当PHP生成JSON时，jQuery不会将JSON对象返回给Firebug

test.json的内容包括：

{"foo": "The quick brown fox jumps over the lazy dog.","bar": "ABCDEFG","baz": [52, 97]}

当我使用以下jQuery.ajax（）调用来处理test.JSON中的静态JSON时

$.ajax({
    url: 'test.json',
    dataType: 'json',
    data: '',
    success: function(data) {
        $('.result').html('<p>' + data.foo + '</p>' + '<p>' + data.baz[1] + '</p>');
    }
});

我在浏览器中获得了正确的输出，但Firebug只显示HTML。在Firebug中展开GET请求时，不存在JSON选项卡。

试试看

<?php
    header('Content-type: application/json');
    $arrCoords = array('foo'=>'The quick brown fox jumps over the lazy dog.','bar'=>'ABCDEFG','baz'=>array(52,97));
    echo json_encode($arrCoords);
?>

试试看

---

您最有可能需要设置内容类型标题：

// this is the "offical" content-type
header('Content-Type: application/json');
// or - this will probably work too
header('Content-Type: text/javascript');

---

您最有可能需要设置内容类型标题：

// this is the "offical" content-type
header('Content-Type: application/json');
// or - this will probably work too
header('Content-Type: text/javascript');

---

查看响应选项卡上的响应选项卡

// this is the "offical" content-type
header('Content-Type: application/json');
// or - this will probably work too
header('Content-Type: text/javascript');