Question

I am trying to access the elements within the text JSON reference from the call, for example, I am trying to access the item (content)
在JSON内，但继续显示错误消息该代码有效，并显示JSON文件的全文没问题是否有类似的PHP lavarel方法可以用作arrylist或用于获取响应中的元素

$client = new Client();
$request = $client->post('http://localhost:4567/user/login', [
                'headers' => ['Content-Type' => 'application/json'],
                'body' => json_encode($data),
          ]);
$request= json_decode($request->getBody(), true);
return $request[0];

Answer 1

您的代码应该像

use GuzzleHttp\Client;

$client = new Client();
    $request = $client->post('http://localhost:4567/user/login', [
                'headers' => ['Content-Type' => 'application/json'],
                'body' => json_encode($data),
            ]);
 return $request->getBody();

处理ErrorException（E_NOTICE）未定义偏移量：0

1 个答案: