blocks|key|120425|text|在Javadoc中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|120426|120427|返回与当前Java虚拟机相关联的唯一控制台对象(如果有)。|blockquote|120428|120429|120430|如果没有与JVM相关联的控制台，则指向的行是对空对象上的方法的调用，因此出现异常。|120431|您如何启动您的应用程序？|120432|120433|虚拟机是否具有控制台取决于底层平台，也取决于虚拟机的调用方式。如果虚拟机是从交互式命令行启动的，而没有重定向标准输入和输出流，那么它的控制台将存在，并且通常将连接到启动虚拟机的键盘和显示器。如果虚拟机是自动启动的，例如由后台作业调度程序启动，则它通常不会有控制台。|120434|120435|120436|如果想要从标准输入中读取用户名，可以使用以下代码：|120437|try+{
++++System.out.print("Enter+Username:+");
++++InputStreamReader+streamReader+=+new+InputStreamReader(System.in);
++++BufferedReader+bufferedReader+=+new+BufferedReader(streamReader);
++++String+username+=+bufferedReader.readLine();
}+catch+(IOException+e)+{
++++e.printStackTrace();
}|code-block|syntax|javascript|120438|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|11|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|G|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|14|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|15|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|E|7|17|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1B|8|@]|9|@]|A|$V|W]]|$1|X|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|Y|$]]

From the Javadoc:
<blockquote>
Returns the unique Console object associated with the current Java
virtual machine, if any.
</blockquote>
If there is no console associated to the JVM, the pointed line is the call of a method on a null object, hence the exception.
How do you launch your application?
<blockquote>
Whether a virtual machine has a console is dependent upon the
underlying platform and also upon the manner in which the virtual
machine is invoked. If the virtual machine is started from an
interactive command line without redirecting the standard input and
output streams then its console will exist and will typically be
connected to the keyboard and display from which the virtual machine
was launched. If the virtual machine is started automatically, for
example by a background job scheduler, then it will typically not have
a console.
</blockquote>
If you want to read the username from the standard input, you could use this code:
<pre><code>try {
 System.out.print(&quot;Enter Username: &quot;);
 InputStreamReader streamReader = new InputStreamReader(System.in);
 BufferedReader bufferedReader = new BufferedReader(streamReader);
 String username = bufferedReader.readLine();
} catch (IOException e) {
 e.printStackTrace();
}
</code></pre>

blocks|key|2887443|text|这是因为System.console()返回null。官方文档声明(我要强调的是粗体)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2887444|2887445|公共静态控制台()|blockquote|2887446|2887447|返回与当前Java虚拟机关联的唯一控制台对象(如果有)。|2887448|返回:系统控制台，如果有，返回，否则返回空。|BOLD|2887449|2887450|你可以在here上看到它。|2887451|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/lang/System.html#console%2528%2529^0|4|G|0|0|0|0|0|F|6|0|0|4|4|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|13|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|14|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|17|8|@$9|18|A|19|B|O]]|D|@]|E|$]]|$1|P|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1B|8|@]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|S|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

That is because <code>System.console()</code>is returning null. The official documentation states (bold is mine to emphasize):
<blockquote>
public static Console console()
Returns the unique Console object associated with the current Java virtual machine, if any.
Returns:
The system console, if any, otherwise null.
</blockquote>
You can see it <a href="https://docs.oracle.com/javase/7/docs/api/java/lang/System.html#console%28%29" rel="nofollow noreferrer">here</a>.

blocks|key|123264|text|您应该使用BufferedReader或Scanner类从控制台读取输入。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|123265|BufferedReader:|BOLD|123266|import+java.io.BufferedReader;
import+java.io.IOException;
import+java.io.InputStreamReader;

public+class+ReadConsoleSystem+{
++public+static+void+main(String[]+args)+{

++++System.out.println("Enter+something+here:+");

++++try+{
++++++++BufferedReader+bufferRead+=+new+BufferedReader(new+InputStreamReader(System.in));
++++++++String+s+=+bufferRead.readLine();

++++++++System.out.println(s);
++++}
++++catch(IOException+e)
++++{
++++++++e.printStackTrace();
++++}
++}
}|code-block|syntax|javascript|123267|扫描程序：|123268|import+java.util.Scanner;

public+class+ReadConsoleScanner+{

++public+static+void+main(String[]+args)+{

++++System.out.println("Enter+something+here:+");

++++String+sWhatever;

++++Scanner+scanIn+=+new+Scanner(System.in);
++++sWhatever+=+scanIn.nextLine();

++++scanIn.close();
++++System.out.println(sWhatever);
++}
}|123269|entityMap^0|5|E|K|7|0|0|F|0|0|0|5|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|11|8|@]|D|@]|E|$L|M]]|$1|N|3|O|5|6|7|12|8|@$9|13|A|14|B|H]]|D|@]|E|$]]|$1|P|3|Q|5|K|7|15|8|@]|D|@]|E|$L|M]]|$1|R|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|S|$]]

You should use a <code>BufferedReader</code> or a <code>Scanner</code> class to read inputs from the console.
BufferedReader:
<pre><code>import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class ReadConsoleSystem {
 public static void main(String[] args) {

 System.out.println(&quot;Enter something here: &quot;);

 try {
 BufferedReader bufferRead = new BufferedReader(new InputStreamReader(System.in));
 String s = bufferRead.readLine();

 System.out.println(s);
 }
 catch(IOException e)
 {
 e.printStackTrace();
 }
 }
}
</code></pre>
Scanner:
<pre><code>import java.util.Scanner;

public class ReadConsoleScanner {

 public static void main(String[] args) {

 System.out.println(&quot;Enter something here: &quot;);

 String sWhatever;

 Scanner scanIn = new Scanner(System.in);
 sWhatever = scanIn.nextLine();

 scanIn.close();
 System.out.println(sWhatever);
 }
}
</code></pre>

blocks|key|123290|text|String+unm+=+System.console().readLine();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|123291|您应该首先在system.console上执行null检查。在没有任何控制台关联的平台上，此应用程序将死于空指针异常。|unstyled|123292|对是否有可用的控制台执行null检查。如果没有，请用扫描仪阅读：|123293|++++String+userName;
++++char[]+password;
++++Scanner+scanner+=+null;
++++Console+console+=+System.console();
++++if(console+==+null){
++++++++scanner+=+new+Scanner(System.in);
++++++++System.out.println("Enter+username:+");
++++++++userName+=+scanner.next();
++++++++//+Read+the+password+as+well
++++}
++++else{
++++++++System.out.println("Enter+username:+");
++++++++userName+=+console.readLine();
++++++++//+Read+the+password+as+well
++++}|123294|备注：|offset|length|style|BOLD|123295|123296|控制台的一个优点是它不会显示在屏幕上，特别是在读取密码时。使用扫描仪，您的密码将显示在屏幕上。|unordered-list-item|123297|以上代码片段仅用于演示目的。|123298|123299|entityMap^0|0|0|0|0|0|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|11|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|12|8|@]|9|@]|A|$B|C]]|$1|K|3|L|5|F|7|13|8|@$M|14|N|15|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|F|7|16|8|@]|9|@]|A|$]]|$1|R|3|S|5|T|7|17|8|@]|9|@]|A|$]]|$1|U|3|V|5|T|7|18|8|@]|9|@]|A|$]]|$1|W|3|-4|5|F|7|19|8|@]|9|@]|A|$]]|$1|X|3|-4|5|F|7|1A|8|@]|9|@]|A|$]]]|Y|$]]

<pre><code>String unm = System.console().readLine();
</code></pre>
You should do a null check at system.console first. On platforms where there isn't any console associate, this application will die with a null pointer exception.
Do a null check on whether any console is available. If not, read with the scanner:
<pre><code> String userName;
 char[] password;
 Scanner scanner = null;
 Console console = System.console();
 if(console == null){
 scanner = new Scanner(System.in);
 System.out.println(&quot;Enter username: &quot;);
 userName = scanner.next();
 // Read the password as well
 }
 else{
 System.out.println(&quot;Enter username: &quot;);
 userName = console.readLine();
 // Read the password as well
 }
</code></pre>
Note:
<ul>
<li>One advantage with the console, specially while reading the password, is that it won't get displayed on the screen. With the scanner, your password will be displayed on the screen.</li>
<li>The above code snippet is for demonstration purposes.</li>
</ul>

blocks|key|2676185|text|如果您只需要输入用户名，提供的答案就足够了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2676186|但是，您还需要安全地输入密码。为此，您提供的代码是完全正确的。您只需编译项目并在控制台或终端中运行它，而不是在IDE中运行。|2676187|例如，如果您使用的是Windows和NetBeans，请单击菜单项Run→build+Project来构建项目。这将创建一个与您的项目同名的.jar文件。|offset|length|style|CODE|2676188|现在打开一个cmd窗口，导航到您的.jar文件所在的路径，然后通过输入java+-jar+YourProjectName.jar命令(将YourProjectName替换为您的项目名称)来执行您的项目。|2676189|entityMap^0|0|0|1Y|4|0|6|3|H|4|Z|T|1W|F|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@$F|Q|G|R|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@$F|T|G|U|H|I]|$F|V|G|W|H|I]|$F|X|G|Y|H|I]|$F|Z|G|10|H|I]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|M|$]]

If you need to input only the username, the answer provided by <a href="https://stackoverflow.com/questions/26078106/system-console-gives-a-nullpointerexception-in-netbeans">System.console() gives a NullPointerException in NetBeans</a> would have been sufficient.
However, you also need to input the password securely. For this, the code you provided is perfectly fine. You just have to compile the project and run it in a Console or Terminal, instead of your IDE.
E.g., if you are using Windows and NetBeans, build your project by clicking on menu item Run → Build Project. This will create a <code>.jar</code> file with the same name as that of your project.
Now open a <code>cmd</code> window, navigate to the path where your <code>.jar</code> file is located and then execute your project by typing <code>java -jar YourProjectName.jar</code> command (replace <code>YourProjectName</code> with the name of your project).

Here is my code:
<pre><code>System.out.println(&quot;Enter Username: &quot;);
String unm=System.console().readLine();

System.out.println(&quot;Enter Password: &quot;);
char[] pwd=System.console().readPassword();

System.out.println(&quot;Welcome: &quot; + &quot;&quot; + &quot; Your password is &quot; + new String(pwd));
</code></pre>
Why am I getting this error?
<blockquote>
Enter Username:
</blockquote>
<blockquote>
Exception in thread &quot;main&quot; java.lang.NullPointerException at Cons.main(Cons.java:13)
</blockquote>

System.console() NullPointerException

虚拟机

Windows

Java

下面是我的代码：System.out.println("Enter Username: ");String unm=System.console().readLine();System.out.println("Enter Password: ");char[] pwd=System.console().readPas...

问System.console() NullPointerException
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