如何在PHP变量中获取JSON响应

Question

我正在获取JSON格式的响应，并且希望它将其转换为PHP变量。

JSON：

{"CreateTransactionResponse":{"CreateTransactionResult":{"TransportKey":"aa900d54-7bfb-47e9-a5de-e423ec34a900"
,"ValidationKey":"fbb28b32-f439-4801-a434-99c70aa388ca","Messages":{}}}}

输出应为PHP：

$transkey = aa900d54-7bfb-47e9-a5de-e423ec34a900;
$vkey = fbb28b32-f439-4801-a434-99c70aa388ca

请告诉我该怎么做。

Answer 1

如果要访问您的json，请尝试先对其进行解码：

$result = json_decode($yourJSON, true);

foreach($result['CreateTransactionResponse'] as $key => $val){
   echo $transkey = 'TransportKey= ' . $val['TransportKey'] . '<br/>;
   echo $vkey = 'ValidationKey= ' . $val['ValidationKey'];
}

或者，如果它是JSON的数组

$result = json_decode($yourJSON, true);

$data = [];
foreach($result['CreateTransactionResponse'] as $key => $val){
   $data[] = [
        'TransportKey' => $val['TransportKey'],
        'ValidationKey' => $val['ValidationKey']
   ];
}
print_r($data);

Answer 2

只需简单地使用json_decode（）;

 $result= json_decode($jSon);

 var_dump($result); // to see the output

Answer 3

json到array（ json_decode ），然后从数组中extract 。

$arr = json_decode($json, true);
extract($arr);
var_dump($CreateTransactionResponse);

输出：

array (size=1)
  'CreateTransactionResult' => 
    array (size=3)
      'TransportKey' => string 'aa900d54-7bfb-47e9-a5de-e423ec34a900' (length=36)
      'ValidationKey' => string 'fbb28b32-f439-4801-a434-99c70aa388ca' (length=36)
      'Messages' => 
        array (size=0)
          empty

有关提取的更多信息

使用$CreateTransactionResult['TransportKey']从JSON访问传输密钥。 类似地， $CreateTransactionResult['ValidationKey']用于验证密钥。

Answer 4

    try this code it will work  
$JSON='{"CreateTransactionResponse":{"CreateTransactionResult":{"TransportKey":"aa900d54-7bfb-47e9-a5de-e423ec34a900" ,"ValidationKey":"fbb28b32-f439-4801-a434-99c70aa388ca","Messages":{}}}}';    

    $arr=json_decode($JSON, TRUE);
    foreach ($arr as  $value) {

    foreach ($arr['CreateTransactionResponse'] as $key => $var) {
        echo 'TransportKey = '.$var['TransportKey'].'<br>';
        echo 'ValidationKey = '.$var['ValidationKey'].'<br>';
        foreach ($var['Messages'] as $key => $msg) {


        echo 'Messages = '.$msg.'<br>';
    }

        }
    }

Answer 5

在这种情况下，如果一次同时使用一个和TransportKey以及一个ValidationKey值（不传递数组/对象），则这是最简单的 。 否则，如果对象包含我们要使用或转换为变量的对象或内部对象，则应使用foreach该对象。

//Debuggig
//The string you provided is converted to a json object 
//In your case if it is a json object already pass directly it to $j
//below is just for debugging and understanding
//$json='{"CreateTransactionResponse":{"CreateTransactionResult":{"TransportKey":"aa900d54-7bfb-47e9-a5de-e423ec34a900","ValidationKey":"fbb28b32-f439-4801-a434-99c70aa388ca","Messages":{}}}}';
//$j=json_decode($json);

$transkey=$j->CreateTransactionResponse->CreateTransactionResult->TransportKey;
$vkey=$j->CreateTransactionResponse->CreateTransactionResult->ValidationKey;

echo $transkey."</br>";
echo $vkey."<br/>";
/*result as said: 
aa900d54-7bfb-47e9-a5de-e423ec34a900
fbb28b32-f439-4801-a434-99c70aa388ca
*/