Question

对于我的生活，我不明白为什么会这样。

基本上我使用以下代码从数据库中恢复表行。

var getCposInputs = {
            'type': 'viewcpos',
            'quoteid': '<?php echo $_GET['id']; ?>'
        };

        $.ajax({
            type: 'POST',
            url: '/Applications/Controllers/Quotes/ajax-add-sin.php',
            data: getCposInputs,
            dataType: 'html',
            encode: true
        }).done(function(data){

            //$('body').html(data);
            buildcotable += data;

        });

因此，您可以看到有一条注释掉的行，当删除时会直接显示详细信息，而不是稍后使用JavaScript中的.html（）函数发送的变量。

所以完整的jQuery代码是：

var buildcotable = '';
    var buildtrs = $('#formentry15').val();
    var coArray = '';
    var coArrayNumber = 1;

    buildcotable += '<div class="table-responsive">';
    buildcotable += '<table class="table table-bordered">';

    buildcotable += '<thead>';
    buildcotable += '<th class="text-center">Number</th>';
    buildcotable += '<th class="text-center">Price</th>';
    buildcotable += '<th class="text-center">Installments</th>';
    buildcotable += '<th class="text-center">Contact Initials</th>';
    buildcotable += '<th class="text-center">Options</th>';
    buildcotable += '</thead>';

    buildcotable += '<tbody id="jquerypotable">';

    //lets do a check and see how many are listed
    if(buildtrs != 'TBC'){

        var getCposInputs = {
            'type': 'viewcpos',
            'quoteid': '<?php echo $_GET['id']; ?>'
        };

        $.ajax({
            type: 'POST',
            url: '/Applications/Controllers/Quotes/ajax-add-sin.php',
            data: getCposInputs,
            dataType: 'html',
            encode: true
        }).done(function(data){

            $('body').html(data);
            buildcotable += data;

        });

    } else {

        buildcotable += '<tr id="jqueryporow1">';
        buildcotable += '<td><input type="hidden" value="No" id="jqueryponumber1" class="form-control">No CPO\'s have been submitted</td>';
        buildcotable += '<td><input type="hidden" value="" id="jquerypovalue1" class="form-control"></td>';
        buildcotable += '<td class="text-center"><input type="hidden" value="" id="jquerypoinstallments1" class="form-control"></td>';
        buildcotable += '<td><input type="hidden" value="" id="jquerypocontactinitials1" class="form-control"></td>';
        buildcotable += '<td class="text-center">&nbsp;</td>';
        buildcotable += '</tr>';

    }

    buildcotable += '</tbody>';

    buildcotable += '</table>';
    buildcotable += '<p><a href="#" class="btn btn-default btn-block" id="addnewpo">Add New CPO Number</a></p>';
    buildcotable += '<p><a href="#" class="btn btn-danger btn-block" id="ubldonepo">Done</a></p>';
    buildcotable += '</div>';

    $('.ubl-section-7').html(buildcotable);

我知道数据回复正常，因为当我删除$（＆＃39; body＆＃39;）。html（数据）的评论时;然后它显示了信息。

但如果试图把它变成一个变量，它就什么都不做。

关于为什么会发生这种情况的任何想法？

由于

Answer 1

您必须确保在ajax请求检索其数据后生成输出。此外，您的最后一行代码甚至在buildcotable += data;执行之前就已运行，因为ajax请求尚未准备好（它是异步的）。尝试这样的事情：

var buildcotable_beg = '<table><tr> ...';
var buildcotable_cnt = '';
var buildcotable_end = '...</table>';
var full_bld_tbl = '';

if (buildtrs != 'TBC') {
    $.ajax(...).done(function(buildcotable_cnt) {
        full_bld_tbl = buildcotable_beg + buildcotable_cnt + buildcotable_end;
        $('.ubl-section-7').html(full_bld_tbl);
    });
} else {
   buildcotable_cnt = '<tr>...</tr>';

   full_bld_tbl = buildcotable_beg + buildcotable_cnt + buildcotable_end;
   $('.ubl-section-7').html(full_bld_tbl);
}

数据回来但无法在jquery ajax中发布

1 个答案: