死锁:仅显示当前登录用户的模板部分 [英] Deadbolt: show parts of template only for current logged user

问题描述

在我的系统中，每个用户都有自己的公开个人资料.我只想在当前登录用户的个人资料页面上显示编辑"按钮. 现在，我通过使用以下代码来做到这一点

In my system every user has an own public profile. I want to show an "Edit" button only on the profile page of the current logged user. Now I'm doing this by using this code

  @subjectPresent() {
    @if(userProfile == userLogged){
      <button>Edit</button>
    }
  }

其中userProfile是当前页面的所有者用户，而userLogged是实际登录的用户.

where userProfile is the owner user of the current page, and userLogged is the actual logged user.

考虑到我将不得不多次进行此检查，Deadbolt或Scala中是否有更好的方法(更清洁)?

Considering that I will have to do this check a lot of times, is there in Deadbolt or Scala a better (cleaner) way to do it?

推荐答案

按照David的建议，您可以将其包装在自己的标签中.标签只是功能，看起来就像其他视图(实际上，它们是其他视图).

As David suggested, you can wrap this up in your own tag. Tags are just functions, and look like other views (in fact, they are other views).

您可以尝试类似的

@(userProfile: User, userLogged: User)(body: => Html)

@subjectPresent() {
  @if(userProfile == userLogged){
    @body
  }
}

并将其保存在名为foo.scala.html

and save this in a file called foo.scala.html

然后可以将其与

@foo(userProfile, userLogged) {
  <button>Edit</button>
}

在必要时，您需要使用正确的类型声明或导入，例如用户，导入标签等.这取决于您的项目的结构.

You'll need to use the correct type declarations or imports where necessary, e.g. User, importing the tag, etc. This depends on the structure of your project.

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