pagination

Zend_Paginator return as objects

由 ぐ巨炮叔叔 提交于 2020-01-23 08:31:41
问题 Zend_Paginator returns results as a standard array but I need to make my results come back as an instance of a class, how do I do this? For example, I want all news articles so would need my items to come back as an instance of News_Model_Article 回答1: You can also create custom zend paginator adapter like: class Application_Paginator_Adapter extends Zend_Paginator_Adapter_DbSelect { public function getItems($offset, $itemCountPerPage) { $this->_select->limit($itemCountPerPage, $offset);

Zend_Paginator return as objects

由 99封情书 提交于 2020-01-23 08:31:09
问题 Zend_Paginator returns results as a standard array but I need to make my results come back as an instance of a class, how do I do this? For example, I want all news articles so would need my items to come back as an instance of News_Model_Article 回答1: You can also create custom zend paginator adapter like: class Application_Paginator_Adapter extends Zend_Paginator_Adapter_DbSelect { public function getItems($offset, $itemCountPerPage) { $this->_select->limit($itemCountPerPage, $offset);

How can we implement Pagination for Mongodb Collection using mongoTemplate

由 偶尔善良 提交于 2020-01-22 19:48:12
问题 I'm a noob in mongoDb i need to implement Pagination for any specific Collection for instance say I have a Collection Foo and i have a Fucntion that returns all the records from the Foo collection public List<Foo> getFoo(){ } But i need to fetch records from the Foo by implementing pagination how can i achieve this by using mongoTemplate Spring data mongodb? 回答1: For general pagination you can use the .skip() and .limit() modifiers on the Query object which you can pass in as arguments to

What is an efficient method of paging through very large result sets in SQL Server 2005?

由 核能气质少年 提交于 2020-01-22 09:51:27
问题 EDIT: I'm still waiting for more answers. Thanks! In SQL 2000 days, I used to use temp table method where you create a temp table with new identity column and primary key then select where identity column between A and B. When SQL 2005 came along I found out about Row_Number() and I've been using it ever since... But now, I found a serious performance issue with Row_Number() . It performs very well when you are working with not-so-gigantic result sets and sorting over an identity column.

Displaying large amount of data with JList?

由 蓝咒 提交于 2020-01-21 22:09:33
问题 I have a JList, wich must display more than 3000 items. I wish to have "visible" around 100 items in the list, and when you scroll and getting close to the end (or begining) of the "visible" items the next portion (around 50) must be loaded in the list. Is there any simple way of doing this? 回答1: no there are no simple way for that, you have to implements Pagination(s) easiest job when is managed by Databases engine, most of then support paginations directly in the Model, but I never seen

Displaying large amount of data with JList?

由 北战南征 提交于 2020-01-21 22:09:10
问题 I have a JList, wich must display more than 3000 items. I wish to have "visible" around 100 items in the list, and when you scroll and getting close to the end (or begining) of the "visible" items the next portion (around 50) must be loaded in the list. Is there any simple way of doing this? 回答1: no there are no simple way for that, you have to implements Pagination(s) easiest job when is managed by Databases engine, most of then support paginations directly in the Model, but I never seen

How to use UI Bootstrap Pagination for a table, generated using ng-repeat

由 可紊 提交于 2020-01-19 12:56:05
问题 I am trying to keep pagination in my Angular application using uib-pagination . I am unable to get proper way to do this. HTML <table id="mytable" class="table table-striped"> <thead> <tr class="table-head"> <th>Name</th> </tr> </thead> <tbody> <tr ng-repeat="person in aCandidates"> <th> <div>{{person}}</div> </th> </tr> </tbody> </table> Controller $scope.totalItems = $scope.aCandidates.length; $scope.currentPage = 1; $scope.itemsPerPage = 10; I am able to see the pagination bar but no

java web项目DAO层通用接口BaseDao与实现类BaseDaoImpl

由 喜你入骨 提交于 2020-01-19 09:19:23
在 spring + hibernate 的web项目中，处理数据层通常会使用Spring框架提供的HibernateTemplate类提供的方法。通常的用法是每一个实体类对应的去写DAO层的接口和实现类。每个实现类中都写hibernateTemp.save(entity)、hibernateTemp.update(entity)、hibernateTemp.get(id)...这样写固然没错，但存在着大量的重复代码。所以懒惰的程序员烦了，他们要写一个通用的实现类来解决这个问题，让DAO层解放出来。如果有特殊的需要则在实现类中写个方法去处理，没有特殊的需要那么我们就不用在写DAO层的代码了。 下面是我写的一个通用DAO层接口和实现类，初次写这种通用型的代码可能会有很多不到位的地方，如果您看到了，请在评论中不吝赐教，谢谢！ BaseDao. Java package org.lxl.mr.common.base.db; import java.util.List; /** * 通用数据层接口 * @author liuXueLiang * @param <Entity> * @param <PK> */ public interface BaseDao<Entity,PK> { /** * 增加 * @param entity */ public void save(Entity

Laravel infinite scroll for pagination output

由 落花浮王杯 提交于 2020-01-17 05:53:23
问题 I'm new to laravel and I'm working on a project based on laravel version 4.2. I've got some problems with loading more results using scroll instead of default pagination. I know there are jQuery plugins that can help me out in this, but none of them or suggested ways in web could help me out well. so here is my code : laravel model //inside a AdGenerator class public function allAds(){ $allAds = DB::table('infos')->paginate(10); return $allAds; } laravel controller //inside controller $ads =

pagination in php [closed]

由 非 Y 不嫁゛ 提交于 2020-01-17 02:48:29
问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 8 years ago . want search result for displaying exactly same result as in : https://www.ippu.purdue.edu/GlobalLinkages/index.cfm?Location=.&Page=search 1 - 25 of 414 records Next > I am using 3 inputs frm user ant it is same
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