2013-03-05 32 views
0

我正面臨一個問題,在我的Zend項目中,我需要在界面上顯示一些DB數據。 我選擇了jGrid(jQuery)作爲可以做子網格甚至是新網格的子網格。如何將jGrid整合到Zend 1中?

我搜索了「Zend + jgrid」,並完成了下載ZendX,Zend的擴展,並將其放入/庫。

我還添加條目的application.ini:

autoloadernamespaces[] = "ZendX" 
resources.view.helperPath.ZendX_JQuery_View_Helper = "ZendX/JQuery/View/Helper" 
pluginPaths.ZendX_Application_Resource = "ZendX/Application/Resource" 

然後在佈局我加入: 回聲$這 - >的jQuery();

終於在視圖中我把:

$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); 

以下從this link 我得到了它是一個錯誤報告用例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

它指的是代碼行:

$this->_helper->jgrid($options); 

我錯過了什麼,以及如何運行簡單的表jGrid從數組中獲取數據?

問候

回答

1

您鏈接到的頁面是一個分量建議,它看起來並不像它曾經有過提案階段。 Zend Framework中沒有jGrid組件。

如果你沒有機會下載第三方實現,$this->_helper->jgrid($options);就是你如何從控制器調用'jgrid'動作助手,以便該行只能在控制器動作中工作。要從視圖中調用視圖幫助程序,語法是$this->jgrid($options)