Question

我坚持使用xml字符串准备。基本上我有一个表，包含一些行。如果我点击提交，我们需要显示xml格式化字符串，如下所示

<OrderSummaryIDs>
<Count>2</Count>
<Row>
    <StartDate>08/01/2015</StartDate>
    <EndDate>12/01/2015</EndDate>
    <remark/>Test1</remark>
</Row>
</OrderSummaryIDs>

我试过下面的代码，它不会返回xml结构格式，

这是我的代码

var xmlStart =  "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\"?>";
var xml = xmlStart;
var xmlFirstRowSet = "<OrderSummaryIDs>";
var xmlLast="</OrderSummaryIDs>";

xmlFirstRowSet = xmlFirstRowSet;
var headerRowSet='';
var bodyRowSet ='';

var elements = $(".mutliple tr");
var size = elements.size();
    bodyRowSet = bodyRowSet + "<Count>" + size + "</Count>";

bodyRowSet = bodyRowSet + "<Row>";   
elements.each(function() {    
        var tdValue1 = $td.closest('table').find('td:first-child').html();        
        bodyRowSet = bodyRowSet + "<StartDate>";          
        bodyRowSet = bodyRowSet +  tdValue1;
        bodyRowSet = bodyRowSet + "</StartDate>";

        /*var tdValue2 = $td.closest('table').find('td:eq(1)').html();
        bodyRowSet = bodyRowSet + "<EndDate>";          
        bodyRowSet = bodyRowSet +  tdValue2;
        bodyRowSet = bodyRowSet + "</EndDate>";

        var tdValue3 = $td.closest('table').find('td:last-child').html();
        bodyRowSet = bodyRowSet + "<remark>";          
        bodyRowSet = bodyRowSet +  tdValue3;
        bodyRowSet = bodyRowSet + "</remark>";*/
});

bodyRowSet = bodyRowSet + "</Row>";
    xml = xmlStart + xmlFirstRowSet + bodyRowSet+xmlLast;
alert(xml);

Fiddle

Answer 1

试试这个：

var elements = $(".mutliple tr");
var xml =  "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\"?>";
xml += '<OrderSummaryIDs><Count>' + elements.length + '</Count>';
elements.each(function() {    
    var tdValue1 = $(this).find('td:first-child').html();        
    xml += "<row><StartDate>" + tdValue1 + "</StartDate>";
    //
    //
    //
    xml += '</row>';
}
xml += '</OrderSummaryIDs>';
alert(xml);

将表数据转换为xml结构格式？

1 个答案: