AngularJS ui-router 可选参数 [英] AngularJS ui-router optional parameters

问题描述

我试着像这样设置我的路由

I tried to set up my routing like this

...
url: '/view/:inboxId?'
...

但是 Angular 会抛出这个错误:

but Angular would throw this error:

Error: Invalid parameter name '' in pattern '/view/:inboxId?'

所以基本上我必须设置两个不同的状态:

so basically I had to set up two different states:

state('view', {
            url: '/view/:inboxId',
            templateUrl: 'templates/view.html',
            controller: 'viewCtrl'
        }).

        state('view_root', {
            url: '/view',
            templateUrl: 'templates/view.html',
            controller: 'viewCtrl'
        })

有没有办法将这些状态合二为一?

Is there any way to combine these states into one?

推荐答案

要有一个可选参数 - 像你一样声明它 - 但不要传递它.这是一个示例.所有这些都可以根据您的喜好使用一种状态(无根) 或两种(根和详细信息).

To have an optional param - declare it as you did - but do not pass it. Here is an example. That all could work with one state (no root) or two (root and detail) as you like.

问题中提到的定义，已准备好处理这些状态调用:

The definition mentioned in the question, is ready to handle these state calls:

// href
<a href="#/view/1">/view/1</a> - id is passed<br />
<a href="#/view/"> /view/ </a> - is is missing - OPTIONAL like <br />
<a href="#/view">  /view  </a> - url matching the view_root

// ui-sref
<a ui-sref="view({inboxId:2})">    - id is passed<br /> 
<a ui-sref="view">                 - is is missing - OPTIONAL like
<a ui-sref="view({inboxId:null})"> - id is explicit NULL <br />
<a ui-sref="view_root()">          - url matching the view_root

我们不必使用 ? 将参数标记为可选.只是两个 url 必须是唯一的(例如 /view/:id 与 /view - 其中第二个没有尾随 /)

We do not have to use ? to mark parameter as optional. Just both url must be unique (e.g. /view/:id vs /view - where the second does not have trailing /)

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