jquery ajax error

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jquery ajax error

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#1 by (0 votes)

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I can not throw the result of the ajax query inside a table. The result is an array (result = Object {BrowseAllCategoriesResult: Array (6)})

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <title></title>
</head>
<body>
<form id="form1">
    <div class="jumbotron">
        <input type="text" id="codUser"/>
        <button onclick="ConsUsuario(); return false;">Consulta Usuario</button>
    </div>

      <div><table id="datagrid"></table></div>
</form>

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script><scripttype="text/javascript">
    function ConsUsuario(){
        $.ajax({
        type: "GET",
        url: "http://food-fast-com.web27.redehost.net/CategoriaService.svc/ConsultarTodosCategorias",
        contentType: "application/json",
        dataType: "json",
        success: function (result) {//result = Object {ConsultarTodosCategoriasResult: Array(6)}
        debugger;

          var tabela = $("#datagrid");
                            var rows = "";
                            tabela.find("tbody td").remove();

          var myData = ConsultarTodosCategoriasResult;
          for (var i = 0; i < myData.length; i++) {
            var obj = myData[i];
            rows += "<tr>";
               rows += " <td>" + result.ConsultarTodosCategoriasResult.Descricao + "</td>";
            rows += "</tr>";
          }

          tabela.html('<tbody>' + rows + '</tbody>');
            }
        });
    }
 </script>

When assigning the result to a variable (var myData = QueryAllCategoriesResult;) it gives the following error: Uncaught ReferenceError: QueryAllCategoriesResult is not defined

jquery ajax

asked by anonymous 22.05.2017 / 23:47

1 answer

Unknown gulp error Identify an optgroup with jQuery

score 0 · Accepted Answer

I discovered the problem, should put result.ConsultAllCategoriesResult in the variable MyData, the modified code looks like this:

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <title></title>
</head>
<body>
<form id="form1">
    <div class="jumbotron">
    <input type="text" id="codUser"/>
    <button onclick="ConsUsuario(); return false;">Consulta Usuario</button>
</div>

  <div><table id="datagrid"></table></div>
</form>

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script><scripttype="text/javascript">
function ConsUsuario(){
    $.ajax({
    type: "GET",
    url: "http://food-fast-com.web27.redehost.net/CategoriaService.svc/ConsultarTodosCategorias",
    contentType: "application/json",
    dataType: "json",
    success: function (result) {//result = Object {ConsultarTodosCategoriasResult: Array(6)}
    console.log(result)
    debugger;

      var tabela = $("#datagrid");
                        var rows = "";
                        tabela.find("tbody td").remove();

    var myData = result.ConsultarTodosCategoriasResult
    for (var i = 0; i < myData.length; i++) {
      var obj = myData[i];
      rows += "<tr>";
      rows += " <td>" + obj.Descricao + "</td>";
      rows += "</tr>";
    }

      tabela.html('<tbody>' + rows + '</tbody>');
        }
    });
}