我正在尝试使用PHP脚本将图像发布到Cheezburger.com,并将URL返回给用户。帖子部分工作正常,我以JSON格式获取链接,ID等,但是当我运行json_decode($var, true)
时,它只返回原始JSON。这是输入脚本的字符串:
{
"items": [
{
"id": 6980805120,
"link": "https://api.cheezburger.com/v1/assets/6980805120",
"created_time": 1358451002,
"updated_time": 1358451002,
"media": [
{
"name": "maxW580",
"url": "https://i.chzbgr.com/maxW580/6980805120/h89D91707/",
"height": 500,
"width": 500,
"is_animated": false
},
{
"name": "maxW320",
"url": "https://i.chzbgr.com/maxW320/6980805120/h89D91707/",
"height": 320,
"width": 320,
"is_animated": false
},
{
"name": "square50",
"url": "https://i.chzbgr.com/square50/6980805120/h89D91707/",
"height": 50,
"width": 50,
"is_animated": false
}
],
"title": "JSA, UR WEBSIET IZ AWSUM. URE HIRD!",
"description": "JSA, UR WEBSIET IZ AWSUM. URE HIRD! -- This image was created by jsa005 from JSiVi using the JSiVi Meme Generator. Try it out at http://jsivi.uni.me!",
"asset_type_id": 0,
"share_url": "http://chzb.gr/10Cg1PS"
}
]
}
当我运行json_decode($jsonstring, true)
时,$jsonstring
是包含上述字符串的cURL返回的变量,我只返回我输入的字符串。我很困惑。
$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);
答案 0 :(得分:2)
组
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
在运行curl_exec($ch);
之前没有它,响应将直接打印到您的浏览器中,因此您看到“原始”JSON,$response
是布尔值(TRUE
或FALSE
)。 See manual page for more details