Search Results

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

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

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

  • cakephp paginate using mysql SQL_CALC_FOUND_ROWS

    - by michael
    I'm trying to make Cakephp paginate take advantage of the SQL_CALC_FOUND_ROWS feature in mysql to return a count of total rows while using LIMIT. Hopefully, this can eliminate the double query of paginateCount(), then paginate(). http://dev.mysql.com/doc/refman/5.0/en/information-functions.html#function_found-rows I've put this in my app_model.php, and it basically works, but it could be done better. Can someone help me figure out how to override paginate/paginateCount so it executes only 1 SQL stmt? /** * Overridden paginateCount method, to using SQL_CALC_FOUND_ROWS */ public function paginateCount($conditions = null, $recursive = 0, $extra = array()) { $options = array_merge(compact('conditions', 'recursive'), $extra); $options['fields'] = "SQL_CALC_FOUND_ROWS `{$this->alias}`.*"; Q: how do you get the SAME limit value used in paginate()? $options['limit'] = 1; // ideally, should return same rows as in paginate() Q: can we somehow get the query results to paginate directly, without the extra query? $cache_results_from_paginateCount = $this->find('all', $options); /* * note: we must run the query to get the count, but it will be cached for multiple paginates, so add comment to query */ $found_rows = $this->query("/* do not cache for {$this->alias} */ SELECT FOUND_ROWS();"); $count = array_shift($found_rows[0][0]); return $count; } /** * Overridden paginate method */ public function paginate($conditions, $fields, $order, $limit, $page = 1, $recursive = null, $extra = array()) { $options = array_merge(compact('conditions', 'fields', 'order', 'limit', 'page', 'recursive'), $extra); Q: can we somehow get $cache_results_for_paginate directly from paginateCount()? return $cache_results_from_paginateCount; // paginate in only 1 SQL stmt return $this->find('all', $options); // ideally, we could skip this call entirely }

    Read the article

  • will paginate, nested routes, ruby, rails

    - by Sam
    I'm trying to get will paginate to link to my nested route instead of the regular posts variable. I know I'm supposed to pass some params to paginate but I don't know how to pass them. Basically there is an array stored in @posts and the other param paginate has access to is category_id. The nested route is /category/1/posts but hitting next and previous on will paginate returns a url like this posts?page=1&category_id=7. <%= will_paginate @most_recent_posts "What do I do here?" %> This is the result of Yannis's answer: In your controller you can do: @posts = @category.posts.paginate And in your view: <%= will_paginate(@post) %> Doing this comes up with the following URL posts?page=2&post_category_id=athlete_management routes.rb #there are more routes but these are the relevant ones map.resources :posts map.resources :post_categories, :has_many => :posts solution map.resources :post_categories do |post_category| post_category.resources :posts end map.resources :posts Had to declare the resource after the block Thanks stephen!

    Read the article

  • Sort by ranking algorithm using will-paginate

    - by bearwithclaws
    I'm creating a digg-like site using Ruby on Rails that ranks the item (based on this algorithm). I'm using the will-paginate gem list the items in pages. The problem is, will-paginate only allows me to insert ':order =' based on the table data. I would like to make will-paginate to sort by a number which is calculated using a function based on different fields on the table (e.g number of votes, age hours). How can I do that?

    Read the article

  • In Django-pagination Paginate does not working...

    - by mosg
    Hello. Python 2.6.2 django-pagination 1.0.5 Question: How to force pagination work correctly? The problem is that {% paginate %} does not work, but other {% load pagination_tags %} and {% autopaginate object_list 10 %} works! Error message appeared, when I add {% paginate %} into html page: TemplateSyntaxError at /logging Caught an exception while rendering: pagination/pagination.html What I have done: Install django-pagination without any problems. When I do in python import pagination, it's work well. Added pagination to INSTALLED_APP in settings.py: INSTALLED_APPS = ( # ..., 'pagination', ) Added in settings.py: TEMPLATE_CONTEXT_PROCESSORS = ( "django.core.context_processors.auth", "django.core.context_processors.debug", "django.core.context_processors.i18n", "django.core.context_processors.media", "django.core.context_processors.request" ) Also add to settings.py middleware: MIDDLEWARE_CLASSES = ( # ... 'pagination.middleware.PaginationMiddleware', ) Add to top in views.py: from django.template import RequestContext And finally add to my HTML template page lines: {% load pagination_tags %} ... {% autopaginate item_list 50 %} {% for item in item_list %} ... {% endfor %} {% paginate %} Thanks. PS: some edits required, because I can't django code style work well here :)

    Read the article

  • Jquery Cycle: How Can I Paginate the Pager?

    - by JohnP
    Hello, I am using the Jquery cycle plug in with the pager. I have a lot of items to page through. Is there a way I can add back and previous arrows to paginate the pager? So I have like: 1 2 3 4 5 6 7 8 9 and I want the same but < 1 2 3 4 5 6 7 8 9 Is there an elegant way to code this? Many thanks for your help!

    Read the article

  • How to use jquery to paginate json data?

    - by Pandiya Chendur
    My json Data looks like this {"Table" : [{"Emp_Id" : "3","Identity_No" : "","Emp_Name" : "Jerome","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Supervisior","Desig_Description" : "Supervisior of the Construction","SalaryBasis" : "Monthly","FixedSalary" : "25000.00"},{"Emp_Id" : "4","Identity_No" : "","Emp_Name" : "Mohan","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Acc ","Desig_Description" : "Accountant","SalaryBasis" : "Monthly","FixedSalary" : "200.00"},{"Emp_Id" : "5","Identity_No" : "","Emp_Name" : "Murugan","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Mason","Desig_Description" : "Mason","SalaryBasis" : "Weekly","FixedSalary" : "150.00"},{"Emp_Id" : "6","Identity_No" : "","Emp_Name" : "Ram","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Mason","Desig_Description" : "Mason","SalaryBasis" : "Weekly","FixedSalary" : "120.00"},{"Emp_Id" : "7","Identity_No" : "","Emp_Name" : "Raja","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Mason","Desig_Description" : "Mason","SalaryBasis" : "Weekly","FixedSalary" : "135.00"},{"Emp_Id" : "8","Identity_No" : "","Emp_Name" : "Raja kumar","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Mason Helper","Desig_Description" : "Mason Helper","SalaryBasis" : "Weekly","FixedSalary" : "105.00"},{"Emp_Id" : "9","Identity_No" : "","Emp_Name" : "Lakshmi","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Mason Helper","Desig_Description" : "Mason Helper","SalaryBasis" : "Weekly","FixedSalary" : "100.00"},{"Emp_Id" : "10","Identity_No" : "","Emp_Name" : "Palani","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Carpenter","Desig_Description" : "Carpenter","SalaryBasis" : "Weekly","FixedSalary" : "200.00"},{"Emp_Id" : "11","Identity_No" : "","Emp_Name" : "Annamalai","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Carpenter","Desig_Description" : "Carpenter","SalaryBasis" : "Weekly","FixedSalary" : "220.00"},{"Emp_Id" : "12","Identity_No" : "","Emp_Name" : "David","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Steel Fixer","Desig_Description" : "Steel Fixer","SalaryBasis" : "Weekly","FixedSalary" : "220.00"},{"Emp_Id" : "13","Identity_No" : "","Emp_Name" : "Chandru","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Steel Fixer","Desig_Description" : "Steel Fixer","SalaryBasis" : "Weekly","FixedSalary" : "220.00"},{"Emp_Id" : "14","Identity_No" : "","Emp_Name" : "Mani","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Steel Helper","Desig_Description" : "Steel Helper","SalaryBasis" : "Weekly","FixedSalary" : "175.00"},{"Emp_Id" : "15","Identity_No" : "","Emp_Name" : "Karthik","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Wood Fixer","Desig_Description" : "Wood Fixer","SalaryBasis" : "Weekly","FixedSalary" : "195.00"},{"Emp_Id" : "16","Identity_No" : "","Emp_Name" : "Bala","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Wood Fixer","Desig_Description" : "Wood Fixer","SalaryBasis" : "Weekly","FixedSalary" : "185.00"},{"Emp_Id" : "17","Identity_No" : "","Emp_Name" : "Tamil arasi","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Wood Helper","Desig_Description" : "Wood Helper","SalaryBasis" : "Weekly","FixedSalary" : "185.00"},{"Emp_Id" : "18","Identity_No" : "","Emp_Name" : "Perumal","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Cook","Desig_Description" : "Cook","SalaryBasis" : "Weekly","FixedSalary" : "105.00"},{"Emp_Id" : "19","Identity_No" : "","Emp_Name" : "Andiappan","Address" : "Madurai","Date_Of_Birth" : "","Desig_Name" : "Watchman","Desig_Description" : "Watchman","SalaryBasis" : "Weekly","FixedSalary" : "150.00"}]} There are 22 records in this json... How to paginate this json data 5 per page using jquery? EDIT: The above image is my summary view of employee list iterated using jquery var jsonObj = JSON.parse(HfJsonValue); for (var i = jsonObj.Table.length - 1; i >= 0; i--) { var employee = jsonObj.Table[i]; $('<div class="resultsdiv"><br /><span class="resultName">' + employee.Emp_Name + '</span><span class="resultfields" style="padding-left:100px;">Category&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee.Desig_Name + '</span><br /><br /><span id="SalaryBasis" class="resultfields">Salary Basis&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee.SalaryBasis + '</span><span class="resultfields" style="padding-left:25px;">Salary&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee.FixedSalary + '</span><span style="font-size:110%;font-weight:bolder;padding-left:25px;">Address&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee.Address + '</span></div>').insertAfter('#ResultsDiv'); } I get 22 records now it may grow how to paginate json date by using jquery pagination.. Any suggestion...

    Read the article

  • Rails paginate array items one-by-one instead of page-by-page

    - by hnovick
    Hi Guys, I have a group of assets, let's call them "practitioners". I'm displaying these practitioners in the header of a calendar interface. There are 7 columns to the calendar. 7 columns = 7 practitioners per view/page. Right now: if the first page shows you practitioners 1-7, when you go the next page you will see practitioners 8-15, next page 16-23, etc. etc. i am wondering how to page the practitioners so that if the first page shows you practitioners 1-7, the next page will show you practitioners 2-8, then 3-9, etc. etc. i would greatly appreciate any help you can offer. here is the rails code i am working with. best regards, harris novick # get the default sort order sort_order = RESOURCE_SORT_ORDER # if we've been given asset ids, start our list with them unless params[:asset_ids].blank? params[:asset_ids] = params[:asset_ids].values unless params[:asset_ids].is_a?(Array) sort_order = "#{params[:asset_ids].collect{|id| "service_provider_resources.id = #{id} DESC"}.join(",")}, #{sort_order}" end @asset_set = @provider.active_resources(:include => {:active_services => :latest_approved_version}).paginate( :per_page => RESOURCES_IN_DAY_VIEW, :page => params[:page], :order => sort_order )

    Read the article

  • Rails paginate existing array of ActiveRecord results

    - by SaoiseK
    Hello, I generally use will_paginate for the pagination in my app, but have hit a stumbler on my search feature. I'm using Thinking Sphinx for doing my full-text search, which returns results paginated. The problem I'm having is that after I've received the results from Thinking Sphinx, I need to merge them with some other results and re-order them. Once I've finished processing them I have an Array of results that is very different from the original from TS. As there could be 1000+ results in this Array Pagination is a necessity. The problem is that I can't figure out how to get will_paginate to play with an existing array. I've done some research and it seems the only solutions to this problem are from several years ago and are based around the old built-in Paginator class. The most recent one I could find that makes use of will_paginate was from devchix from mid-2007: http://www.devchix.com/2007/07/23/will_paginate-array/comment-page-1/ - I've given this a go but it doesn't seem to do anything for me. Are there any current methods for applying pagination (preferably via will_paginate) for existing arrays of AR results?

    Read the article

  • CakePHP - Paginate / Query: find lowest of multiple fields & order by it

    - by Atea
    I am using CakePHP for a price-comparison website. I have a table products with fields: id, pride_regular, price_action. I would like to combine the fields price_regular and price_action into a dynamic field: price. The lowest value of these two fields should be the value of the new price field. Also I want to order on it Ascending. Should I use a custom MySQL-query?

    Read the article

  • jquery toggle+paginate+multiple instances

    - by mark
    Hi, I'm a jquery newbie - wanted to ask what might be the best strategy for achieving what I am after as I think it uses a mix of jquery functions : If list of items exceeds 5 items a 'more' link appears which when toggled will reveal the rest of the items. If items list is 5 or less no 'more' link is shown. Hide button also at bottom of long full revealed list. (perhaps this is toggle+pagination?) And then also that this can be used in multiple instances, as it is for multiple category menu's.(to be used in typical indexhibit websites structure like http://mikeyburton.com/) link 1 link 2 link 3 link 4 link 5 link 6 link 7 link 8 more Any help or links greatly appreciated.

    Read the article

  • cakephp pagination overide paginate() method

    - by islam
    iam using cakephp pagination and to paginate using custome query i override the paginate() method in my controller the problem is that every other action that use paginate() method not work any more how i can override this method without conflict with other that use this method from the same model

    Read the article

  • Laravel - Paginate and get()

    - by Bajongskie
    With the code below, what I wanted was paginate the query I created. But, when I try to add paginate after get, it throws an error. I wanted to remain get since I want to limit to columns that was set on $fields. What would should be the better idea to paginate this thing? or what's a good substitute for get and limit the columns? What I tried: ->get($this->fields)->paginate($this->limit) Part of my controller: class PhonesController extends BaseController { protected $limit = 5; protected $fields = array('Phones.*','manufacturers.name as manufacturer'); /** * Display a listing of the resource. * * @return Response */ public function index() { if (Request::query("str")) { $phones = Phone::where("model", 'LIKE', '%'. Request::query('str') . '%') ->join('manufacturers', 'manufacturers_id', '=', 'manufacturers.id') ->get($this->fields); } else { $phones = Phone::join('manufacturers', 'manufacturers_id', '=', 'manufacturers.id') ->get($this->fields); } return View::make('phones.index')->with('phones', $phones); } }

    Read the article

  • Grails g:paginate tag and custom URL

    - by aboxy
    Hello, I am trying to use g:paginate in a shared template where depending on the controller, url changes e.g. For my homepage url should be : mydomain[DOT]com/news/recent/(1..n) For search Page: www[DOT]mydomain[DOT]com/search/query/"ipad apps"/filter/this month and my g:paginate looks like this: g:paginate controller=${customeController} action=${customAction} total:${total} For the first case, I was able to provide controller as 'news' and action as 'recent' and mapped url /news/recent/$offset to my controller. But for the search page, I am not able to achieve what I want to do. I have a URL mapping defined as /search/$filter**(controller:"search",action:"fetch") $filter can be /query/"ipad apps"/filter/thismonth/filter/something/filter/somethingelse. I want to be able to show the url as above rather than ?query="ipad apps"&filter=thismonth&filter=something&filter=somethingelse. I believe I can pass all the parameters in params attribute of g:paginate but that will not give me pretty URL. What would be the best way to achieve this? Please feel free to ask questions If i missed anything.Thanks in advance.

    Read the article

  • cakePHP paginate with post data without sessions, serialization or post to get

    - by openprojdevel
    I have created a small search and filter form method post in controller/index, which posts to it self the conditions and fields to paginate ( $this-paginate($conditions) ) However that is good for the first page, the subsequent pages the filer conditions are lost. pagination passArgs supports get variables well. Is there an un complex way to pass the post conditions to the other paginated pages? The method I have looked at is pass the $conditions in session , which isnt without complexity of assigning session and unset the session on submitting the form again (more refinements to the filter criteria by the user ) The other method is passing the $conditions as serialized string url_encode as an get parameter. Is there an good cake way to do this more like passArgs, sessions and url encode do not look like cake style. Thanks

    Read the article

  • cakephp paginate multiple habtm

    - by izmanromli
    hi guys, i have multiple habtm like these : // User model var $hasMany = array('Post'); // Post model var $hasAndBelongsToMany = array('Category', 'Tag'); // Category model var $hasAndBelongsToMany = array('Post'); // Tag model var $hasAndBelongsToMany = array('Post'); I tried to fetch all post along with its user and tags (within a certain category), somehow if i fetch tags, the result was wrong. $this->paginate = array ( 'Post' => array ( 'limit' => 2, 'fields' => array( 'Post.title', 'Post.content', 'Post.slug', 'Post.created', 'Tag.name', 'User.username', 'User.created', 'User.post_count', 'User.avatar_file_name'), 'joins' => array ( array( 'table' => 'categories_posts', 'alias' => 'CategoriesPost', 'type' => 'inner', 'conditions'=> array('CategoriesPost.post_id = Post.id') ), // FETCH USER array( 'table' => 'users', 'alias' => 'User', 'type' => 'inner', 'conditions'=> array('Post.user_id = User.id') ), // FETCH TAGS array( 'table' => 'posts_tags', 'alias' => 'PostsTag', 'type' => 'inner', 'conditions'=> array('PostsTag.post_id = Post.id') ), array( 'table' => 'tags', 'alias' => 'Tag', 'type' => 'inner', 'conditions'=> array('Tag.id = PostsTag.tag_id') ), array( 'table' => 'categories', 'alias' => 'Category', 'type' => 'inner', 'conditions'=> array('Category.id = CategoriesPost.category_id', 'Category.slug' => $slug) ) ) ) ); $posts = $this->paginate(); could anyone gimme a solution since i'm a newbie? many thanks...

    Read the article

  • will_paginate with named_scopes

    - by Lowgain
    I'm using will_paginate for pagination, which has been working well so far, except for this one thing. If I try to paginate a scope, for instance class User < ActiveRecord::Base named_scope :scope, lambda { etc } end User.scope.paginate({:page => params[:page], :per_page => 10}) That will tell me paginate is an undefined method. I'd rather not have to use a second solution for only this scope, is there something I can do here?

    Read the article

  • Filter, sort and paginate in Codeigniter

    - by Babak
    Ive recently started using CodeIgniter as I was searching for a very lightweight framework and it seemed to come up as a top choice. I am new to the whole MVC thing, enjoying it but am stuck with something which seems very simple. I am coding a CMS and need a way to filter, sort and paginate the results. I am used to doing it with querystrings, so I would have something along the lines of: articles.php?order=title&sort=desc&filter=articletitle&page=5 I have no idea how I would go about doing this in CI so I just turned on the EnableQueryStrings in the config and it works fine, but i am getting the feeling its probably not the most elegant solution. I suppose I could have index.php/articles/index/order/title/sort/desc/filter/articletitle/page/5 but to me this seems very inflexible, what if for example i dont need to sort, how would i make sure i am looking at the correct uri segment? any ideas?

    Read the article

  • How to paginate json data?

    - by Pandiya Chendur
    I used json data and iterated div elements to get my result.... function Iteratejsondata(HfJsonValue) { //var jsonObj = eval('(' + HfJsonValue + ')'); var jsonObj = JSON.parse(HfJsonValue); for (var i = jsonObj.Table.length - 1; i >= 0; i--) { var employee = jsonObj.Table[i]; $('<div class="resultsdiv"><br /><span class="resultName">' + employee.Emp_Name + '</span><span class="resultfields" style="padding-left:100px;">Category&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee.Desig_Name + '</span><br /><br /><span id="SalaryBasis" class="resultfields">Salary Basis&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee.SalaryBasis + '</span><span class="resultfields" style="padding-left:25px;">Salary&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee.FixedSalary + '</span><span style="font-size:110%;font-weight:bolder;padding-left:25px;">Address&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee.Address + '</span></div>').insertAfter('#ResultsDiv'); } $(".resultsdiv:even").addClass("resultseven"); $(".resultsdiv").hover(function() { $(this).addClass("resultshover"); }, function() { $(this).removeClass("resultshover"); }); } I get 22 records in total... Now i want to paginate these divs (i.e) 5 divs per page... Any suggestion...

    Read the article

  • PHP $_GET and $_POST are returning empty arrays--trying to paginate SQL data

    - by George88
    I have set up the following: Database class ($db) Pagination class ($paginator) I am attempting to write a basic system to let me administrate pages. I have a page "page_manager.php" in which I include both my database class (database.php) and my pagination class (paginate.php). In my pagination class I have a function which echoes my SQL data. I've come up with a way to echo an HTML < select element with the necessary IDs, which allows me to successfully echo the corresponding results (10 per page), based on the value of the < select element. So, "1" will echo the first 10 results in the database, "2" will echo from 11-20, "3" will echo from 21-30, etc., etc.. I have added an onChange event to the < select element which will copy its value (using "this.value") to a hidden form field. I then submit this form using document.getElementById().submit(); This will then add the $_GET variable to the URL, so the URL becomes ".../?pagenumber_form=X". However, when I try to grab this value back from the URL, the $_GET['pagenumber_form'] is empty. Some code: <span style='font-family: tahoma; font-size: 10pt;'>Page #</span> <select id="page_number_selection" onchange='javascript: document.getElementById("pagenumber_form").value = this.value; document.getElementById("pagenumber").submit();'> <?php for($i = 1; $i <= $this->num_pages; $i++) echo"<option id='" . $i . "'>" . $i . "</option>"; ?> </select> <form name="pagenumber" id="pagenumber" action="" method="get"> <input type="text" name="pagenumber_form" id="pagenumber_form" /> </form> So, I've tried using $_POST as well, but the same thing happens. I want to use $_GET, for a couple of reasons: it's easier to see what is happening with my values and the data I'm using doesn't need to be secure. To recap: the $_GET variable is being added to the URL when I change the < select element, and the corresponding value gets added to the URL as: ".../?pagenumber_form=X", but when I try to use the value in PHP, for example... $page_number = $_GET['pagenumber_form']; ... I get a NULL value. :-( Can anybody help me out please? Thank you. EDIT: I've just made a discovery. If I move my print_r($_GET) to my main index page, then the superglobals are returning as expected. My site structure is like this: index.php - JavaScript buttons use AJAX HTTP requests to include the "responseText" as the .innerHTML of my main < div . The "responseText" is the contents the page itself, in this case page_manager.php, which in turn includes pagination.php. So in other words, my site is built from PHP includes, which doesn't seem to be compatible with HTTP superglobals. Any idea how I can get around this problem? Thank you :-).

    Read the article

  • Is it possible to filter by conditions before paginating?

    - by Brian Roisentul
    I'm using ruby on rails 2.3.8 and will_paginate plugin. I've just noticed that if I write something like this: Announcement.paginate :page => params[:page], :per_page => 10, :conditions => some_condition it will work. But, if I write something like this: announcements = Announcement.all :conditions => some_condition @ann = announcements.paginate :page => params[:page], :per_page => 10 it won't recognize conditions. Maybe you're wondering "why would he want to do that?", well, it's a long story and I really need to first filter the conditions and then paginate the results.

    Read the article

  • Using will_paginate with AJAX live search with jQuery in Rails

    - by Mark Richman
    I am using will_paginate to successfully page through records. I am also using AJAX live search via jQuery to update my results div. No problem so far. The issue I have is when trying to paginate through those live search results. I simply get "Page is loading..." with no div update. Am I missing something fundamental? # index.html.erb <form id="searchform" accept-charset="utf-8" method="get" action="/search"> Search: <input id="search" name="search" type="text" autocomplete="off" title="Search location, company, description..." /> <%= image_tag("spinner.gif", :id => "spinner", :style =>"display: none;" ) %> </form> # JobsController#search def search if params[:search].nil? @jobs = Job.paginate :page => params[:page], :order => "created_at desc" elsif params[:search] and request.xhr? @jobs = Job.search params[:search], params[:page] end render :partial => "jobs", :layout => false, :locals => { :jobs => @jobs } end # Job#search def self.search(search, page) logger.debug "Job.paginate #{search}, #{page}" paginate :per_page => @@per_page, :page => page, :conditions => ["description LIKE ? or title LIKE ? or company LIKE ?", "%#{search}%", "%#{search}%", "%#{search}%"], :order => 'created_at DESC' end # search.js $(document).ready(function(){ $("#search").keyup(function() { $("#spinner").show(); // show the spinner var form = $(this).parents("form"); // grab the form wrapping the search bar. var url = form.attr("action"); // grab the URL from the form's action value. var formData = form.serialize(); // grab the data in the form $.get(url, formData, function(html) { // perform an AJAX get, the trailing function is what happens on successful get. $("#spinner").hide(); // hide the spinner $("#jobs").html(html); // replace the "results" div with the result of action taken }); }); });

    Read the article

  • How to paginate Django with other get variables?

    - by vagabond
    I am having problems using pagination in Django. Take the URL below as an example: http://127.0.0.1:8000/users/?sort=first_name On this page I sort a list of users by their first_name. Without a sort GET variable it defaults to sort by id. Now if I click the next link I expect the following URL: http://127.0.0.1:8000/users/?sort=first_name&page=2 Instead I lose all get variables and end up with http://127.0.0.1:8000/users/?page=2 This is a problem because the second page is sorted by id instead of first_name. If I use request.get_full_path I will eventually end up with an ugly URL: http://127.0.0.1:8000/users/?sort=first_name&page=2&page=3&page=4 What is the solution? Is there a way to access the GET variables on the template and replace the value for the page? I am using pagination as described in Django's documentation and my preference is to keep using it. The template code I am using is similar to this: {% if contacts.has_next %} <a href="?page={{ contacts.next_page_number }}">next</a> {% endif %}

    Read the article

  • paginate the class

    - by small
    please help me with pagination of this method my method is this one controller # def show @topic = Topic.find(params[:id]) @posts = @topic.posts.find(:all ,:order=> 'id') end views # %div{:style=>"margin: 0 auto;"} %table.sortable.style2{:cellpadding=>5} %thead %tr %td{:width => "25%",:align => "center",:style => "font-weight: bold;"}Posted By %td{:width => "75%",:style => "font-weight: bold;"}Comments %tbody - for post in @posts %tr %td{:align => "center"} &nbsp %div{:width=>"5px" , :style=>"border: 1px solid rgb(232, 232, 232);background-color: rgb(248, 248, 248);width: 60px; height:60px;" } - if post.posted_by.image = image_tag(post.posted_by.image.public_filename(),:width => "60px", :height => "60px",:align=>"center") %div{:style => "font-weight: bold; font-style: italic;"} = post.posted_by ? post.posted_by.firstname : "<em>Unknown User</em>" %br %div{:style => "font-style: italic;"} Posted on = post.created_at.strftime('%d of %B %Y ') = post.created_at.strftime('at %H:%M') %td{:valign => "top"} =post.body

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

1 2 3 4 5 6 7  | Next Page >