Laravel 5.5 PersonalAccessToken table 新列

Question

我已经设法在 oauth_access_tokens table 中添加了一个名为 callback_url 的新列。我遇到的问题是当我试图将 callback_url 值传递给 createToken 函数时。

public function store(Request $request)
{

    $this->validation->make($request->all(), [
        'name' => 'required|max:255',
        'scopes' => 'array|in:'.implode(',', Passport::scopeIds()),
    ])->validate();

    return $request->user()->createToken(
        $request->name, $request->callback_url, $request->scopes ?: []
    );
}

VUEJS 组件：

store() {
            this.accessToken = null;

            this.form.errors = [];

            axios.post('/user/token', this.form)
                    .then(response => {
                        this.form.name = '';
                        this.form.callback_url = '';
                        this.form.scopes = [];
                        this.form.errors = [];

                        this.tokens.push(response.data.token);

                        this.showAccessToken(response.data.accessToken);
                    })
                    .catch(error => {
                        if (typeof error.response.data === 'object') {
                            this.form.errors = _.flatten(_.toArray(error.response.data));
                        } else {
                            this.form.errors = ['Something went wrong. Please try again.'];
                        }
                    });
        },

它在没有 $request->callback_url 的情况下工作正常，但是有了那个额外的参数，这是一个错误，我正在返回。

Whoops! Something went wrong!

Type error: Argument 2 passed to Models\User::createToken() must be of the type array, string given, called in UserAccessTokenController.php on line 62 Symfony\Component\Debug\Exception\FatalThrowableError/vendor/laravel/passport/src/HasApiTokens.php 64 { "file": "/UserAccessTokenController.php", "line": 62, "function": "createToken", "class": "Models\User", "type": "->" } { "function": "store", "class": "UserAccessTokenController", "type": "->" }

有没有人遇到过这个问题或知道如何处理这个问题。提前致谢。

Answer 1

尝试将 $request->callback_url 转换为这样的数组：

(array)$request->callback_url

Answer 2

函数createToken() 只接受两个参数。我通过创建 accessToken 然后用新的 callback_url 值更新结果来解决我的问题。希望对您有所帮助。

 public function store(Request $request)
{

    $token = $request->user()->createToken(
        $request->name, $request->scopes ?: []
    );

    if ($request->callback_url){

        $url = $request->callback_url;

        $request->user()->tokens()->where( 'id', $token->token->id )->update( [
            'callback_url' => $url,
        ] );

    }

    return $token;

}

Laravel 5.5 PersonalAccessToken table 新列

Laravel 5.5 PersonalAccessToken table new column

php

mysql

laravel-5

vuejs2

laravel-passport