我想在登錄后使用令牌傳遞用戶的角色。而不是像下面這樣傳遞 return response()->json([ 'success' => true, 'access_token' => $token, 'token_type' => 'bearer', 'role' => auth('api')->user()->getRoleNames() /* 'expires_in' => auth('api')->factory()->getTTL() * 60 */ ]);我想使用 JWT 令牌對用戶的角色進行編碼。我試過的 public function login() { $credentials = request(['email', 'password']); if (! $token = auth()->guard('api')->attempt($credentials)) { return response()->json(['errors' => 'In-valid username and Password'], 401); } return $this->respondWithToken($token); } protected function respondWithToken($token) { $role =auth('api')->user()->getRoleNames(); $payload = JWTFactory::make($role); $token = JWTAuth::encode($payload); return response()->json([ 'success' => true, 'access_token' => $token, 'token_type' => 'bearer', ]); }當我嘗試這個時，我沒有得到令牌。我已經看過這個文檔https://github.com/tymondesigns/jwt-auth/wiki/Creating-Tokens來做到這一點。
1 回答
慕妹3146593
TA貢獻1820條經驗 獲得超9個贊
在這里，我們需要使用用戶角色創建自己的令牌。對我有用的代碼可能對其他人有幫助。
public function login()
{
$credentials = request(['email', 'password']);
if (!auth()->guard('api')->claims(['role' => 'bar'])->attempt($credentials)) {
return response()->json(['errors' => 'In-valid username and Password'], 401);
}
$token = auth('api')->claims(['role' => str_replace(['["', '"]', '/', '*'], '',auth('api')->user()->getRoleNames())])->attempt($credentials);
return $this->respondWithToken($token);
}
protected function respondWithToken($token)
{
auth('api')->payload($token);
return response()->json([
'success' => true,
'access_token' => $token,
'token_type' => 'bearer',
]);
}
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