在 java 中获取登录用户名

如何在 Java 中获得用户名/登录名？

这是我试过的密码。

try{
LoginContext lc = new LoginContext(appName,new TextCallbackHandler());
lc.login();
Subject subject = lc.getSubject();
Principal principals[] = (Principal[])subject.getPrincipals().toArray(new Principal[0]);


for (int i=0; i<principals.length; i++) {
if (principals[i] instanceof NTUserPrincipal || principals[i] instanceof UnixPrincipal) {
String loggedInUserName = principals[i].getName();
}
}


}
catch(SecurityException se){
System.out.println("SecurityException: " + se.getMessage());
}

当我尝试运行这个代码时，我得到一个 SecurityException。谁能告诉我，我的方向是否正确，并帮助我理解这个问题。

192831 次浏览
小开
最佳答案 
System.getProperty("user.name");
小开

in Unix:

new com.sun.security.auth.module.UnixSystem().getUsername()

in Windows:

new com.sun.security.auth.module.NTSystem().getName()

in Solaris:

new com.sun.security.auth.module.SolarisSystem().getUsername()
小开

System.getProperty("user.name") is not a good security option since that environment variable could be faked: C:\ set USERNAME="Joe Doe" java ... // will give you System.getProperty("user.name") You ought to do:

com.sun.security.auth.module.NTSystem NTSystem = new com.sun.security.auth.module.NTSystem();
System.out.println(NTSystem.getName());

JDK 1.5 and greater.

I use it within an applet, and it has to be signed. info source

小开

inspired by @newacct's answer, a code that can be compiled in any platform:

String osName = System.getProperty( "os.name" ).toLowerCase();
String className = null;
String methodName = "getUsername";


if( osName.contains( "windows" ) ){
className = "com.sun.security.auth.module.NTSystem";
methodName = "getName";
}
else if( osName.contains( "linux" ) ){
className = "com.sun.security.auth.module.UnixSystem";
}
else if( osName.contains( "solaris" ) || osName.contains( "sunos" ) ){
className = "com.sun.security.auth.module.SolarisSystem";
}


if( className != null ){
Class<?> c = Class.forName( className );
Method method = c.getDeclaredMethod( methodName );
Object o = c.newInstance();
System.out.println( method.invoke( o ) );
}
小开

The 'set Username="Username" ' is a temporary override that only exists as long as the cmd windows is still up, once it is killed off, the variable loses value. So i think the

System.getProperty("user.name");

is still a short and precise code to use.

小开

System.getenv().get("USERNAME"); - works on windows !

In environment properties you have the information you need about computer and host! I am saying again! Works on WINDOWS !

小开

Using JNA its simple:

String username = Advapi32Util.getUserName();
System.out.println(username);


Advapi32Util.Account account = Advapi32Util.getAccountByName(username);
System.out.println(account.accountType);
System.out.println(account.domain);
System.out.println(account.fqn);
System.out.println(account.name);
System.out.println(account.sidString);

https://github.com/java-native-access/jna

小开

Below is a solution for WINDOWS ONLY

In cases where the application (like Tomcat) is started as a windows service, the System.getProperty("user.name") or System.getenv().get("USERNAME") return the user who started the service and not the current logged in user name.

Also in Java 9 the NTSystem etc classes will not be accessible

So workaround for windows: You can use wmic, so you have to run the below command

wmic ComputerSystem get UserName

If available, this will return output of the form: 

UserName
{domain}\{logged-in-user-name}

Note: For windows you need to use cmd /c as a prefix, so below is a crude program as an example:

    Process exec = Runtime.getRuntime().exec("cmd /c wmic ComputerSystem get UserName".split(" "));
System.out.println(exec.waitFor());
try (BufferedReader bw = new BufferedReader(new InputStreamReader(exec.getInputStream()))) {
System.out.println(bw.readLine() + "\n" + bw.readLine()+ "\n" + bw.readLine());
}
小开

I tested in linux centos

Map<String, String> env = System.getenv();
for (String envName : env.keySet()) {
System.out.format("%s=%s%n", envName, env.get(envName));
}


System.out.println(env.get("USERNAME"));

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