如何將'content-type application / json'的數據發布到苗條的php rest服務

Question

我在將來自Chrome Advance rest Client的mime類型content-type application/json數據發布到苗條框架Web服務時遇到問題。 我嘗試了這些代碼以發送application/json

$app->post('/register', function() use ($app) {
            $app->add(new \Slim\Middleware\ContentTypes());            
            $params = $app->request->getBody();
            $name = $params->name;
            $email = $params->email;
            $password = $params->password;
            ...});

也嘗試了這個

$params = json_decode($app->request()->getBody());


 var_dumb($params); //get NULL value here

得到錯誤

Trying to get property of non-object to this `$name = $params->name;`

請幫我如何捕獲application / json格式的數據？ 謝謝

Answer 1

根據上面的詳細信息，假設您的原始JSON看起來像這樣

{"name":"John Smith", "mail":"jhon@mail.com", "password":"foobar"}

您可以像這樣訪問您的params 數組

$app->post('/register', function () use ($app) {
$params =  $app->request->getBody() ;
$params = array_filter($params);

if(!empty($params)){
    $name = $params['name'];
    $mail = $params['mail'];
    $pass = $params['password'];

   // print $name;
  }

})->name("register");

或者，如果您通過Content-Type: application/x-www-form-urlencoded在Advanced Rest客戶端中發布Content-Type: application/x-www-form-urlencoded ，則可以使用$app->request->post(); 訪問您的數組

$app->post('/register/', function () use ($app) {

   $userInfo = $app->request()->params() ;
   //or
   $userInfo =  $app->request->post() ;
   $name = $userInfo['name'];
   $mail = $userInfo['email'];
   $pass = $userInfo['password'];

 // print $name

})->name("register");