Question

如何在JavaScript中访问此json数据。当我提醒它时，结果是undefined

这是jQuery代码

$.ajax({
    type: "POST",
    url: "frmMktHelpGridd.php",
    data: {
        labNo: secondElement
    },
    dataType: "json",
    beforeSend: function () {
        // Do something before sending request to server
    },
    error: function (jqXHR, textStatus, errorThrown) {
        alert('error has occured');
        alert(errorThrown);
    },
    success: function (data) {

        //Here is the problem

        alert(data[0]['Result']);
    }
});

这是PHP代码

            $data=array($no);
    for($i=0;($i<$no && ($row=mysql_fetch_array($result)));$i++)
    {
        $data[$i]=array();
        $data[$i]['Result']         =   $row['Result'];         
        $data[$i]['TestCode']       =   $row['TestCode'];           
        $data[$i]['TestStatus']     =   $row['TestStatus'];         
        $data[$i]['SrNo']           =   $row['SrNo'];               
    }

    $data1=json_encode($data);

    echo $data1;
      exit;

我已经独立测试了PHP文件， json数据输出如下：

      [{"Result":"1","TestCode":"22","TestStatus":"0","SrNo":"1"},{"Result":"1","TestCode":"23","TestStatus":"1","SrNo":"2"}]

Answer 1

$.ajax({
    type: "POST",
    url: "frmMktHelpGridd.php",
    data: {
        labNo: secondElement
    },
    dataType: "json",
    beforeSend: function () {
        // Do something before sending request to server
    },
    error: function (jqXHR, textStatus, errorThrown) {
        alert('error has occured');
        alert(errorThrown);
    },
    success: function (data) {
        //Added parse json
        var data = jQuery.parseJSON(data)

        alert(data[0]['Result']);
    }
});

Answer 2

您可以通过

访问您的数据

data[0].Result

它是一个对象，而不是一个数组。

所以data[0]['Result']这不是正确的方式

修改：由于你有更多的对象，你必须以这种方式循环：

$.each(data, function(key, val){ console.log(val.Result); console.log(val.TestCode); //... });

当您看到类似
的内容时
{ "foo":"bar", ... }

您可以像上面一样访问它：

name_of_the_object.foo

将具有值“bar”

Answer 3

尝试添加解析JSON。我已经添加了。现在可能有用了。

$.ajax({
    type: "POST",
    url: "frmMktHelpGridd.php",
    data: {
        labNo: secondElement
    },
    dataType: "json",
    beforeSend: function () {
        // Do something before sending request to server
    },
    error: function (jqXHR, textStatus, errorThrown) {
        alert('error has occured');
        alert(errorThrown);
    },
    success: function (data) {
        //Added parse json
        var data = $.parseJSON(data)

        alert(data[0]['Result']);
    }
});

无法使用jquery $ .ajax访问从php页面检索的json数据

3 个答案: