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copied!<p>Instead of <code>scan.nextInt()</code>, I would use <code>scan.nextLine()</code> again, and then parse the input. If you get a <code>NumberFormatException</code>, you can bark at the user that the input was invlaid.</p> <p>The reason why you need to do something like this is that you still have a newline sitting at the end of your serverPort input. When you use <code>nextInt()</code>, the scanner grabs all of the next token and tries to convert it into an <code>Integer</code>. Anything after the Integer token remains to be scanned.</p> <p>To prove this, with your original code, if you put a token separator, like a space, after your server port input and then add some text, you will find that your subsequent <code>userName</code> assignment will grab from the end of the number to the next newline. </p> <pre><code> System.out.println("\nEnter port number:"); Integer serverPort = scan.nextInt(); System.out.println("\nEnter your username:"); String userName = scan.nextLine(); System.out.println("\nserverPort: " + serverPort + "\nuserName: " + userName); </code></pre> <p>with the input:</p> <pre><code> Enter port number: 123 abc Enter your username: </code></pre> <p>will output </p> <pre><code> serverPort: 123 userName: abc </code></pre> <p>This is because the space is acting as a token separator, and after the <code>scan.nextInt()</code> is called, the space and the <code>abc</code> are still left in the buffer. When the subsequent <code>scan.nextLine()</code> is called, the space and the <code>abc</code> are scanned from the buffer.</p>

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