Question

我遇到将Chrome预设休息客户端的mime类型content-type application/json数据发布到精简框架网络服务的问题。我尝试使用这些代码发送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;
            ...});

也尝试了这个

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


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

错误

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

请帮我看看如何捕获application / json格式的数据？谢谢

Answer 1

根据上面的细节，假设您的原始JSON看起来像这样

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

您可以像这样

访问你的参数数组

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

或者如果您通过Content-Type: application/x-www-form-urlencoded在Advanced Rest客户端发帖，则可以使用$app->request->post();访问您的阵列

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

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

 // print $name

})->name("register");

如何发布＆＃39;内容类型应用程序/ json＆＃39;瘦身php休息服务

1 个答案: