将JSON传递给PHP后进行解码

Parsing JSON in PHP

在PHP中解析JSON

My variable jsonvar is in the form of {"rating":"good"}

我的变量jsonvar采用{“rating”：“good”}的形式

Once I push it through with this $.ajax code to submit, I'm a bit confused at what my PHP (jsonproc.php) should look like.

一旦我用这个$ .ajax代码提交它，我对我的PHP（jsonproc.php）看起来有点困惑。

$.ajax({
                url: 'jsonproc.php',
                data: {js:jsonvar},
                dataType: 'json',
                type: 'post',
                success: function (j) {
                        if (j.ok){
                            alert(j.msg);
                        } else {
                            alert(j.msg);
                        }
                    }
                });

I have it set up as

我把它设置为

$decoded = $_GET['js'];
$response = array(
   'ok' => true, 
   'msg' => $decoded['rating']);

However when I echo it back,

然而，当我回复它，

   echo json_encode($response);

using alert(j.msg) shows a "null" value.

使用alert（j.msg）显示“null”值。

Assuming that I am passing JSON in correctly, how can I point to the rating and get the value of "good"?

假设我正确地传递JSON，我如何指向评级并获得“好”的值？

Thanks

谢谢

EDIT

编辑

SOLVED, USING $_REQUEST I was able to get the JSON, however $_GET did not work.

已解决，使用$ _REQUEST我能够获得JSON，但$ _GET无效。

Also, the key was using $decoded->{'rating'} as $decoded is no longer just an array i don't think or rather it's a diff type of one?

另外，关键是使用$ decoding - > {'rating'}因为$ decode不再仅仅是一个我不认为的数组，或者它是一个diff类型的？

2 个解决方案

#1

It looks like you're mixing data types here:

看起来你在这里混合数据类型：

 data: "js="+jsonvar,

jQuery will convert JSON if you pass an object, but you're mixing query string with JSON.

如果传递一个对象，jQuery将转换JSON，但是您将查询字符串与JSON混合。

Try:

尝试：

data: {js: jsonvar},

You may also need to do json_decode($_GET['js']).

您可能还需要执行json_decode（$ _ GET ['js']）。

edit: You can double check what jQuery is POSTing with Firebug/Web Inspector. Easiest way to know for sure.

编辑：您可以使用Firebug / Web Inspector仔细检查jQuery正在POST的内容。最简单的方式来确定。

#2

-2

Try adding this to the top of your PHP file:

尝试将其添加到PHP文件的顶部：

header('Content-type: application/json');

#1