Laravel 5.2路由模型绑定 [英] Laravel 5.2 route model binding

问题描述

Laravel有一个有关路由模型绑定的文档，可以在此处.但是，没有关于这种情况的示例:

Laravel has a documentation regarding route model binding which could be found here. But there is no example with regards to this kind of scenario:

Route::get('search/', 'ArticleController@search');

如何隐式将模型绑定到路由中?我知道我可以直接在控制器的方法上做类似的事情.

How to I implicitly bind a model into the route? I know I could do something like this directly on the controller's method.

public function search(Model $model) {
    // some code here
}

但是我只是好奇如何在路线上做到这一点.

But I'm just curious on how to do it on the routes instead.

我正在采用这种方法

Route::get('search/{article}', function(ArticlesModel $articlesModel) {
    // this should be calling 'ArticleController@search'
});

谢谢！

推荐答案

由于您的变量名为$model，因此Laravel会在网址中查找通配符段，写为{model}:

Because your variable is called $model, Laravel will look for a wildcard segment of the url written as {model}:

在routes.php中:

In routes.php:

Route::get('search/{article}', 'ArticleController@search');

在控制器中:

function search(Article $article) {
    //$article is the Article with the id from {article}, ie. articles/2 is article 2
}

编辑...您建议的方式实际上没有任何意义.那只是一个额外的步骤，仅使用"ArticleController@search"可以完全跳过该步骤.我认为这段代码可以运行，尽管我不建议这样做:

Edit... the way that you are suggesting doesn't really make sense. That would just be an extra step that is skipped entirely by just using "ArticleController@search". I think this code would function although I don't recommend it:

Route::get('search/{article}', function(Article $article)
{
    $controller = App::make(ArticleController::class);
    return App::call([$controller, 'search'], compact('article'));
}

这篇关于Laravel 5.2路由模型绑定的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！