Question

请帮助我基本上想要使用表单过滤我的数据表。

每个输入的值都会影响提交表单的查询。

这是我的JQUERY。

$("#submit").click(function (e) {               

    $('#table').dataTable
    ({  

        "sAjaxSource": "index.php/report/get_report",       
        "sServerMethod": "POST",        
        'fnServerData': function (url, data, callback) {
        // Add new data 
            dataString = $("#myform").serialize();
            $.ajax({
                'url': "index.php/report/get_report",
                'data': dataString,
                'type': 'POST',
                'success': callback,
                'dataType': 'json',
                'cache': true
            });
        },                          
        'bServerSide'    : true,
        "aaSorting": [[ 3, "desc" ]],
        "bPaginate": true,                       
        "bSortClasses": false,
        "bAutoWidth": true,
        "bInfo": true,          
        "iDisplayLength"    : 3,        
        "bScrollCollapse": true,                                        
        "oLanguage": {
            "sSearch": "Search:"
        },
        "bDestroy": true        
    });     
});

这是我的HTML表格

<form name="myform">                                        
    <label>Employee:</label>
    <input type="text" name="employeeid" id="employeeid" title="Type Employee" />
    <label>Training Type: </label>
        <select name="trainingtype" id="trainingtype" >
        <option value="" selected="selected">All</option>
        <option value="1">Externally Facilitated Training</option>
        <option value="3">Internally Facilitated Training</option>
        <option value="2">Webcast/E-Learning</option>
        </select>                                       
    <label>Datestart</label>
    <input type="text" class="field size3" name="datestart" id="datepicker_s" />                    
    <label>Dateend </label>
    <input type="text" class="field size3" name="dateend" id="datepicker_e" />              
    <input type="hidden" id="txtsearchid" name="txtsearchid">
    <input type="button" class="button" value="Submit" id="submit" />

当我提交表格时，我什么都没得到。

我是以正确的方式做的吗？

请帮助。

GOT IT

"fnServerData": function ( sSource, aoData, fnCallback ) {
                //REQUIRED: Add a Post variable with the object value                   
                aoData.push( 
                    { "name": "txtsearchid", "value": $( "#txtsearchid" ).val() },
                    { "name": "datestart", "value": $( "#datepicker_s" ).val() },
                    { "name": "dateend", "value": $( "#datepicker_e" ).val() },
                    { "name": "trainingtype", "value": $( "#trainingtype" ).val() }

                );

                $.ajax( {
                        dataType: 'json',
                        type: "POST",
                        url: sSource,
                        data: aoData ,
                        success: fnCallback
                } );
        },

这一定是我问题的解决方案。我没有使用序列化而不是推送

Answer 1

使用Mozilla中的Firebug添加来检查错误

codeigniter数据表从表单中获取数据

1 个答案: