读jQuery的JSON结构 - 不能得到它的工作它的、结构、工作、jQuery

2023-09-10 18:17:26 作者:失心

我有以下的JSON结构,正在从一个AJAX调用返回(伪值)

I have the following JSON structure, being returned from an AJAX call (pseudo values)

[
  {
    "id":"14",
    "product_title":"Foo Bar v1.2",
    "discount":"10% Off - New price: $97",
    "vendor":"foobar"
  }
]

下面是我的AJAX程序。

Here is my AJAX routine.

    $.post('Ajax.php',function(res){
        alert(res.product_title);
      });
    },'json');

然而,警报说未定义,即使它显然不是(萤火虫)。

However, the alert says "Undefined", even though it is clearly not (Firebug).

我试着做一个 $每(); 上的资源,它出现的唯一钥匙中有0

I tried doing a $.each(); on the res, and it appears the only "key" in there is "0".

我在做什么错在这里?是它的方式,我处理的结构?干杯!

What am I doing wrong here? Is it the way I am processing the structure? Cheers!

推荐答案

尝试:

var j = res.pop(); // this will extract the `object` form the `array`
console.log(j.product_title);

您也可以这样写:

res.pop().product_title;