PHP努力解码JSON
我的脚本通过cURL调用script。看起来是这样的,PHP努力解码JSON,php,laravel,json,Php,Laravel,Json,我的脚本通过cURL调用script。看起来是这样的, Route::get('login-redirect', function() { if (Input::has('error')) { return Input::get('error_description'); } if (Input::has('code')) { $fields = array( 'grant_type' => 'password', 'username'
Route::get('login-redirect', function() {
if (Input::has('error')) {
return Input::get('error_description');
}
if (Input::has('code')) {
$fields = array(
'grant_type' => 'password',
'username' => 'admin@local.com',
'password' => 'passwohrd',
'client_id' => 'testclient'
);
$fieldstring = http_build_query($fields, "\n");
$url = "http://apitest.local/api/v1/get-token";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_POST, count($fields));
curl_setopt($ch, CURLOPT_POSTFIELDS, $fieldstring);
$result = curl_exec($ch);
$json = json_decode($result);
curl_close($ch);
$fields = array('access_token' => '3c1e6b099f172fc01304403939edf8e56904ab61');
$fieldstring = http_build_query($fields, "\n");
$url = "http://apitest.local/api/v1/me";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_POST, count($fields));
curl_setopt($ch, CURLOPT_POSTFIELDS, $fieldstring);
$result = curl_exec($ch);
curl_close($ch);
dd($result);
}
如果我执行dd($json)
{“内容”:null,“错误”:true,“错误描述”:“无效用户名和密码组合”}int(1)
我觉得在运行了json\u decode
之后,我应该能够只输出$json->error
,但是没有
JSON是在下面的类中生成的,但是我在这里也看不到任何奇怪的东西,我做的不正确,或者我误解了JSON_解码
<?php
namespace Shaunpersad\ApiFoundation\Http;
use App;
use Response;
class ErrorResponse
{
public static function make($message = '', $status = 200, array $headers = array(), $options = 0)
{
$response = App::make(
'api_response_array',
array(
'content' => null,
'error' => true,
'error_description' => $message
)
);
return Response::json($response, $status, $headers, $options);
}
}
更新
如其他答案中所述,您实际上没有收到输出,因为您没有设置CURLOPT\u RETURNTRANSFER
。因此,当curl
请求成功运行时,curl\u exec()
通过在某个地方的curl请求中设置此选项,您可以运行以下内容:
curl_setop(CURLOPT_RETURNTRANSFER, true);
dd()
是一个函数,文档中是这么说的:
转储给定变量并结束脚本的执行
我假设它只是一个包装器函数,用于外观更漂亮的var\u dump()
(因为我没有使用laravel,所以我不知道它的确切输出)
您需要解码从cUrl
返回的$result
。这样的东西应该足够了:
$data = json_decode($result);
echo $data->error_description;
成功解码的对象如下所示:
stdClass Object
(
[content] =>
[error] => 1
[error_description] => Invalid username and password combination
)
您甚至可以像现在这样测试布尔值错误值:
if($data->error) {
//....true
} else {
//....false
}
首先,您没有CURLOPT_RETURNTRANSFER—您的curl_exec将输出缓冲区直接返回到屏幕
其次,看起来您在某个地方有var_转储,我看不到在哪里:)
第三,你没有问任何直接的问题
编辑
好的,我已经读了几遍了,下面回答。dd()函数实际上是一个var_dump包装器,但它将var_dump数据转储到json格式的afaics中。作为输出的不是来自dd($json)
:
原因如下:
// no CURLOPT_RETURNTRANSFER, so curl_exec() outputs result and returns true:
$result = curl_exec($ch);
// thus $result = true;
// so here $json = 1, since this is what json_decode(true) will return
$json = json_decode($result);
// then you did dd($json), so it just appended var_dump(1) to the output:
{"content":null,"error":true,"error_description":"Invalid username and password combination"}int(1)
如果int(1)
实际上包含在响应中,而不是复制/粘贴错误,则响应无效。
// no CURLOPT_RETURNTRANSFER, so curl_exec() outputs result and returns true:
$result = curl_exec($ch);
// thus $result = true;
// so here $json = 1, since this is what json_decode(true) will return
$json = json_decode($result);
// then you did dd($json), so it just appended var_dump(1) to the output:
{"content":null,"error":true,"error_description":"Invalid username and password combination"}int(1)