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php中使用easyui datagrid的问题

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2016-06-23 13:59:52839browse

<table id="adg"  class="easyui-datagrid" style="width:410px;height:auto" data-options="  url: 'daily_all_area.php',  singleSelect: true " > <thead>  <tr><th field="area_no" width="100"  formatter="getArea"><strong>Region Name</strong></th> <th field="daily_quantity_area" width="80" align='right'><strong>Quantity</strong></th> </tr> </thead> </table>

一般在使用datagrid时url填写的是个生成json的php文件。这个url如果想换成个生成json的方法该怎么写?


回复讨论(解决方案)

散分咯 来接分

解决啦?
潜水那么久

解决啦?
潜水那么久
嗯 用的很原始的方法 url: 'daily_all_area.php?type=xxx'
再在后台用$_request['type'], 根据不同type调用不同方法

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