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Laravel 如何在请求中向重定向路由传递参数_Laravel_Parameters_Laravel 5_Routes_Request - Fatal编程技术网

Laravel 如何在请求中向重定向路由传递参数

Laravel 如何在请求中向重定向路由传递参数,laravel,parameters,laravel-5,routes,request,Laravel,Parameters,Laravel 5,Routes,Request,语言是路由文件中的模型绑定 路线 我喜欢声明一个扩展请求(FormRequest)的请求(DictionaryRequest)文件,它负责控制器上的请求参数 方法原型为: public function store(DictionaryRequest $request, Language $lang) 请求类中的重定向路由设置为: protected $redirectRoute = "dictionary.create"; 如何将参数传递到路由??(朗格模型) 我检查了FormReques

语言是路由文件中的模型绑定

路线

我喜欢声明一个扩展请求(FormRequest)的请求(DictionaryRequest)文件,它负责控制器上的请求参数

方法原型为:

public function store(DictionaryRequest $request, Language $lang)
请求类中的重定向路由设置为:

protected $redirectRoute = "dictionary.create";
如何将参数传递到路由??(朗格模型)

我检查了FormRequest类,但redirectRoute只是传递到UrlGenerator,没有参数

/**
 * Get the URL to redirect to on a validation error.
 *
 * @return string
 */
protected function getRedirectUrl()
{
    $url = $this->redirector->getUrlGenerator();

    if ($this->redirect) {
        return $url->to($this->redirect);
    } elseif ($this->redirectRoute) {
        return $url->route($this->redirectRoute);
    } elseif ($this->redirectAction) {
        return $url->action($this->redirectAction);
    }

    return $url->previous();
}

您是否尝试覆盖
getRedirectUrl

/**
 * Get the URL to redirect to on a validation error.
 *
 * @return string
 */
protected function getRedirectUrl()
{
    $url = $this->redirector->getUrlGenerator();

    return $url->route($this->redirectRoute, [ /*your parameters*/ ]);

}

顺便说一句,我不想使用$url->previous();也是。你需要传递整个模型实例来代替id吗?@huuuk没关系
/**
 * Get the URL to redirect to on a validation error.
 *
 * @return string
 */
protected function getRedirectUrl()
{
    $url = $this->redirector->getUrlGenerator();

    return $url->route($this->redirectRoute, [ /*your parameters*/ ]);

}