Search Results

Search found 156 results on 7 pages for 'paginate'.

Page 5/7 | < Previous Page | 1 2 3 4 5 6 7  | Next Page >

  • What is the best way of doing this?

    - by NewSOuser
    Hey, I am wondering how exactly the below pagination systems are set up? In terms of the how the data is displayed, for instance, does BuySellAds load all results in from database and then paginate that data so filter can be done easier? or, do they call the database every time a new page is loaded and then grab the info and display it? I really would like to know how both BuySellAds and Nike make their pagination systems, since I have to make a similar system that has many filters and options to easily sort through the data from a database. I want to accomplish this with php and jquery if that's what I even need to use. Here is what I am talking about so you can experience it for yourself: BuySellAds store.nike.com Any info/suggestions would be great. Thanks for your input.

    Read the article

  • Django pagination | get current index of paginated item in page index, (not the page index range its

    - by cka
    I am trying to build a photo gallery with Django. It is set up by category. I have paginated the results of a category by n amount of images per page. I want to also use the paginator on the page that shows just the single image and have a prev/next button for the prev/next image in that category. My thought was to get the current index for the image itself and have that be the link to the /category/CUR_IMG_ID_PAGINATION_LIST/ as the result of paginating the entire set would yield the same index as the current image index in the paginated results. For instance if the image i want is image 45 out of 150 images total for a category, then when i paginate the 150 images the 45 will be the actual number of the page I want. If there's an easier way to do this, let me know. Django 1.1

    Read the article

  • will_paginate undefined method error - Ruby on Rails

    - by bgadoci
    I just installed the gem for will_paginate and it says that it was installed successfully. I followed all the instructions listed with the plugin and I am getting an 'undefined method `paginate' for' error. Can't find much in the way of Google search and haven't been able to fix it myself (obviously). Here is the code: PostsController def index @tag_counts = Tag.count(:group => :tag_name, :order => 'updated_at DESC', :limit => 10) @posts = Post.paginate :page => params[:page], :per_page => 50 respond_to do |format| format.html # index.html.erb format.xml { render :xml => @posts } format.json { render :json => @posts } format.atom end end /model/post.rb class Post < ActiveRecord::Base validates_presence_of :body, :title has_many :comments, :dependent => :destroy has_many :tags, :dependent => :destroy cattr_reader :per_page @@per_page = 10 end /posts/views/index.html.erb <%= will_paginate @posts %>

    Read the article

  • Stubbing ActiveRecord result arrays

    - by Matthias
    Using Ruby 1.8.6, stubbing a finder method to return an Array does not work when the code under test calls the 'count' method on the result: User.any_instance.stubs(:friends).returns([user1, user2]) That's because Array.count has only been added in Ruby 1.8.7. Rails also adds a count method dynamically, I believe through the ActiveRecord::Calculations module. Is there any way to turn this result array into something that behaves exactly as the special Array kind returned by a Rails finder method? When paginating results, it's easy: I can simply call [].paginate. But that doesn't work with normal finder results. I tried [].extend(ActiveRecord::Calculations) but that doesn't work either.

    Read the article

  • Error when pushing to Heroku - ...appear in group - Ruby on Rails

    - by bgadoci
    I am trying to deploy my first rails app to Heroku and seem to be having a problem. After git push heroku master, and heroku rake db:migrate I get an error saying: SELECT posts.*, count(*) as vote_total FROM "posts" INNER JOIN "votes" ON votes.post_id = posts.id GROUP BY votes.post_id ORDER BY created_at DESC LIMIT 5 OFFSET 0): I have included the full error below and also included the PostControll#index as it seems that is where I am doing the grouping. Lastly I included my routes.rb file. I am new to ruby, rails, and heroku so sorry for simple/obvious questions. Processing PostsController#index (for 99.7.50.140 at 2010-04-21 12:50:47) [GET] ActiveRecord::StatementInvalid (PGError: ERROR: column "posts.id" must appear in the GROUP BY clause or be used in an aggregate function : SELECT posts.*, count(*) as vote_total FROM "posts" INNER JOIN "votes" ON votes.post_id = posts.id GROUP BY votes.post_id ORDER BY created_at DESC LIMIT 5 OFFSET 0): vendor/gems/will_paginate-2.3.12/lib/will_paginate/finder.rb:82:in `send' vendor/gems/will_paginate-2.3.12/lib/will_paginate/finder.rb:82:in `paginate' vendor/gems/will_paginate-2.3.12/lib/will_paginate/collection.rb:87:in `create' vendor/gems/will_paginate-2.3.12/lib/will_paginate/finder.rb:76:in `paginate' app/controllers/posts_controller.rb:28:in `index' /home/heroku_rack/lib/static_assets.rb:9:in `call' /home/heroku_rack/lib/last_access.rb:25:in `call' /home/heroku_rack/lib/date_header.rb:14:in `call' thin (1.0.1) lib/thin/connection.rb:80:in `pre_process' thin (1.0.1) lib/thin/connection.rb:78:in `catch' thin (1.0.1) lib/thin/connection.rb:78:in `pre_process' thin (1.0.1) lib/thin/connection.rb:57:in `process' thin (1.0.1) lib/thin/connection.rb:42:in `receive_data' eventmachine (0.12.6) lib/eventmachine.rb:240:in `run_machine' eventmachine (0.12.6) lib/eventmachine.rb:240:in `run' thin (1.0.1) lib/thin/backends/base.rb:57:in `start' thin (1.0.1) lib/thin/server.rb:150:in `start' thin (1.0.1) lib/thin/controllers/controller.rb:80:in `start' thin (1.0.1) lib/thin/runner.rb:173:in `send' thin (1.0.1) lib/thin/runner.rb:173:in `run_command' thin (1.0.1) lib/thin/runner.rb:139:in `run!' thin (1.0.1) bin/thin:6 /usr/local/bin/thin:20:in `load' /usr/local/bin/thin:20 PostsController def index @tag_counts = Tag.count(:group => :tag_name, :order => 'count_all DESC', :limit => 20) conditions, joins = {}, :votes @ugtag_counts = Ugtag.count(:group => :ugctag_name, :order => 'count_all DESC', :limit => 20) conditions, joins = {}, :votes @vote_counts = Vote.count(:group => :post_title, :order => 'count_all DESC', :limit => 20) conditions, joins = {}, :votes unless(params[:tag_name] || "").empty? conditions = ["tags.tag_name = ? ", params[:tag_name]] joins = [:tags, :votes] end @posts=Post.paginate( :select => "posts.*, count(*) as vote_total", :joins => joins, :conditions=> conditions, :group => "votes.post_id", :order => "created_at DESC", :page => params[:page], :per_page => 5) @popular_posts=Post.paginate( :select => "posts.*, count(*) as vote_total", :joins => joins, :conditions=> conditions, :group => "votes.post_id", :order => "vote_total DESC", :page => params[:page], :per_page => 3) respond_to do |format| format.html # index.html.erb format.xml { render :xml => @posts } format.json { render :json => @posts } format.atom end end routes.rb ActionController::Routing::Routes.draw do |map| map.resources :ugtags map.resources :wysihat_files map.resources :users map.resources :votes map.resources :votes, :belongs_to => :user map.resources :tags, :belongs_to => :user map.resources :ugtags, :belongs_to => :user map.resources :posts, :collection => {:auto_complete_for_tag_tag_name => :get } map.resources :posts, :sessions map.resources :posts, :has_many => :comments map.resources :posts, :has_many => :tags map.resources :posts, :has_many => :ugtags map.resources :posts, :has_many => :votes map.resources :posts, :belongs_to => :user map.resources :tags, :collection => {:auto_complete_for_tag_tag_name => :get } map.resources :ugtags, :collection => {:auto_complete_for_ugtag_ugctag_name => :get } map.login 'login', :controller => 'sessions', :action => 'new' map.logout 'logout', :controller => 'sessions', :action => 'destroy' map.root :controller => "posts" map.connect ':controller/:action/:id' map.connect ':controller/:action/:id.:format' end UPDATE TO SHOW MODEL AND MIGRATION FOR POST class Post < ActiveRecord::Base has_attached_file :photo validates_presence_of :body, :title has_many :comments, :dependent => :destroy has_many :tags, :dependent => :destroy has_many :ugtags, :dependent => :destroy has_many :votes, :dependent => :destroy belongs_to :user after_create :self_vote def self_vote # I am assuming you have a user_id field in `posts` and `votes` table. self.votes.create(:user => self.user) end cattr_reader :per_page @@per_page = 10 end migrations for post class CreatePosts < ActiveRecord::Migration def self.up create_table :posts do |t| t.string :title t.text :body t.timestamps end end def self.down drop_table :posts end end _ class AddUserIdToPost < ActiveRecord::Migration def self.up add_column :posts, :user_id, :string end def self.down remove_column :posts, :user_id end end

    Read the article

  • Get last n lines of a file with Python, similar to tail

    - by Armin Ronacher
    I'm writing a log file viewer for a web application and for that I want to paginate through the lines of the log file. The items in the file are line based with the newest item on the bottom. So I need a tail() method that can read n lines from the bottom and supports an offset. What I came up with looks like this: def tail(f, n, offset=0): """Reads a n lines from f with an offset of offset lines.""" avg_line_length = 74 to_read = n + offset while 1: try: f.seek(-(avg_line_length * to_read), 2) except IOError: # woops. apparently file is smaller than what we want # to step back, go to the beginning instead f.seek(0) pos = f.tell() lines = f.read().splitlines() if len(lines) >= to_read or pos == 0: return lines[-to_read:offset and -offset or None] avg_line_length *= 1.3 Is this a reasonable approach? What is the recommended way to tail log files with offsets?

    Read the article

  • Removing a result from Queryset

    - by Enrico
    Is there a simple way to discard/remove the last result in a queryset without affecting the db? I am trying to paginate results in Django, but don't know the total number of objects for a given query. I was planning on using next/previous or older/newer links, so I only need to know if this is the first and/or last page. First is easy to check. To check for the last page I can compare the number of results with the pagesize or make a second query. The first method fails to detect the last page when the number of results in the last set equals the pagesize (ie 100 records get broken into 10 pages with the last page containing exactly 10 results) and I would like to avoid making a second query. My current thought is that I should fetch pagesize + 1 results from the db. If the queryset length equals 11, I know this is not the last page and I want to discard the last result in the queryset before passing the queryset to the template.

    Read the article

  • Scraping paginated items from a website using scrapy

    - by Mridang Agarwalla
    I'm using scrapy to scrape items from a site. I'm not being able to implement this scraping pattern. The site I'm trying to scrape is a forum and I scrape the site once a day. Each page has a table containing posts. New posts are added to the top of the table and as more and more posts are posted to the site, the older posts go further into the pages due to pagination. This is a very simple scenario and we will assume that the order of the posts never change. I would like to scrape this site and scrape all the "new" records until the last scraped post from yesterday is encountered. I have configured my spider to paginate endlessly and when it encounters yesterday's last scraped post, it should stop. How can implement this? (My Scrapy installation works with my Django installation using django-dynamic-scraper )

    Read the article

  • Adding custom columns to Propel model?

    - by Hard-Boiled Wonderland
    At the moment I am using the below query: $claims = ClaimQuery::create('c') ->leftJoinUser() ->withColumn('CONCAT(User.Firstname, " ", User.Lastname)', 'name') ->withColumn('User.Email', 'email') ->filterByArray($conditions) ->paginate($page = $page, $maxPerPage = $top); However I then want to add columns manually, so I thought this would simply work: foreach($claims as &$claim){ $claim->actions = array('edit' => array( 'url' => $this->get('router')->generate('hera_claims_edit'), 'text' => 'Edit' ) ); } return array('claims' => $claims, 'count' => count($claims)); However when the data is returned Propel or Symfony2 seems to be stripping the custom data when it gets converted to JSON along with all of the superflous model data. What is the correct way of manually adding data this way?

    Read the article

  • Difficulty creating a paging function with MySQL and ColdFusion

    - by Mel
    I'm trying to create pagination for search results using MySQL and ColdFusion. My intention is to only retrieve the queries that can be displayed on a single page, thus making the process efficient. I tried using two queries in my function, but I could not return two variables to the cfinvoke. The following code does not paginate, but it displays the result search results using a CFC: <!---DEFINE DEFAULT STATE---> <cfparam name="variables.searchResponse" default=""> <cfparam name="URL.titleName" default=""> <cfparam name="URL.genreID" default=""> <cfparam name="URL.platformID" default=""> <!---TitleName can only be blank if one or both genre and platform are selected---> <cfif StructKeyExists(URL, "searchQuery") AND (Len(Trim(URL.titleName)) LTE 2 AND Len(URL.genreID) IS 0 AND Len(URL.platformID) IS 0)> <cfset variables.searchResponse = "invalidString"> <cfelseif StructKeyExists(URL, "searchQuery")> <cfinvoke component="gz.cfcomp.test" method="searchGames" returnvariable="resultData" argumentcollection="#URL#"> <cfset variables.searchResponse = "hasResult"> </cfif> <cfif searchResponse EQ "hasResult" AND resultData.RecordCount EQ 0> <cfset variables.searchResponse = "noResult"> </cfif> Using this logic, I can display what I need to display on the page: <cfif searchResponse EQ "invalidString"> <cfoutput>Invalid search</cfoutput> </cfif> <cfif searchResponse EQ "noResult"> <cfoutput>No results found</cfoutput> </cfif> <cfif searchResponse EQ "hasResult"> <cfoutput>Display Results</cfoutput> </cfif> If I were executing the queries on the same page, it would be easy to follow the many tutorials out there. But the queries are executing in a function. Displaying the data is easy, but paginating it has become a nightmare for me. Here is my function: <cffunction name="searchGames" access="public" output="false"> <cfargument name="titleName" required="no" type="string"> <cfargument name="genreID" required="no" type="string"> <cfargument name="platformID" required="no" type="string"> <!--- DEFINE LOCAL VARIABLES---> <cfset var resultData = ""> <!---GET DATA---> <cfquery name="resultData" datasource="myDSN"> SELECT * <!---JOINS FOR GENRE/PLATFORM GO HERE---> WHERE <!---CONDITIONS GO HERE---> </cfquery> <!---RETURN VARIABLE---> <cfreturn resultData> </cffunction> To paginate, I thought about modifying my function to the following (a new query using a count statement): <!--- DEFINE LOCAL VARIABLES---> <cfset var resultCount = ""> <!---GET DATA---> <cfquery name="resultCount" datasource="myDSN"> SELECT COUNT(gameID) AS rowsFound FROM GAMES <!---JOINS FOR GENRE/PLATFORM GO HERE---> WHERE <!---CONDITIONS GO HERE---> </cfquery> <!---RETURN VARIABLE---> <cfreturn resultCount> Then I figured if there is a result to return, I would execute a nested query and create the pagination variables: <cfif resultCount.rowsFound GTE 0> <cfparam name="pageNumber" default="1"> <cfset var recordsPerPage = 5> <cfset var numberOfPages = Int(resultCount.RecordCount / recordsPerPage)> <cfset var recordsToSkip = pageNumber * recordsPerPage - recordsPerPage> <!---DEFINE LOCAL VARIABLE---> <cfset var resultData = ""> <cfquery name="resultData" datasource="myDSN"> <!---GET DATA AND SEND IT BACK USING LIMIT WITH #recordsToSkip# and #RecordsPerPage#---> </cfquery> <!---RETURN VARIABLE---> <cfreturn resultData> </cffunction> I figured I would return two variables: resultCount and resultData. I would use #resultCount# to build my pagination, and #resultData# to display the output. The problem is I can't return two variables in the same cfinvoke tag. Any ideas of how to approach the the right way? I'm totally lost as to the logic I need to follow.

    Read the article

  • ActionLink sending a model of a complex type

    - by Xenph Yan
    I'm trying to paginate the results of a "advanced search", I have a complex model that represents the search options; int ZipCode int MinAge int MaxAge Availability bool Monday bool Tuesday ... bool Friday Requirements bool FirstAid bool DriversLicense I'm using; <%: Html.ActionLink("Next »", "Save", "Notification", Model.options)%> Which correctly sends all the data at the first level, but anything that is a sub-object (Availability or requirements) isn't expanded in the URL, all I get is the class name and so I lose most of the search options when I click the link to change to a different page. Any thoughts?

    Read the article

  • Devise and cancan gems: has_many association

    - by tiktak
    I use devise and cancan gems and have simple model association: user has_many subscriptions, subscription belongs_to :user. Have following SubscriptionsController: class SubscriptionsController < ApplicationController load_and_authorize_resource :user load_and_authorize_resource :subscription, through: :user before_filter :authenticate_user! def index @subscriptions = @user.subscriptions.paginate(:page => params[:page]).order(:created_at) end #other actions end And Cancan Ability.rb: class Ability include CanCan::Ability def initialize(user) user ||=User.new can [:index, :show], [Edition, Kind] if user.admin? can :manage, :all elsif user.id can [:read, :create, :destroy, :pay], Subscription, user_id: user.id can [:delete_from_cart, :add_to_cart, :cart], User, id: user.id end end end The problem is that i cannot use subscriptions actions as a user but can as a admin. And have no problems with UsersController. When i delete following lines from SubscriptionsController: load_and_authorize_resource :user load_and_authorize_resource :subscription, through: :user before_filter :authenticate_user! Have no problems at all. So the issue in these lines or in Ability.rb. Any suggestions?

    Read the article

  • NameError when using act_as_ferret

    - by manish nautiyal
    Hi all I am getting this error when I am using acts_as_ferret :fields =[:competitor], :remote = true NameError in PartController#index uninitialized constant PartController::Competitor My Model class Competitor < ActiveRecord::Base validates_presence_of :fee_earner_id, :notes belongs_to :fee_earner belongs_to :country belongs_to :state belongs_to :user acts_as_ferret :fields =[:competitor], :remote = true end My controller class PartController < ApplicationController def index @proscribeds = Competitor.paginate(:all, :order = sort , :page = params[:page], :per_page = 70 ) end end Its working fine in localhost but when I deploy it in the server than I get this error. act_as_ferret is working good with other models. I don't know why this is not working with only Competitor model.

    Read the article

  • How to break up HTML documents into pages for ebook?

    - by radnoise
    For an iPhone ebook application I need to break arbitrarily long HTML documents up into pages which fit exactly on one screen. If I simply use UIWebView for this, the bottom-most lines tend to get displayed only partly: the rest disappears off the edge of the view. So I assume I would need to know how many complete lines (or characters) would be displayed by the UIWebView, given the source HTML, and then feed it exactly the right amount of data. This probably involves lots of calculation, and the user also needs to be able to change fonts and sizes. I have no idea if this is even possible, although apps like Stanza take HTML (epub) files and paginate them nicely. It's a long time since I looked at JavaScript, would that be an option worth looking at? Any suggestions very much appreciated!

    Read the article

  • QT4, paginated showing elements

    - by matiit
    I am going to write an application that uses QT4 (with C++ or python it isnt important in that moment). One of functionality is "Showing all items in database". One item has a Title, author, description and photo (constant size) And there could be very many items. Let's say 400. There won't be enough space to show'em all at once time. One row will have 200px, so i need at most 4 for once time. How to paginate them? I have no idea. I can use limit and offset in SQL queries, but how to tell window: "that's 5th page"? Any solutions?

    Read the article

  • Paging in SQL Server problems

    - by Manh Trinh
    I have searched for paging in SQL Server. I found most of the solution look like that What is the best way to paginate results in SQL Server But it don't meet my expectation. Here is my situation: I work on JasperReport, for that: to export the report I just need pass the any Select query into the template, it will auto generated out the report EX : I have a select query like this: Select * from table A I don't know any column names in table A. So I can't use Select ROW_NUMBER() Over (Order By columsName) And I also don't want it order by any columns. Anyone can help me do it? PS: In Oracle , it have rownum very helpful in this case. Select * from tableA where rownum > 100 and rownum <200 Paging with Oracle

    Read the article

  • Django template tag basic question

    - by ninja123
    It looks like this template tag works like a charm for most people: http://blog.localkinegrinds.com/2007/09/06/digg-style-pagination-in-django/ For some reason I get this error: Caught an exception while rendering: 'is_paginated' I use this template tag in my template like so: {% load digg_paginator %} {% digg_paginator %} Where digg_paginator.py is in my app/templatetags folder and the included template context digg_paginator.html is in my app/templates folder. The queryset that needs pagination is called 'destinations'. If i just specify {% digg_paginator %}, how does it know what variable to paginate?? I feel I am missing something important here or just plain stupid :P Someone please help, or explain to me how this should be done. Thanks

    Read the article

  • Need help to format the result page after searching

    - by kshama
    Hi, I have built a small text based search engine on ROR which will display relevant records having a specified search word in it.since few of the records has more than 1000 words i have truncated each result set to 200 characters.My views file search.html.erb looks like this <% @results_with_ranks.each do |result| -%> <% content_id = rtable.find(result[0]).content_id %> <% content= Content.find(content_id) %> <%= truncate content.body, :length => 200 %><br/> <p> Record id <%= content.id %></p> <hr style="color:blue"> <% end -%> I want to provide an option so that whenever any truncated record is selected its entire body has to be displayed. I also want to paginate the result page displaying some fixed number of records per page.Can any body help me in doing this? Thanks in advance.

    Read the article

  • ferret,multiple model search -undefined method `aaf_index' for #<Class:>

    - by jissy
    ferret,multiple model search - I have 2 models A and B.I want to perform a text search by using 3 fields; title, description(part of A) and comment(part of B). Where I want to include the comment field to perform the ferret search.Then,what other changes needed. class A < ActiveRecord::Base has_one :b acts_as_ferret :fields => [:title, :description], :additional_fields => [:comment_text] def comment_text return b.comment end In a_controller, i wrote: @search = A.find_with_ferret( params[:st][:text_search], :limit => :all, :multi => [B] ).paginate :per_page =>10, :page=>params[:page] The second mosel is given below: class B < ActiveRecord::Base belongs_to :a while using :multi[B] option with the find_with_ferret,the following error is getting: undefined method `aaf_index' for #ClassName

    Read the article

  • any simple jquery <table> paginators

    - by Mike
    I'm looking for a jquery plugin so I can paginate my html tables (the content of which is generated by a jsp c:forEach). But so far I haven't found anything that worked. The plugin doesn't need to be fancy, in fact something like the first example on http://dl.dropbox.com/u/4151695/html/pajinate-0.2/examples/example1.html would do fine. I'm still very new to jquery so I don't have the knowledge yet to write one myself. So does anyone have any pointers where I could find one?

    Read the article

  • jquery - How do i call the same function with different div ids?

    - by Nahom
    Hi, I have two divs with different ids (#washing,#bleaching). How can i use a function for different id's. I have tried adding both the ids together $("#washing, #bleaching"), but the function is not working correctly on the divs. Can anyone please help me... Here is the code $(function() { $("#washing").paginate({ count: 10, start: 1, display: 7, border: true, border_color: '#fff', onChange: function(page) { $('._current', '#paginationdemo').removeClass('_current').hide(); $('#p' + page).addClass('_current').show(); } }); });

    Read the article

  • Rails: show some examples of code from controllers, models and views

    - by Totty
    Hy, my controller example: class FriendsController < ApplicationController before_filter :authorize, :except => [:friends] ############## ############## ## REQUESTS ## ############## ############## ################## # GET MY FRIENDS # ################## # Get my friends. def friends @friends = @my_profile.friends.paginate({:page => params[:page], :per_page => 3}) @profile = @my_profile end ################### # REMOVED FRIENDS # ################### # Get my deleted friends. def removed_friends @removed_friends = @my_profile.friends('removed_friends', params[:page]) end ################### # PENDING FRIENDS # ################### # Friend requests made by other profiles to me. def pending_friends @pending_friends = @my_profile.friends('pending_friends', params[:page]) end ############################ # REJECTED PENDING FRIENDS # ############################ # Rejected friend requests made by other profiles to me. def rejected_pending_friends @rejected_pending_friends = @my_profile.friends('rejected_pending_friends', params[:page]) end ##################### # REQUESTED FRIENDS # ##################### # The friend requests I've sent to others profiles. def requested_friends @requested_friends = @my_profile.friends('requested_friends', params[:page]) end ############################# # DELETED REQUESTED FRIENDS # ############################# # The requests I've sent to others # profiles and then canceled. def deleted_requested_friends @deleted_requested_friends = @my_profile.friends('deleted_requested_friends', params[:page]) end ############# ############# ## ACTIONS ## ############# ############# ########################## # ADD FRIENDSHIP REQUEST # ########################## # Add a friendship request. def add_friendship_request friendship = @my_profile.add_friendship_request(params[:profile_id]) render :json => friendship end ############################# # REMOVE FRIENDSHIP REQUEST # ############################# # Removes a friendship request I've done. def remove_friendship_request friendship = @my_profile.remove_friendship_request(params[:profile_id]) render :json => friendship end ###################### # PROCESS FRIENDSHIP # ###################### # Process friendship: accept or reject a friend. # This will make a new friend or # will make a new rejected pending friend. def process_friendship friendship = @my_profile.process_friendship(params[:profile_id].to_i, params[:accepted].to_i) render :json => friendship end ################### # REMOVE A FRIEND # ################### # Remove a friend from my friends by id. def remove_friend friendship = @my_profile.remove_friend(params[:profile_id]) render :json => friendship end end

    Read the article

  • Pagination in Joomla!

    Hi every one. I'm a bit new to Joomla! and I'm building my new web site using Joomla! 1.5.8 with a template bought from RocketTheme. The long article in question is a gallery of photos, and I wanted to create a horizontal pagination nav-bar in the footer of the gallery, I just clicked pagebreak where I wanted to paginate in my article in the WYSIWYG editor and thought that it's actually all I have to do. What I got was ugly; a vertically directed navigation bar placed in the top right side of my gallery with the links to each part of the article, the all without the minimum styling. And as I see it's a table! Why not outputting a list instead of a table! (Well this is not actually my concern right now). I can understand CSS and html, but I'm not really a PHP guy, so I don't understand how to use pagination.php or pagenavigation.php! Do I have to customize one of them to have my nav-bar? What I have to do to get that horizontal nav-bar? Really thanks in advance. Wassim

    Read the article

  • Ordering view by highest group count question - Ruby on Rails

    - by bgadoci
    I've read the couple of questions about this on stack overflow but can't seem to find the answer. I am trying to display the tags in my blog by the ones with the highest count in the tags table. Thanks to KandadaBoggu for helping me get the tags feature of the blog I am designing working. Here is the basics and my question. Tag belongs_to :post and Post has_many :tags. The tags table is simple really, consisting of the normal scaffolded fields plus post_id and tag_name (I actually called the column 'tag_name' instead of just 'name'). in my /views/posts/index.html/erb file I correctly am displaying the tags by group and the amount of times they are being used (appearing in the tags table). I just want to know how to order them by the highest count. Here is the code, and I currently have it set to updated_at: PostsController def index @tag_counts = Tag.count(:group => :tag_name, :order => 'updated_at DESC', :limit => 10) conditions, joins = {}, nil unless(params[:tag_name] || "").empty? conditions = ["tags.tag_name = ? ", params[:tag_name]] joins = :tags end @posts=Post.all(:joins => joins, :conditions=> conditions, :order => 'created_at DESC').paginate :page => params[:page], :per_page => 5 respond_to do |format| format.html # index.html.erb format.xml { render :xml => @posts } format.json { render :json => @posts } format.atom end end

    Read the article

  • django url user id versus userprofile id problem

    - by dana
    hello there, i have a mini comunity where each user can search and find another user profile. Userprofile is a class model, indexed differently compared to user model class (user id is not equal to userprofile id) But i cannot see a user profile by typing in the url the corresponding id. I only see the profile of the currently logged in user. Why is that? I'd also want to have in my url the username (a primary key of the user table also) and NOT the id (a number). The guilty part of the code is: what can i replace that request.user with so that it wil actually display the user i searched for, and not the currently logged in? def profile_view(request, id): u = UserProfile.objects.get(pk=id) cv = UserProfile.objects.filter(created_by = request.user) blog = New.objects.filter(created_by = request.user) return render_to_response('profile/publicProfile.html', { 'u':u, 'cv':cv, 'blog':blog, }, context_instance=RequestContext(request)) in urls (of the accounts app): url(r'^profile_view/(?P<id>\d+)/$', profile_view, name='profile_view'), and in template: <h3>Recent Entries:</h3> {% load pagination_tags %} {% autopaginate list 10 %} {% paginate %} {% for object in list %} <li>{{ object.post }} <br /> Voted: {{ vote.count }} times.<br /> {% for reply in object.reply_set.all %} {{ reply.reply }} <br /> {% endfor %} <a href=''> {{ object.created_by }}</a> <br /> {{object.date}} <br /> <a href = "/vote/save_vote/{{object.id}}/">Vote this</a> <a href="/replies/save_reply/{{object.id}}/">Comment</a> </li> {% endfor %} thanks in advance!

    Read the article

< Previous Page | 1 2 3 4 5 6 7  | Next Page >