similar to: Has And Belongs To Many and Pagination

Hi, Why can''t I use a variable to specify the :per_page attribute when using the most basic form of pagination? (if I use a hardcode number, it runs) What did I do? Thx. My code: def list_orders page_size = 20 ... order_pages, @orders = paginate(:orders, :per_page => page_size) end My Errors: undefined method `>'' for false:FalseClass RAILS_ROOT:

hi everybody, i am new to rails.. plz clear my doubt... this is my question In my table(public_topics), i have 11 rows of data.. whenever i run the following view, it shows the entire rows in the current page.. and i have set the '':per_page'' option to ''2''.so it must show two data per page..but it shows all the rows in a single page..and this case continue

Hi Guys, I am trying to paginate the results from a search. When I use actual constants in the search conditions, it works fine: def query @k2_result_pages, @k2_results = paginate (:k2_results, :conditions => [''e_date >= ? AND e_date <= ?'', ''2006-07-12'', ''2006-08-12''],

I am having this problem with pagination query in my controller class. I am using SQL server and following code to create pagination fails, complains about column name ''program_name'' (which is valid column/attribute of program) The error message is: DBI::DatabaseError: S0022 (207) [Microsoft][ODBC SQL Server Driver][SQL Server]Invalid column name

I am trying to use pagination with the order option. I am following examples I''ve seen all over the web, but I am getting an error. My code seems simple: def list @upload_pages, @uploads = paginate :uploads, :per_page => 20, :order => ''id'' end Yet I keep getting this error: ___________________________________________

I''m using will_paginate in a Post class method; [code]def self.list(page) paginate :per_page => 10, :page => page, :include => :user end[/code] I can happily use the method on my posts controller index action; [code] class PostsController < ApplicationController def index @posts = Post.list(params[:page]) respond_to do |format|

I have the following: def index begin @restaurant_pages, @restaurants = paginate :restaurants, :order => (params[ :sort ] || "name"), :per_page => 2 @cuisines = Cuisine.find_all rescue redirect_to :action => :index end end # sort by cuisine def

I have searched for days for an example that demonstrates what i would like to do, and this morning i thought i''d worked it out, but no....here''s what i have: Controller ========== class ProfilesController < ApplicationController def friends current_user.profile.friends end def list_friends @profile_pages, @profiles = paginate( friends, :per_page => 10) end end

> Unfortunately I still get the same error. Here is my code now (the > commented line works): > > def list > @upload_pages, @uploads = paginate(:uploads, :per_page => > 20, :order > => ''id'') > # @upload_pages, @uploads = paginate(:uploads, :per_page => 20) > end You may have an older version, try using the order_by clause

Hi everyone, I''m at the end of my rope on this. I can''t get pagination to work with anything but a standard find on a model. If I try to do a search and customize the pagination, I get lots of different variations. My thought was to have the list action do what it does, but to pass it a list of search conditions from the search action. So, if search determines that we need

Hello, I want to know how can I paginate a set of records. It seems there is no direct way of doing this, and I am not able to understand the documentation of paginate, so any pointers on how to do this will be much appreciated. To be specific what I want is that the records returned by user.recipes, the recipes of the particular user be paginated. My attempt to do this failed because

Is it possible to paginate rss feeds with FeedTools. The following code generates an error about constants. latest_feeds = FeedTools::Feed.open(''http://feeds.feedburner.com/Eribium'') @news_pages, @pages = paginate :latest_feeds, :per_page => 5 -- Posted via http://www.ruby-forum.com/.

I am a Rails newbie and I am trying to build a test site in which users will have their own sets of data. I understand how to manipulate shared data very well. Most of the Rails examples out there seem to deal with such shared data. I am struggling with the "best practice" method for a user to work with their own data however (and not be able to view/edit other users''

Hi, please help... In controller... def index @customers = Customer.find(:all) pagination code here.... end In index.html.erb Displayes list of customers and has pagination 1,2,3,4...etc. On clicking page 2, it will again run this(Customer.find(:all)) query in controller. So, there will be lack of performance. So, how to increase the mysql performace? Can i use the index as below or any

There must be a better way to write this code: @project_pages, @projects= paginate :project, :per_page => 10, :conditions => ["account_id = ?", account] ?! If only I could pass the sub-collection account = ... @projects = account.project to paginate, instead of letting it extract it with a find :all + sql conditions Alain. -- Posted via

hi i know i can do: @customer_pages, @customers = paginate :customers, :per_page => 8 but these customers have helpdesk calls which are child records. I hoped i could do something like paginate customers.calls, but dat doesn''t work what is the best way to get a paginated list of child records? remco -- Posted via http://www.ruby-forum.com/.

hi; i use will_paginate the first page come true but all other page get the same data of first page i checked in controller "page" parameter works true but paginate get only first page''s data and shows these data in all pages forexample first page seen A,B second page also seen A,B but second page must be C,D my code is; controller : @list = Company::Env.paginate :page =>

I am trying to do a simple search by product name. The first page returns fine, but when I try to go to any other pages of the search results, I lose the original search parameters from my search form and get NilClass errors. Here is the code in my controller: namesearch = "%" + params[:name].strip + "%" @product_pages, @products = paginate :product,

I have my list view filtering rows shown within the table <tr> tags like most of the introductory tutorials do. As I''m adding a few dozen entries I''m noticing the standard scaffolding pagination doesn''t pick up the number of rows from the filter since it is not being passed the number of rows returned within the scope of the filter statement. >From a

Hi, So I''m using @post_pages, @posts = paginate :posts, :per_page => 10 to paginate. But I have rails 2.0.2 on my VPS. Now this version of paginate throws an error. What happened to classic pagination and what is done now? Thanks, Ron --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Ruby on