分页对于基于数据库的应用来讲,是很常见的一个问题。新的zentaoPHP框架中,提供了内置的分页功能。使用如下:
我们以查询用户列表为例,在应用中建立一个user的模块,在其中的control文件中,定义一个browse方法,来完成分页的功能:
一、browse方法关于分页的三个参数
browse方法需要定义三个参数:recTotal, recPerPage, pageID,变量名是固定的。
public function browse($recTotal, $recPerPage, $pageID) { /* 加载分页类,并生成pager对象。*/ $this->app->loadClass('pager', $static = true); $pager = new pager($recTotal, $recPerPage, $pageID); /* 将分页类传给model,进行分页。*/ $users = $this->user->getList($pager); }
二、model方法中调用pager对象
model中定义一个getList方法,接收pager对象,并在dao查询的时候,调用pager($pager)方法来生成分页语句。
public function getList($pager) { return $this->dao->select(*)->from('user')->page($pager)->fetchAll(); }
三、control中将pager对象赋值给模板
再回到control的browse方法中,将pager对象赋值给模板。
public function browse($recTotal, $recPerPage, $pageID) { /* 加载分页类,并生成pager对象。*/ $this->app->loadClass('pager', $static = true); $pager = new pager($recTotal, $recPerPage, $pageID); /* 将分页类传给model,进行分页。*/ $users = $this->user->getList($pager); /* 赋值到模板。*/ $this->view->users = $users; $this->view->pager = $pager; }
模板中显示分页链接:show()方法有两个参数,$align: left, center, right,默认是居右对齐。$type: full|short|shortest
<?php $pager->show();?>
$data = $this->dao->select('DISTINCT t1.*,t2.realname AS handler,t3.bugNum')
->from(TABLE_TASK)->alias('t1')
->leftJoin(TABLE_USER)->alias('t2')->on('t1.handler = t2.account')
->leftJoin("(SELECT task,COUNT(*) AS bugNum FROM zt_bug WHERE task > 0 GROUP BY task)")->alias('t3')->on('t1.id = t3.task')
->page($pager)
->fetchAll();
18:48:39 ERROR: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 <p>The sql is: SELECT DISTINCT t1.*,t2.realname AS handler,t3.bugNum FROM `zt_task` AS t1 LEFT JOIN `zt_user` AS t2 ON t1.handler = t2.account LEFT JOIN (SELECT task,COUNT(*) AS bugNum FROM zt_bug WHERE task > 0 GROUP BY task) AS t3 ON t1.id = t3.task </p> in /var/www/html/zentao/lib/base/dao/dao.class.php on line 1381, last called by /var/www/html/zentao/lib/base/dao/dao.class.php on line 379 through function sqlError.
in /var/www/html/zentao/framework/base/router.class.php on line 2280 when visiting
请求地址:http://www.yc-zentao.com:8146/itservice-getList.json?page=2
改成这样:http://www.yc-zentao.com:8146/itservice-getList-page-2.json $_GET 中也没有page 但是在入口文件index.php 中打印$_GET 是有page的 ,请问 有什么方法可以获取到么?