在刀片模板中使用命名URL

本文介绍了在刀片模板中使用命名URL的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

在Django中，我可以这样做:

In Django, I can do this:

<a href="{% url 'account_login' %}">Account Link</a>

这将给我domain/account/login在我的urls.py中命名该URL的位置

which would give me domain/account/login where I have that URL named in my urls.py

url(r'^account/login/$', views.Login.as_view(), name='account_login'),

我想在Laravel 5.2中做类似的事情

I want to do something similar in Laravel 5.2

我目前有这样的东西:

Route::get('/survey/new', ['as' => 'new.survey', 'uses' => 'SurveyController@new_survey']);

如何在模板中使用以及传递参数?

How do I use in my template, plus passing in parameters?

我遇到了这个问题: laravel/docs/5.1/helpers ，但这只是一张白纸，没有实际使用方法的相关内容.

I came across this: laravel/docs/5.1/helpers, but it was just a piece of a white page without relevant content of how to actually use it.

推荐答案

您可以使用记录在route()帮助器. rel ="noreferrer">此处.

You can use the route() helper, documented here.

<a href="{{ route('new.survey') }}">My Link</a>

如果包含laravelcollective/html，则可以使用link_to_route().它最初是核心的一部分，但已在Laravel 5中删除.在此处

If you include laravelcollective/html you could use link_to_route(). This was originally part of the core but removed in Laravel 5. It's explained here

{!! link_to_route('new.survey', 'My Link') !!}

Laravel集体在此处中记录了上述帮助器.函数原型如下

The Laravel Collective have documented the aforementioned helper here. The function prototype is as follows

link_to_route($routeName, $title = null, $parameters = [], $attributes = []);

例如，如果您想使用参数，则它接受键值对的数组，这些键值对对应于路由URI中的命名段.

If for example you wanted to use parameters, it accepts an array of key value pairs which correspond to the named segments in your route URI.

例如，如果您有路线

Route::get('surveys/{id}', 'SurveyController@details')->name('detail.survey');

您可以使用参数中的以下内容来生成此路线的链接.

You can generate a link to this route using the following in the parameters.

['id' => $id]

完整的示例，回显标记，其中包含指向指定路线的锚点.

A full example, echoing markup containing an anchor to the named route.

{!! link_to_route('new.survey', 'My Link', ['id' => $id]) !!}