使用PHP将图像发布到Cheezburger，json_decode将不会处理响应 [英] Using PHP to post an image to Cheezburger, json_decode won't handle the response

问题描述

我试图使用PHP脚本将图像发布到Cheezburger.com，并将URL返回给用户。 post部分工作正常，我得到的链接，ID等以JSON格式，但是当我运行 json_decode（$ var，true）它只给我原始的JSON背部。下面是送入脚本的字符串：

I am trying to post an image to Cheezburger.com with a PHP script, and return the URL to the user. The post part works fine, I get the links, IDs etc. back in JSON format, but when I run json_decode($var, true) it only gives me the raw JSON back. Here is the string that got fed into the script:

{
    "items": [
        {
            "id": 6980805120,
            "link": "https://api.cheezburger.com/v1/assets/6980805120",
            "created_time": 1358451002,
            "updated_time": 1358451002,
            "media": [
                {
                    "name": "maxW580",
                    "url": "https://i.chzbgr.com/maxW580/6980805120/h89D91707/",
                    "height": 500,
                    "width": 500,
                    "is_animated": false
                },
                {
                    "name": "maxW320",
                    "url": "https://i.chzbgr.com/maxW320/6980805120/h89D91707/",
                    "height": 320,
                    "width": 320,
                    "is_animated": false
                },
                {
                    "name": "square50",
                    "url": "https://i.chzbgr.com/square50/6980805120/h89D91707/",
                    "height": 50,
                    "width": 50,
                    "is_animated": false
                }
            ],
            "title": "JSA, UR WEBSIET IZ AWSUM. URE HIRD!",
            "description": "JSA, UR WEBSIET IZ AWSUM. URE HIRD! -- This image was created by jsa005 from JSiVi using the JSiVi Meme Generator. Try it out at http://jsivi.uni.me!",
            "asset_type_id": 0,
            "share_url": "http://chzb.gr/10Cg1PS"
        }
    ]
}

当我运行 json_decode $ jsonstring，true）上， $ jsonstring 是由cURL返回的包含上面的字符串的变量，我只得到回馈的字符串。我很困惑。

When I run json_decode($jsonstring, true) on that, $jsonstring being the variable returned by cURL containing the string above, I only get back the string I fed in. I am confused.

$fields = array(
'access_token' => $this->getToken(),
'title' => $title,
'description' => $description,
'content' => $base64data,
'anonymous' => 'true');
$url = 'https://api.cheezburger.com/v1/assets';
$fields_string = http_build_query($fields);
//open connection
$ch = curl_init();

//set the url, number of POST vars, POST data
curl_setopt($ch,CURLOPT_URL, $url);
curl_setopt($ch,CURLOPT_POST, count($fields));
curl_setopt($ch,CURLOPT_POSTFIELDS, $fields_string);

//execute post
$result = curl_exec($ch);

//close connection
curl_close($ch);

$jsonstring = json_decode($result, TRUE);

推荐答案

设定

curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);

<没有它，响应直接打印到您的浏览器，所以你看到原始JSON和 $响应是布尔值（ TRUE 或 FALSE ）。 有关详情，请参阅手册页

before running curl_exec($ch); Without it, response is printed directly into your browser, so you are seeing "raw" JSON and $response is boolean value (TRUE or FALSE). See manual page for more details

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