使用PHP从JSON响应中提取值

我将JSON数据发送到URL,然后我也收到了JSON.
我正在使用PHP CURL来接收数据并在我的页面上显示它.
我的问题是我只能完全获得json,而且我不能只显示该JSON数据中的选定值.
我已经得到了这个：

//API Url
$url = "https://auth.unilinxx.com/test";

//Initiate cURL.
$ch = curl_init($url);

//The JSON data.
$jsonData = array("value" => "Gideon lol");

//Encode the array into JSON.
$jsonDataEncoded = json_encode($jsonData);

curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
//Tell cURL that we want to send a POST request.
curl_setopt($ch, CURLOPT_POST, 1);

//Attach our encoded JSON string to the POST fields.
curl_setopt($ch, CURLOPT_POSTFIELDS, $jsonDataEncoded);

//Set the content type to application/json
curl_setopt($ch, CURLOPT_HTTPHEADER, array("Content-Type: application/json")); 

//Execute the request
$result = curl_exec($ch);

$values = json_decode($result, true);

echo $values["value"];

我的结果是：{“value”：“U heeft Gideon lol gestuurd.”}

我喜欢只有这个：U heeft Gideon lol gestuurd.

我怎样才能做到这一点？

解决方法:

您的整个代码都是正确的,只需在代码中将CURLOPT_RETURNTRANSFER添加为true即可.其他明智的结果不会存储在变量中.并直接输出到屏幕

<?PHP
//API Url
$url = "https://auth.unilinxx.com/test";

//Initiate cURL.
$ch = curl_init($url);

//The JSON data.
$jsonData = array("value" => "Gideon lol");

//Encode the array into JSON.
$jsonDataEncoded = json_encode($jsonData);

curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
//Tell cURL that we want to send a POST request.
curl_setopt($ch, CURLOPT_POST, 1);

//Attach our encoded JSON string to the POST fields.
curl_setopt($ch, CURLOPT_POSTFIELDS, $jsonDataEncoded);

//Set the content type to application/json
curl_setopt($ch, CURLOPT_HTTPHEADER, array("Content-Type: application/json")); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); //<---------- Add this line
//Execute the request
$result = curl_exec($ch);

$values = json_decode($result, true);

echo $values["value"];