如何通过在GSP中使用Spring Security Grails插件来获取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。

我想这样...

   编辑此帖子
< / g：link> 
< / g：if> 
  

这里我想要显示编辑此帖子链接，由signed_in用户创建。但它显示错误 -

 错误500：内部服务器错误
 
 URI 
 / groovypublish / post / list 
类
 java.lang.NullPointerException 
消息
无法在空对象上获取属性'principal'
  

这是我的Post.groovy -

  class Post { 
 
 static hasMany = [评论：评论] 
 
字符串标题
字符串预告片
字符串内容
日期lastUpdated 
布尔已发布= false 
 SortedSet comment 
 Person author 
 
 ....... more code .... 
  

这是我的Person.groovy域类文件 -

  class Person {
 
 transient springSecurityService 
 
字符串实数名称
字符串用户名
字符串密码
布尔启用
布尔accountExpired 
布尔accountLocked 
布尔passwordExpired 
字节[]头像
字符串头像类型
 
静态hasM any = [follow：Person，posts：Post] 
 static searchable = [only：'realName'] 
 ........ more code ...... 
  

请帮助。

解决方案

<尝试使用springSecurity插件提供的标签，如：

 < sec：isLoggedIn> 
 
 编辑此帖子
< / g：link> 
 
< / sec：isLoggedIn> 
  

其实你试图在你的GSP页面上注入一个服务，你可以用一些import语句在页面上，但我会说这不会是很好的编程习惯，我认为你应该将当前登录的用户实例从控制器发送到GSP页面，然后执行检查：

假设你有控制器方法：

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

以及您的GSP页面：

 < g：if test =$ {post.author == currentLoggedInUser}> 
 编辑此帖子
< / g：link> 
< / g：if> 
  

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.

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>

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

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 ....

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 ......

Please help.

解决方案

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>

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:

let say you have the controller method:

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

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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