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jQuery("#assignfixtureTable").jqGrid({     
    url : successdata,
    datatype: "xml",
    height: 250,

我想将 xml 数据分配给 jqgrid ,我不想调用 url 并将其分配给它。

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1 回答 1

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您可以使用数据类型并使用该选项xmlstring传递您的 xml 数据。datastr这是文档页面中的一个示例:

<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>
于 2012-10-02T13:55:17.570 回答