2012-10-02 39 views

回答

2

您可以使用xmlstring的數據類型,並使用datastr選項傳遞xml數據。以下是來自documentation page的一個示例:

<script> 
var mystr = 
"<?xml version='1.0' encoding='utf-8'?> 
<invoices> 
    <rows> 
     <row> 
      <cell>data1</cell> 
      <cell>data2</cell> 
      <cell>data3</cell> 
      <cell>data4</cell> 
      <cell>data5</cell> 
      <cell>data6</cell>  
     </row> 
    </rows> 
</invoices>"; 

jQuery(document).ready(function(){ 
    jQuery("#list").jqGrid({ 
    datatype: 'xmlstring', 
    datastr : mystr, 
    colNames:['Inv No','Date', 'Amount','Tax','Total','Notes'], 
    colModel :[ 
     {name:'invid', index:'invid', width:55, sorttype:'int'}, 
     {name:'invdate', index:'invdate', width:90, sorttype:'date', datefmt:'Y-m-d'}, 
     {name:'amount', index:'amount', width:80, align:'right', sorttype:'float'}, 
     {name:'tax', index:'tax', width:80, align:'right', sorttype:'float'}, 
     {name:'total', index:'total', width:80, align:'right', sorttype:'float'}, 
     {name:'note', index:'note', width:150, sortable:false} ], 
    pager: '#pager', 
    rowNum:10, 
    viewrecords: true, 
    caption: 'My first grid' 
    }); 
}); 
</script> 
+0

謝謝。我的問題已解決。現在的問題是我想進行ajax調用並將xml對象作爲帖子傳遞並以json格式接收響應 – user1671219