我试图从php $ _GET获取数据，然后使用rest api解析它，但它会导致错误

Question

我试图从php $ _GET获取数据然后使用rest api解析它，但它会导致错误，所以请任何人帮助..

<?php  
$number=$_GET['name'];
$media=$_GET['media'];

$url = 'https://api.parse.com/1/classes/AppTo';  
$appId = 'ccccc';  
$restKey = 'ccccc';  
$headers = array(  
"Content-Type: application/json",  
"X-Parse-Application-Id: " . $appId,  
"X-Parse-REST-API-Key: " . $restKey  
);  

$objectData = '{"name":(json_encode($number)), "age":(json_encode($media))}';  
$rest = curl_init();  

curl_setopt($rest,CURLOPT_URL,$url);  
curl_setopt($rest,CURLOPT_POST,1);  
curl_setopt($rest,CURLOPT_POSTFIELDS,$objectData);  
curl_setopt($rest,CURLOPT_HTTPHEADER,$headers);  
curl_setopt($rest,CURLOPT_SSL_VERIFYPEER, false);  
curl_setopt($rest,CURLOPT_RETURNTRANSFER, true);  
$response = curl_exec($rest);  
echo $response;  
print_r($response);  
curl_close($rest);  
?> 

Answer 1

您实际上是在发布文字{"name":(json_encode($number)), "age":(json_encode($media))}。你可以这样构建你的json：

$objectData = json_encode(array('name' => $number, 'age' => $media));

Answer 2

您发布了错误的数据更改

$objectData = '{"name":(json_encode($number)), "age":(json_encode($media))}';

到

$objectData = json_encode(array('name' => $number, 'age' => $media));