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因为 AJAX 是异步的。您需要放入$("#list").jqGrid({...成功回调中:

// No need to define the variables outside
$.post('procstring.php', { data: window.location.hash },
    function(result)
        var grade = result.grade;
        var section = result.section;
        var from = result.from;
        var until = result.until;
        var user = result.user;

        $("#list").jqGrid({...
},
'json');
于 2010-08-26T09:28:25.370 回答