使用AJAX jQuery排序表

Question

我正在尝试使用JSON填充的JS tablesort对表进行排序，并且出现错误提示

“无法读取未定义的属性'0'”

看起来tablesort无法识别ajax数据。

这是我的jQuery

 $.get(url, function(response){

    serverResponse = response;

    for(i in response.content){

        totalclientes++;

        var status = response.content[i].LojaStatus;
            if(status == 0){
                LojaStatus = "Ativo";
                botao = '\ ' +
                    '<div class = "btn-group"> <button type = "button" class = "btn btn-primary" onclick="redirect(' + response.content[i].LojaId + ')" data-toggle="modal" data-target="#myModal">Editar</button>       <button type = "button" class = "btn btn-primary dropdown-toggle" data-toggle = "dropdown"> <span class = "caret"></span> <span class = "sr-only">Toggle Dropdown</span> </button> <ul class = "dropdown-menu" role = "menu"> <li><a href = "javascript:deletar(' + response.content[i].LojaId + ')">Desativar</a></li> <li><a href = "http://amovitrine.devmaker.com.br/relatorioloja.php?LojaId='+ response.content[i].LojaId +'">Relatorio</a></li> </ul> </div>';
            }else{
                LojaStatus = "Inativo";
                botao = '\ ' +
                    '<div class = "btn-group"> <button type = "button" class = "btn btn-primary" onclick="redirect('+response.content[i].LojaId+')" data-toggle="modal" data-target="#myModal">Editar</button>       <button type = "button" class = "btn btn-primary dropdown-toggle" data-toggle = "dropdown"> <span class = "caret"></span> <span class = "sr-only">Toggle Dropdown</span> </button> <ul class = "dropdown-menu" role = "menu"> <li><a href = "javascript:ativar('+response.content[i].LojaId+')">Ativar</a></li><li><a href = "http://amovitrine.devmaker.com.br/relatorioloja.php?LojaId='+ response.content[i].LojaId +'">Relatorio</a></li> </ul> </div>';
            }

        var botao =


            data +='\
        <tr>\
            <td>'+response.content[i].LojaNome+'</td>\
            <td>'+response.content[i].LojaBairro+'</td>\
            <td>'+response.content[i].LojaTelefone1+'</td>\
            <td>'+LojaStatus+'</td>\
            <td>'+response.content[i].PlanoNome+'</td>\
            <td>'+response.content[i].LojaInicioPlano+'</td>\
            <td>'+response.content[i].LojaFimPlano+'</td>\
            <td>'+botao+'</td>\
        </tr>';
    }

    $('#corpotabela').empty();
    $('#corpotabela').append(data);

这是我的桌子

 <table class="table table-bordered table-hover" id="table">
            <thead style="border: 1px solid #ddd;" >
            <tr style="cursor: pointer;">
                <th>Nome</th>
                <th>Bairro</th>
                <th>Telefone</th>
                <th>Status</th>
                <th>Plano</th>
                <th>Data Inicio</th>
                <th>Data Fim</th>
                <th>Ações</th>
            </tr>
            </thead>
            <tbody id="corpotabela">

            </tbody>
        </table>

有什么建议么？

*编辑这是我的桌子

*编辑2

这是我回应中对象的一个​​例子

编辑3-解决方案

基本上，我是在getData函数中使用tablesorter的，后者将表格填充到了我。

因此，我在$('#corpotabela').append(data);之后应用表$('#corpotabela').append(data); 线，效果很好。

Answer 1

我建议您检查收到的响应是否实际上是JSON。 然后，尝试使用Firebug获取具体错误，然后将其发布在此处。