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我想輸出Username如果他登錄 從@Controller我有機會:接取從JSP用戶名中使用Spring Security的
@RequestMapping("/success")
public String success(Model model) {
Authentication auth = SecurityContextHolder.getContext().getAuthentication();
String name = auth.getName(); //get logged in username
model.addAttribute("name", name);
return "success";
}
它工作的,如果我在JSP中使用的名字,我看到鍵入名稱。
但是,如果在這個JSP我寫
<%@ taglib prefix="sec" uri="http://www.springframework.org/security/tags" %>
<sec:authentication property="principal.username"/>
我得到堆棧跟蹤:
INFO : com.epam.hhsystem.util.CustomAuthentificationProvider - User with name 'Nikolay_Tkachev' log in
07.08.2013 17:00:57 org.apache.jasper.compiler.TldLocationsCache tldScanJar
INFO: At least one JAR was scanned for TLDs yet contained no TLDs. Enable debug logging for this logger for a complete list of JARs that were scanned but no TLDs were found in them. Skipping unneeded JARs during scanning can improve startup time and JSP compilation time.
07.08.2013 17:00:57 org.apache.catalina.core.ApplicationDispatcher invoke
SEVERE: Servlet.service() for servlet jsp threw exception
org.springframework.beans.NotReadablePropertyException: Invalid property 'principal.username' of bean class [org.springframework.security.authentication.UsernamePasswordAuthenticationToken]: Bean property 'principal.username' is not readable or has an invalid getter method: Does the return type of the getter match the parameter type of the setter?
at org.springframework.beans.BeanWrapperImpl.getPropertyValue(BeanWrapperImpl.java:707)
at org.springframework.beans.BeanWrapperImpl.getPropertyValue(BeanWrapperImpl.java:699)
...
爲什麼無處不在寫? –
user2645679
AFAIK,如果你有一個簡單的用戶名和密碼,用戶名可以作爲一個字符串直接存儲在principal中。但是,如果您有用戶名等公司名稱,部門等額外詳細信息,則可以將自定義對象分配給委託人,因此在這種情況下,您將使用委託人,用戶名 – coder