Rails - Jul 2006 - Accessing @org.id yields internal number, not record id

With apologies, I can''t figure out the simplest thing: How to reference
a record id instead of the internal memory location of that value. I''m 
new to Ruby but otherwise a veteran VB6/SQL programmer.

My code:

<snip>

     sSQL = "SELECT id, org_name FROM organizations WHERE user_id = 
''#{sUserId}'' AND user_password =
''#{sPassword}''"

     @org = Organization.find_by_sql sSQL

     if @org

        write_log("Org found. Id = #{@org.id}")


</snip>

My log file contains:

   Org found. Id = 30382648

I am expecting Id to be 1.
There are only two records in the organizations table: 1 and 2.

Thanks for putting me straight!

Michael

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Posted via http://www.ruby-forum.com/.

Michael Wilkes wrote:
> With apologies, I can''t figure out the simplest thing: How to
reference
> a record id instead of the internal memory location of that value.
I''m
> new to Ruby but otherwise a veteran VB6/SQL programmer.
> 
> My code:
> 
> <snip>
> 
>      sSQL = "SELECT id, org_name FROM organizations WHERE user_id = 
> ''#{sUserId}'' AND user_password =
''#{sPassword}''"
> 
>      @org = Organization.find_by_sql sSQL
> 
>      if @org
> 
>         write_log("Org found. Id = #{@org.id}")
> 
> 
> </snip>
find_by_sql returns an Array (since you can select multiple records).

You should also consider using parameters to avoid the potential for SQL 
injection with the above code:

@org = Organization.find_by_sql(["SELECT id, org_name FROM organizations 
WHERE user_id = ? AND user_password = ?", sUserId, sPassword]).first

-- 
Philip Ross
http://tzinfo.rubyforge.org/ -- DST-aware timezone library for Ruby

org = Organization.find :first, :conditions = ["user_id = ? and password
?", sUserId, sPassword]

logger.info "Org found. Id = #{@org.id}"


On 7/31/06, Michael Wilkes <vbsql7@gmail.com>
wrote:
>
> With apologies, I can''t figure out the simplest thing: How to
reference
> a record id instead of the internal memory location of that value.
I''m
> new to Ruby but otherwise a veteran VB6/SQL programmer.
>
> My code:
>
> <snip>
>
>     sSQL = "SELECT id, org_name FROM organizations WHERE user_id >
''#{sUserId}'' AND user_password =
''#{sPassword}''"
>
>     @org = Organization.find_by_sql sSQL
>
>     if @org
>
>        write_log("Org found. Id = #{@org.id}")
>
>
> </snip>
>
> My log file contains:
>
>   Org found. Id = 30382648
>
> I am expecting Id to be 1.
> There are only two records in the organizations table: 1 and 2.
>
> Thanks for putting me straight!
>
> Michael
>
> --
> Posted via http://www.ruby-forum.com/.
> _______________________________________________
> Rails mailing list
> Rails@lists.rubyonrails.org
> http://lists.rubyonrails.org/mailman/listinfo/rails
>
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> logger.info "Org found. Id = #{@org.id}"
Thanks for the help to both responders. There''s something I still
don''t
understand tho because attempts to access @org as an array fail for me.

==== code =====
     sSQL = "SELECT id, org_name FROM organizations WHERE user_id = 
''#{sUserId}'' AND user_password =
''#{sPassword}''"

 write_log(sSQL)

     @org = Organization.find_by_sql(sSQL).first

 if @org.nil?
    write_log("Org is nil")
 else
    write_log("Org list returned: " + @org.inspect)
 end

===============
Results in log:

SELECT id, org_name FROM organizations WHERE user_id = ''kevin''
AND
user_password = ''password''

Org list returned: #<Organization:0x38130e0
@attributes={"id"=>"1",
"org_name"=>"Carlie Couches"}>

==============
When I try to access @org.id, however, I get errors.
For instance, this line:
   write_log("Org list returned: #{@org.id}")
throws this error:
   Called id for nil, which would mistakenly be 4 -- if you really 
wanted the id of nil, use object_id

This line:
   write_log("Org list returned: #{@org.org_name}")
throws the same error.

This line:
   write_log("Org list returned: #{@org[0].org_name}")
throws this error:
   You have a nil object when you didn''t expect it!
   The error occured while evaluating nil.org_name

I''ve looked at it with inspect and to_xml and there is always data 
there.
How do I get to it? I''m missing something basic here.

Thanks for helping,
Michael

P.S. I couldn''t figure out how to pass quoted strings as parameters, 
that''s why i''m using the sSQL statement instead of parameters.
Any
example of passing a quoted string using the ? parameter method would be 
helpful too. Passing the strings without accounting for the quotes gave 
SQL syntax errors because of empty strings coming through.



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