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  • laravel学习之路2: jwt集成

    "tymon/jwt-auth": "^1.0@dev",
    执行 composer update
    'providers' => [ .... TymonJWTAuthProvidersLaravelServiceProvider::class, // 上文已经提到过,这里的provider已经不是JWTauthServiceProvider ], 'aliases' => [ .... 'JWTAuth' => TymonJWTAuthFacadesJWTAuth::class ],
    发布配置文件#
    php artisan vendor:publish --provider="TymonJWTAuthProvidersLaravelServiceProvider"
    php artisan jwt:secret
    /**
    * Get the value of the model's primary key.
    *
    * @return mixed
    */
    public function getKey()
    {
    return $this->getAttribute($this->getKeyName());
    }
    这个一般是得到id
    调用Auth::guard('your jwt guard name')->attempt($credentials)
    实际是调用了JWTGuard.php里面的attempt方法
    oh yeah,终于生成了jwt token了
    {"register_result1":"eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJpc3MiOiJodHRwOi8vbGFyYXZlbF9hcGkuYXBwL2FwaS9yZWdpc3RlciIsImlhdCI6MTUwMjM1NjE1MCwiZXhwIjoxNTAyMzU5NzUwLCJuYmYiOjE1MDIzNTYxNTAsImp0aSI6ImJSSHZsUXB5ZzN1WGtTR2MiLCJzdWIiOjksInBydiI6IjM3ODdmYmExNjE4YTkzMDUyNmFjYTZjOGJiOWI0NGI4M2YyOTc3MjYifQ.JpqCVjZggb2BHsCEXzITdnX70HbYIAfQY-iYSpkfHSw"}
    $credentials我感觉不应该先用bcrypt,不然attempt会验证失败,只有插入数据库的时候才需要bcrypt
    protected function getClaimsForSubject(JWTSubject $subject)
    {
    return [
    'sub' => $subject->getJWTIdentifier(), //主键id
    'prv' => $this->hashProvider($subject),
    ];
    }
    如何自定义customClaims
    public function getJWTCustomClaims()
    {
    // TODO: Implement getJWTCustomClaims() method.
    return ['key1' => 'elesos', 'key2' => 'test'];
    }
    上面是静态的,如何动态的加呢?
    $client = DB::select('select * from clients where email = ?', [$email]);
    //return $client;
    //var_dump($client);return;//数组里面是对象元素
    $customClaims = ['name' => $client[0]->name, 'vip_level' => '1'];
    Auth::guard('client')->customClaims($customClaims);
    接口如何访问
    Route::get('/test1', function () {
    return ['state' => 1, 'data' => 'sucess'];
    })->middleware('auth:client');
    或者
    http://api.mysite.com/me?token={yourtokenhere}
    验证token信息。
    public function validate_test(){
    //echo 'validate_test';
    //$token = JWTAuth::getToken();
    //return $token;
    try {
    if (! $user = JWTAuth::parseToken()->authenticate()) {
    return response()->json(['user_not_found'], 404);
    }
    } catch (TymonJWTAuthExceptionsTokenExpiredException $e) {
    return response()->json(['token_expired'], $e->getStatusCode());
    } catch (TymonJWTAuthExceptionsTokenInvalidException $e) {
    return response()->json(['token_invalid'], $e->getStatusCode());
    } catch (TymonJWTAuthExceptionsJWTException $e) {
    return response()->json(['token_absent'], $e->getStatusCode());
    }
    // the token is valid and we have found the user via the sub claim
    return response()->json(compact('user'));
    }
    下一步要实现错误时返回json,而不是错误页面
    Add the following code to the render method within app/Exceptions/Handler.php
    public function render($request, Exception $e) { if ($e instanceof TymonJWTAuthExceptionsTokenExpiredException) { return response()->json(['token_expired'], $e->getStatusCode()); } else if ($e instanceof TymonJWTAuthExceptionsTokenInvalidException) { return response()->json(['token_invalid'], $e->getStatusCode()); } return parent::render($request, $e); }
    或全部
    // 这是我自己错定义的错误
    return response()->json(array('error_code' => $e->getStatusCode()));
    // 这是默认的错误返回,已注释了
    //return parent::render($request, $e);
    return response()->json(['errcode' => 4000, 'errmsg' => $exception->getMessage()], 200);
    开发环境,当 APP_DEBUG = true 时,使用默认错误页面;
    生产环境,当 APP_DEBUG = false 时,使用自定义错误页面,异步请求返回json异常信息
    修改app/Exceptions/Handler.php
    1. public function render($request, Exception $exception)
    2. {
    3. $debug = config('app.debug', false);
    4. if($debug) {
    5. return parent::render($request, $exception);
    6. }
    7. if ($exception instanceof HttpException) {
    8. $code = $exception->getStatusCode();
    9. $message = $exception->getMessage();
    10. if ($request->expectsJson()) {
    11. return response()->json(['error' => $message], $code);
    12. }
    13. if (view()->exists('errors.custom' . $code)) {
    14. return response()->view('errors.custom' . $code, ['message'=>$message], $code);
    15. }
    16. }
    17. return parent::render($request, $exception);
    18. }
    已知laravel5的默认ExceptionsHandler会优先匹配404异常,所以建议在Handler进行处理。
    修改app/Exceptions/Handler.phprender方法如下
    /** * Render an exception into an HTTP response. * * @param IlluminateHttpRequest $request * @param Exception $exception * @return IlluminateHttpResponse */ public function render($request, Exception $exception) { if (is_a($exception, SymfonyComponentHttpKernelExceptionNotFoundHttpException::class) && $request->expectsJson()) { return response()->json(['msg'=>'NotFound']); } else { return parent::render($request, $exception); } }
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  • 原文地址:https://www.cnblogs.com/elesos/p/7344963.html
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