如何在 GSP 中使用 Spring Security Grails 插件获取 current_user [英] How to get current_user by using Spring Security Grails plugin in GSP

问题描述

我是 Grails 的新手.我正在使用 Spring Security Grails 插件 进行身份验证.我想在我的视图 gsp 文件中获取当前用户.

I am newbie in Grails. I am using Spring Security Grails plugin for Authentication purpose. I want to get current user in my view gsp file.

我正在尝试这样......

I am trying like this ...

<g:if test="${post.author == Person.get(springSecurityService.principal.id).id }">
      <g:link controller="post" action="edit" id="${post.id}">
            Edit this post
      </g:link>
</g:if>

在这里，我想显示编辑此帖子链接，仅指向由登录用户创建的帖子.但它显示错误 -

Here I want to show Edit this post link to only those posts who created by signed_in user. But It showing ERROR -

Error 500: Internal Server Error

 URI
    /groovypublish/post/list
 Class
   java.lang.NullPointerException
 Message
   Cannot get property 'principal' on null object

这是我的 Post.groovy --

Here is my Post.groovy --

class Post {

static hasMany = [comments:Comment]

String title
String teaser
String content
Date lastUpdated
Boolean published = false
SortedSet comments
Person author

....... more code ....

这是我的 Person.groovy 域类文件 --

Here is my Person.groovy Domain Class File --

class Person {

transient springSecurityService

String realName
String username
String password
boolean enabled
boolean accountExpired
boolean accountLocked
boolean passwordExpired
byte[] avatar
String avatarType

static hasMany = [followed:Person, posts:Post]
static searchable = [only: 'realName']
    ........ more code ......

请帮忙.

推荐答案

试试springSecurity插件提供的标签，比如:

Try tags provided by springSecurity plugin, something like:

<sec:isLoggedIn>

  <g:link controller="post" action="edit" id="${post.id}">
            Edit this post
      </g:link>

</sec:isLoggedIn>

实际上你是想在你的 GSP 页面上注入一个服务，你可以在页面上使用一些 import 语句来完成，但我会说这不是一个好的编程习惯，我认为你应该发送当前登录用户的实例从控制器到 GSP 页面，然后对其进行检查:

Actually you are trying to inject a service on your GSP page, you can do it with some import statement on the page, but I would say it will not be good programming practice, I think you should send current logged In user's instance from the controller to the GSP page, and then perform a check on it:

假设你有控制器方法:

def showPostPage(){
Person currentLoggedInUser = springSecurityService.getCurrentUser();
[currentLoggedInUser:currentLoggedInUser]
}

并在您的 GSP 页面上:

and on your GSP page:

<g:if test="${post.author == currentLoggedInUser }">
      <g:link controller="post" action="edit" id="${post.id}">
            Edit this post
      </g:link>
</g:if>

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