我正面临一个问题,在我的Zend项目中,我需要在界面上显示一些DB数据。 我选择了jGrid(jQuery)作为可以做子网格甚至是新网格的子网格。如何将jGrid整合到Zend 1中?
我搜索了“Zend + jgrid”,并完成了下载ZendX,Zend的扩展,并将其放入/库。
我还添加条目的application.ini:
autoloadernamespaces[] = "ZendX"
resources.view.helperPath.ZendX_JQuery_View_Helper = "ZendX/JQuery/View/Helper"
pluginPaths.ZendX_Application_Resource = "ZendX/Application/Resource"
然后在布局我加入: 回声$这 - >的jQuery();
终于在视图中我把:
$options = array(
"colModel" => array(
array(
"name" => "Inv No",
"id" => "id",
"index" => "id",
"width" => 75,
"align" => "center"
),
array(
"name" => "Date",
"id" => "invdate",
"index" => "invdate"
),
array(
"name" => "Client",
"id" => "name",
"index" => "name"
),
),
"rowNum" => 10,
"autowidth" => "true",
"rowList" => array(10, 20, 30),
"sortorder" => "desc",
"caption" => "Example"
);
$this->_helper->jgrid($options);
以下从this link 我得到了它是一个错误报告用例UC-1:
Fatal error: Call to a member function jgrid() on a non-object in D:\PROJEKTY\wtms_gui\application\views\scripts\jgrid\index.phtml on line 28
它指的是代码行:
$this->_helper->jgrid($options);
我错过了什么,以及如何运行简单的表jGrid从数组中获取数据?
问候