paginate

grails paginate doesn't work

倖福魔咒の 提交于 2019-12-08 04:39:04
问题 I'm trying to use the paginate tag in grails but it isn't working. in controller: def show(Integer max) { params.max = Math.min(max ?: 10, 100) def etpse def total if (params.data == 'all') { etpse = Enterprise.findAll() total = Enterprise.count() } else { def paramsLike = "%" + params.data + "%" etpse = Enterprise.findAllByKeywordLike(paramsLike) total = Enterprise.countByKeywordLike(paramsLike) } [etpseList: etpse, instanceTotal: total] } in gsp: <div id='pagination'> <g:paginate total="$

Paginate results filtered by condition on associated model (HABTM) using Containable

橙三吉。 提交于 2019-12-08 03:04:07
问题 I need to paginate list of Product s belonging to specific Category (HABTM association). In my Product model I have var $actsAs = array('Containable'); var $hasAndBelongsToMany = array( 'Category' => array( 'joinTable' => 'products_categories' ) ); And in ProductsController $this->paginate = array( 'limit' => 20, 'order' => array('Product.name' => 'ASC'), 'contain' => array( 'Category' => array( 'conditions' => array( 'Category.id' => 3 ) ) ) ); $this->set('products', $this->paginate());

Paginate Dynamically created table in Cakephp?

为君一笑 提交于 2019-12-06 12:20:46
问题 Please look at the sample code mysql_query('SET AUTOCOMMIT=0;'); mysql_query('START TRANSACTION;'); $tableName = rand().'_table;' $this->loadModel('Home'); // Model for homes_table $sql = 'CREATE TABLE '.$tableName.'_table LIKE homes_table'; mysql_query($sql); // FEW INSERT STATEMENTS ON THE NEW TABLE $tableName // /* Here I want to paginate the new table using $this->paginate(); HOW? */ mysql_query('TRUNCATE table '.$tableName); mysql_query('COMMIT;'); I want to paginate the new table

Paginate results filtered by condition on associated model (HABTM) using Containable

别等时光非礼了梦想. 提交于 2019-12-06 05:40:59
I need to paginate list of Product s belonging to specific Category (HABTM association). In my Product model I have var $actsAs = array('Containable'); var $hasAndBelongsToMany = array( 'Category' => array( 'joinTable' => 'products_categories' ) ); And in ProductsController $this->paginate = array( 'limit' => 20, 'order' => array('Product.name' => 'ASC'), 'contain' => array( 'Category' => array( 'conditions' => array( 'Category.id' => 3 ) ) ) ); $this->set('products', $this->paginate()); However, resulting SQL looks like this: SELECT COUNT(*) AS `count` FROM `products` AS `Product` WHERE 1 = 1

Jekyll pagination on multiple pages

霸气de小男生 提交于 2019-12-02 10:25:24
问题 I am new to html/css but am attempting to create a blog using Jekyll and this theme I found here https://github.com/rosario/kasper The homepage index.html has the all the posts in a paginated list. This is cool. However i would like to group my posts into different categories and have an additional page for each group which would have a paginated list of just posts of that groups. I can create the additional pages but can't get the lists using any sort of variant of the code in index.html but

Jekyll pagination on multiple pages

我的未来我决定 提交于 2019-12-02 04:32:05
I am new to html/css but am attempting to create a blog using Jekyll and this theme I found here https://github.com/rosario/kasper The homepage index.html has the all the posts in a paginated list. This is cool. However i would like to group my posts into different categories and have an additional page for each group which would have a paginated list of just posts of that groups. I can create the additional pages but can't get the lists using any sort of variant of the code in index.html but specifying a group. Is this possible? You should share your code with your answer if you want a more

Laravel 4 - paginate ignore distinct in Fluent

人盡茶涼 提交于 2019-12-02 04:26:59
问题 I make fluent request with distinct and paginate. My problem is that the paginate request is execute before disinct request My Fluent request : $candidates = DB::table('candidates') ->select('candidates.*') ->distinct() ->join('candidate_region', 'candidates.id', '=', 'candidate_region.candidate_id') ->join('candidate_job', 'candidates.id', '=', 'candidate_job.candidate_id') ->whereIn('candidate_region.region_id', $inputs['region']) ->whereIn('candidate_job.job_id', $inputs['job']) ->where(

Laravel5学生成绩管理系统-05-分页

本小妞迷上赌 提交于 2019-12-01 18:39:37
在其它的框架中,分页是非常让人苦恼的。而在 Laravel 中却是很轻而易举的。 Laravel 可以快速生成基于当前页面链接的智能「范围」,并且生成的 HTML 兼容于 Bootstrap CSS 框架。本项目中,我们将利用laravel的分页功能,来轻松处理学生列表的分页。 强烈建议新手或不熟练的朋友在做分页时先仔细的阅读下官方文档。 官方文档-分页 #基本使用 ##对查询语句构造器进行分页 有几种方法可以对项目进行分页。最简单的是使用 paginate 方法。在使用 查询语句构造器 或是 Eloquent 查找 时。由 Laravel 提供的 paginate 方法能够自动判定当前页面正确的数量限制和偏移数。默认状况下,当前页数由 HTTP 请求所带的** ?page **参数来决定。当然,该值由 Laravel 自动检测,并自动插入由分页器生成的链接。 首先,让我们来看看如何在数据库查找时使用 paginate 方法。在这个例子中,传递给 paginate 唯一的参数是你想在「每页」显示的数据数。我们在此指定每页显示 15 条数据: <?php namespace App\Http\Controllers; use DB; use App\Http\Controllers\Controller; class UserController extends Controller

Laravel sortBy paginate

余生颓废 提交于 2019-12-01 05:40:02
I have a posts table and comments table, comment belongs to post, and I have the relationship setup in Post and Comment model. I did sort posts by the number of comments of each post like this: $posts = Post::with('comments')->get()->sortBy(function($post) { return $post->comments->count(); }); What I wonder is how I can paginate these sorted posts? $posts = Post::with('comments')->get()->sortBy(function($post) { return $post->comments->count(); })->paginate(20); doesn't work and gives me error that says paginate is an undefined method. I don't know if you can do it using Eloquent but you can

Add some data in laravel paginate

守給你的承諾、 提交于 2019-11-30 19:44:25
As the tittle said, Here is my Controller $book = Data::where('userId','1')->paginate(3); return response()->json($book); And get the json data like this: data:[{id: 1, userId: 1, vendorId: 1, name: "Alfreda Berge PhD", phone: "1-850-813-5950 x96169",…},…] from:1 last_page:2 next_page_url: "http:/localhost/XX/public/user/book/list/1?page=2" perpage:4 prev_page_url:null to:4 total:5 // if I want to add a new column and value here ,what should I do? I tried to do this : $book = Data::where('userId','1')->paginate(3); $book->printWord = 'Hellow!'; return response()->json($book); But, It seem will