunch of queries > > >> that you don''t need..I''m pretty sure it''s possible to do what you > > >> want > > >> in a single SQL statement with a join, but unfortunately I don''t know > > >> enough SQL to do it :( > > >> > > >> Pat > > >> > > >> > > >> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: > > >>> Pat: > > >>> > > >>> Thanks. That''s great. > > >>> > > >>> However, (if only life were that simple) I also want to get a subset > > >>> of all the rows, so I don''t think ".size" will work. I know how to > > >>> do it with a join, I was just trying to do it by associations > > >>> because > > >>> it seemed neater. > > >>> > > >>> Anyhow, I appreciate your help so far and will apply it immediately. > > >>> > > >>> bruce > > >>> > > >>> > > >>> On 28-Jan-06, at 8:16 PM, Pat Maddox wrote: > > >>> > > >>>> client.lists.size ? > > >>>> > > >>>> Does each list object have more than one element? That''d make > > >>>> sense > > >>>> given the name, but I''m not entirely sure. If that''s the case, you > > >>>> could do it with a join. > > >>>> > > >>>> Pat > > >>>> > > >>>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: > > >>>>> Hi: > > >>>>> > > >>>>> I have a table called clients and each client has many lists (a > > >>>>> has_many and belongs_to has been created in the class. > > >>>>> > > >>>>> What I need to do is get a count of the rows in each list for each > > >>>>> client. I was hoping I could do something like > > >>>>> > > >>>>> client.lists.count but it does not seem to work. Any > > >>>>> suggestions? > > >>>>> > > >>>>> bruce > > >>>>> > > >>>>> > > >>>>> _______________________________________________ > > >>>>> Rails mailing list > > >>>>> Rails@lists.rubyonrails.org > > >>>>> http://lists.rubyonrails.org/mailman/listinfo/rails > > >>>>> > > >>>> _______________________________________________ > > >>>> Rails mailing list > > >>>> Rails@lists.rubyonrails.org > > >>>> http://lists.rubyonrails.org/mailman/listinfo/rails > > >>> > > >>> _______________________________________________ > > >>> Rails mailing list > > >>> Rails@lists.rubyonrails.org > > >>> http://lists.rubyonrails.org/mailman/listinfo/rails > > >>> > > >> > > > _______________________________________________ > > > Rails mailing list > > > Rails@lists.rubyonrails.org > > > http://lists.rubyonrails.org/mailman/listinfo/rails > > > > _______________________________________________ > > Rails mailing list > > Rails@lists.rubyonrails.org > > http://lists.rubyonrails.org/mailman/listinfo/rails > > >
Pat Maddox wrote:> I need to stop sending so quickly. > > num_orgs = Organization.find_by_sql(["SELECT lists.id, > lists.cliend_id, organizations.list_id FROM clients, lists, > organizations WHERE lists.client_id = ? AND organizations.list_id > lists.id", client.id]).sizeDon''t forget about counter caching. Some associations will let you add a column that gets incremented and decremented when association items are added or deleted. It''s probably much faster, but not as flexible as the queries. _Kevin -- Posted via http://www.ruby-forum.com/.
On 1/29/06, Kevin Olbrich <kevin.olbrich@duke.edu> wrote:> Pat Maddox wrote: > > I need to stop sending so quickly. > > > > num_orgs = Organization.find_by_sql(["SELECT lists.id, > > lists.cliend_id, organizations.list_id FROM clients, lists, > > organizations WHERE lists.client_id = ? AND organizations.list_id > > lists.id", client.id]).size > > Don''t forget about counter caching. Some associations will let you add > a column that gets incremented and decremented when association items > are added or deleted. It''s probably much faster, but not as flexible as > the queries. > > _Kevin > > -- > Posted via http://www.ruby-forum.com/. > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/railsAny idea what would be faster? Looping with counter caching or one monster SQL statement?
Pat Maddox wrote:> On 1/29/06, Kevin Olbrich <kevin.olbrich@duke.edu> wrote: >> are added or deleted. It''s probably much faster, but not as flexible as >> the queries. >> >> _Kevin >> >> -- >> Posted via http://www.ruby-forum.com/. >> _______________________________________________ >> Rails mailing list >> Rails@lists.rubyonrails.org >> http://lists.rubyonrails.org/mailman/listinfo/rails > > Any idea what would be faster? Looping with counter caching or one > monster SQL statement?I''m sure it would depend on the monster SQL query, but I''m guessing that retrieving and summing a few integers would be faster than finding and counting a bunch of records. In fact, I''m fairly certain that you could write a monster SQL query to sum the relevant counter cache columns for you and get the best performance from that. _Kevin -- Posted via http://www.ruby-forum.com/.
On Jan 28, 2006, at 8:44 PM, Pat Maddox wrote:> Hey Bruce, > > I thought that was what you might want. Not sure what the association > in the list in named, but if you want to try associations, do > something like: > > count = 0 > client.lists.each { |list| count += list.elements.size }clients.lists.inject(0){|count, list| count += list.elements.size } -Ezra> > The problem with that is that I''m sure it will do a bunch of queries > that you don''t need..I''m pretty sure it''s possible to do what you want > in a single SQL statement with a join, but unfortunately I don''t know > enough SQL to do it :( > > Pat > > > On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >> Pat: >> >> Thanks. That''s great. >> >> However, (if only life were that simple) I also want to get a subset >> of all the rows, so I don''t think ".size" will work. I know how to >> do it with a join, I was just trying to do it by associations because >> it seemed neater. >> >> Anyhow, I appreciate your help so far and will apply it immediately. >> >> bruce >> >> >> On 28-Jan-06, at 8:16 PM, Pat Maddox wrote: >> >>> client.lists.size ? >>> >>> Does each list object have more than one element? That''d make sense >>> given the name, but I''m not entirely sure. If that''s the case, you >>> could do it with a join. >>> >>> Pat >>> >>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>>> Hi: >>>> >>>> I have a table called clients and each client has many lists (a >>>> has_many and belongs_to has been created in the class. >>>> >>>> What I need to do is get a count of the rows in each list for each >>>> client. I was hoping I could do something like >>>> >>>> client.lists.count but it does not seem to work. Any suggestions? >>>> >>>> bruce >>>> >>>> >>>> _______________________________________________ >>>> Rails mailing list >>>> Rails@lists.rubyonrails.org >>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>> >>> _______________________________________________ >>> Rails mailing list >>> Rails@lists.rubyonrails.org >>> http://lists.rubyonrails.org/mailman/listinfo/rails >> >> _______________________________________________ >> Rails mailing list >> Rails@lists.rubyonrails.org >> http://lists.rubyonrails.org/mailman/listinfo/rails >> > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails >-Ezra Zygmuntowicz WebMaster Yakima Herald-Republic Newspaper ezra@yakima-herald.com 509-577-7732
Ezra: Thanks. bruce PS. If I should not be clogging things up with "thank you''s" I will stop. That said, as long as threads are on, it does not seem too bad. Your advice will be acted on (although, I do think it helps maintain a positive tone) On 29-Jan-06, at 1:28 PM, Ezra Zygmuntowicz wrote:> > On Jan 28, 2006, at 8:44 PM, Pat Maddox wrote: > >> Hey Bruce, >> >> I thought that was what you might want. Not sure what the >> association >> in the list in named, but if you want to try associations, do >> something like: >> >> count = 0 >> client.lists.each { |list| count += list.elements.size } > > clients.lists.inject(0){|count, list| count += list.elements.size } > > -Ezra > > >> >> The problem with that is that I''m sure it will do a bunch of queries >> that you don''t need..I''m pretty sure it''s possible to do what you >> want >> in a single SQL statement with a join, but unfortunately I don''t know >> enough SQL to do it :( >> >> Pat >> >> >> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>> Pat: >>> >>> Thanks. That''s great. >>> >>> However, (if only life were that simple) I also want to get a subset >>> of all the rows, so I don''t think ".size" will work. I know how to >>> do it with a join, I was just trying to do it by associations >>> because >>> it seemed neater. >>> >>> Anyhow, I appreciate your help so far and will apply it immediately. >>> >>> bruce >>> >>> >>> On 28-Jan-06, at 8:16 PM, Pat Maddox wrote: >>> >>>> client.lists.size ? >>>> >>>> Does each list object have more than one element? That''d make >>>> sense >>>> given the name, but I''m not entirely sure. If that''s the case, you >>>> could do it with a join. >>>> >>>> Pat >>>> >>>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>>>> Hi: >>>>> >>>>> I have a table called clients and each client has many lists (a >>>>> has_many and belongs_to has been created in the class. >>>>> >>>>> What I need to do is get a count of the rows in each list for each >>>>> client. I was hoping I could do something like >>>>> >>>>> client.lists.count but it does not seem to work. Any >>>>> suggestions? >>>>> >>>>> bruce >>>>> >>>>> >>>>> _______________________________________________ >>>>> Rails mailing list >>>>> Rails@lists.rubyonrails.org >>>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>>> >>>> _______________________________________________ >>>> Rails mailing list >>>> Rails@lists.rubyonrails.org >>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>> >>> _______________________________________________ >>> Rails mailing list >>> Rails@lists.rubyonrails.org >>> http://lists.rubyonrails.org/mailman/listinfo/rails >>> >> _______________________________________________ >> Rails mailing list >> Rails@lists.rubyonrails.org >> http://lists.rubyonrails.org/mailman/listinfo/rails >> > > -Ezra Zygmuntowicz > WebMaster > Yakima Herald-Republic Newspaper > ezra@yakima-herald.com > 509-577-7732 > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails
Bruce- I think its fine to say thanks. No worries here ;-) -Ezra On Jan 29, 2006, at 3:54 PM, Bruce Balmer wrote:> Ezra: > > Thanks. > > bruce > > PS. If I should not be clogging things up with "thank you''s" I > will stop. That said, as long as threads are on, it does not seem > too bad. Your advice will be acted on (although, I do think it > helps maintain a positive tone) > > > On 29-Jan-06, at 1:28 PM, Ezra Zygmuntowicz wrote: > >> >> On Jan 28, 2006, at 8:44 PM, Pat Maddox wrote: >> >>> Hey Bruce, >>> >>> I thought that was what you might want. Not sure what the >>> association >>> in the list in named, but if you want to try associations, do >>> something like: >>> >>> count = 0 >>> client.lists.each { |list| count += list.elements.size } >> >> clients.lists.inject(0){|count, list| count += list.elements.size } >> >> -Ezra >> >> >>> >>> The problem with that is that I''m sure it will do a bunch of queries >>> that you don''t need..I''m pretty sure it''s possible to do what you >>> want >>> in a single SQL statement with a join, but unfortunately I don''t >>> know >>> enough SQL to do it :( >>> >>> Pat >>> >>> >>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>>> Pat: >>>> >>>> Thanks. That''s great. >>>> >>>> However, (if only life were that simple) I also want to get a >>>> subset >>>> of all the rows, so I don''t think ".size" will work. I know how to >>>> do it with a join, I was just trying to do it by associations >>>> because >>>> it seemed neater. >>>> >>>> Anyhow, I appreciate your help so far and will apply it >>>> immediately. >>>> >>>> bruce >>>> >>>> >>>> On 28-Jan-06, at 8:16 PM, Pat Maddox wrote: >>>> >>>>> client.lists.size ? >>>>> >>>>> Does each list object have more than one element? That''d make >>>>> sense >>>>> given the name, but I''m not entirely sure. If that''s the case, >>>>> you >>>>> could do it with a join. >>>>> >>>>> Pat >>>>> >>>>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>>>>> Hi: >>>>>> >>>>>> I have a table called clients and each client has many lists (a >>>>>> has_many and belongs_to has been created in the class. >>>>>> >>>>>> What I need to do is get a count of the rows in each list for >>>>>> each >>>>>> client. I was hoping I could do something like >>>>>> >>>>>> client.lists.count but it does not seem to work. Any >>>>>> suggestions? >>>>>> >>>>>> bruce >>>>>> >>>>>> >>>>>> _______________________________________________ >>>>>> Rails mailing list >>>>>> Rails@lists.rubyonrails.org >>>>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>>>> >>>>> _______________________________________________ >>>>> Rails mailing list >>>>> Rails@lists.rubyonrails.org >>>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>> >>>> _______________________________________________ >>>> Rails mailing list >>>> Rails@lists.rubyonrails.org >>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>> >>> _______________________________________________ >>> Rails mailing list >>> Rails@lists.rubyonrails.org >>> http://lists.rubyonrails.org/mailman/listinfo/rails >>> >> >> -Ezra Zygmuntowicz >> WebMaster >> Yakima Herald-Republic Newspaper >> ezra@yakima-herald.com >> 509-577-7732 >> _______________________________________________ >> Rails mailing list >> Rails@lists.rubyonrails.org >> http://lists.rubyonrails.org/mailman/listinfo/rails > > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails >-Ezra Zygmuntowicz WebMaster Yakima Herald-Republic Newspaper ezra@yakima-herald.com 509-577-7732
Thanks :-) bruce On 29-Jan-06, at 6:43 PM, Ezra Zygmuntowicz wrote:> Bruce- > > I think its fine to say thanks. No worries here ;-) > > -Ezra > > On Jan 29, 2006, at 3:54 PM, Bruce Balmer wrote: > >> Ezra: >> >> Thanks. >> >> bruce >> >> PS. If I should not be clogging things up with "thank you''s" I >> will stop. That said, as long as threads are on, it does not seem >> too bad. Your advice will be acted on (although, I do think it >> helps maintain a positive tone) >> >> >> On 29-Jan-06, at 1:28 PM, Ezra Zygmuntowicz wrote: >> >>> >>> On Jan 28, 2006, at 8:44 PM, Pat Maddox wrote: >>> >>>> Hey Bruce, >>>> >>>> I thought that was what you might want. Not sure what the >>>> association >>>> in the list in named, but if you want to try associations, do >>>> something like: >>>> >>>> count = 0 >>>> client.lists.each { |list| count += list.elements.size } >>> >>> clients.lists.inject(0){|count, list| count += list.elements.size } >>> >>> -Ezra >>> >>> >>>> >>>> The problem with that is that I''m sure it will do a bunch of >>>> queries >>>> that you don''t need..I''m pretty sure it''s possible to do what >>>> you want >>>> in a single SQL statement with a join, but unfortunately I don''t >>>> know >>>> enough SQL to do it :( >>>> >>>> Pat >>>> >>>> >>>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>>>> Pat: >>>>> >>>>> Thanks. That''s great. >>>>> >>>>> However, (if only life were that simple) I also want to get a >>>>> subset >>>>> of all the rows, so I don''t think ".size" will work. I know >>>>> how to >>>>> do it with a join, I was just trying to do it by associations >>>>> because >>>>> it seemed neater. >>>>> >>>>> Anyhow, I appreciate your help so far and will apply it >>>>> immediately. >>>>> >>>>> bruce >>>>> >>>>> >>>>> On 28-Jan-06, at 8:16 PM, Pat Maddox wrote: >>>>> >>>>>> client.lists.size ? >>>>>> >>>>>> Does each list object have more than one element? That''d make >>>>>> sense >>>>>> given the name, but I''m not entirely sure. If that''s the >>>>>> case, you >>>>>> could do it with a join. >>>>>> >>>>>> Pat >>>>>> >>>>>> On 1/28/06, Bruce Balmer <brucebalmer@mac.com> wrote: >>>>>>> Hi: >>>>>>> >>>>>>> I have a table called clients and each client has many lists (a >>>>>>> has_many and belongs_to has been created in the class. >>>>>>> >>>>>>> What I need to do is get a count of the rows in each list for >>>>>>> each >>>>>>> client. I was hoping I could do something like >>>>>>> >>>>>>> client.lists.count but it does not seem to work. Any >>>>>>> suggestions? >>>>>>> >>>>>>> bruce >>>>>>> >>>>>>> >>>>>>> _______________________________________________ >>>>>>> Rails mailing list >>>>>>> Rails@lists.rubyonrails.org >>>>>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>>>>> >>>>>> _______________________________________________ >>>>>> Rails mailing list >>>>>> Rails@lists.rubyonrails.org >>>>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>>> >>>>> _______________________________________________ >>>>> Rails mailing list >>>>> Rails@lists.rubyonrails.org >>>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>>> >>>> _______________________________________________ >>>> Rails mailing list >>>> Rails@lists.rubyonrails.org >>>> http://lists.rubyonrails.org/mailman/listinfo/rails >>>> >>> >>> -Ezra Zygmuntowicz >>> WebMaster >>> Yakima Herald-Republic Newspaper >>> ezra@yakima-herald.com >>> 509-577-7732 >>> _______________________________________________ >>> Rails mailing list >>> Rails@lists.rubyonrails.org >>> http://lists.rubyonrails.org/mailman/listinfo/rails >> >> _______________________________________________ >> Rails mailing list >> Rails@lists.rubyonrails.org >> http://lists.rubyonrails.org/mailman/listinfo/rails >> > > -Ezra Zygmuntowicz > WebMaster > Yakima Herald-Republic Newspaper > ezra@yakima-herald.com > 509-577-7732 > _______________________________________________ > Rails mailing list > Rails@lists.rubyonrails.org > http://lists.rubyonrails.org/mailman/listinfo/rails