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Simple stuff, I am learning URLs/Networking in my class and I am trying to display something on a webpage. Later I am going to connect it to a MySQL DB... anyway here is my program:

import java.net.*; import java.io.*;


public class asp {

    public static URLConnection
connection;

    public static void main(String[] args) {

        try {

        System.out.println("Hello World!"); // Display the string.
        try {
        URLConnection connection = new URL("post.php?players").openConnection();
    }catch(MalformedURLException rex) {}
        InputStream response =
connection.getInputStream();
        System.out.println(response);
    }catch(IOException ex) {}

    } }

It compiles fine... but when I run it I get:

Hello World!
Exception in thread "main" java.lang.NullPointerException at asp.main(asp.java:17)

Line 17: InputStream response = connection.getInputStream();

Thanks, Dan

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    Please replace all code inside catch blocks by e.printStackTrace() or just throw e. Suppressing exceptions isn't going to help you much since they contains information about the cause of the problem. You know, once the cause is understood, the solution is obvious ;) Commented Jul 7, 2010 at 23:45
  • Please use Java naming convention, and use class names starting with a capital! And be aware that a catch block is for situations when something went wrong. If you have to put code in a catch block, then you have done something really awful, and need to rethink the whole thing! This is not a simple stuff, given that you have problems with fundamental concepts in Java, like visibility scope... Commented Oct 12, 2012 at 12:51

2 Answers 2

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3

You have a malformed URL, but you wouldn't know because you swallowed its exception!

URL("post.php?players")

This URL is not complete, it misses the host (maybe localhost for you?), and the protocol part, say http so to avoid the malformed URL exception you have to provide the full URL including the protocol

new URL("http://www.somewhere-dan.com/post.php?players")

Use the Sun tutorials on URLConnection first. That snippet is at least known to work, if you substitute the URL in that example with a valid URL you should have a working piece of code.

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1 Comment

That'll be because the connection variable you set inside the try is not in scope of connection.getInputStream(). It is using the wider scoped static connection variable which is and has always remained null.
2

It's because your URL is not valid. You need to put the full address to the page you are trying to open a connection to. You are catching the malformedurlexception but that means that there is no "connection" object at that point. You have an extra closed bracket after the first catch block it appears as well. You should put the line that you are getting the null pointer for and the system.out.println above the catch blocks

import java.net.*; import java.io.*;

public class asp {

    public static URLConnection connection;

    public static void main(String[] args) {

        try {
        System.out.println("Hello World!"); // Display the string.
            try {
            URLConnection connection = new URL("http://localhost/post.php?players").openConnection();
            InputStream response = connection.getInputStream();
            System.out.println(response);

            }catch(MalformedURLException rex) {
                System.out.println("Oops my url isn't right");
        }catch(IOException ex) {}
    }    
}
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7 Comments

I didn't get this part: You should put the line that you are getting the null pointer for and the system.out.println above the catch blocks
OK I replaced it with: localhost/mystikrpg/post.php?players and I am getting NullPointerException on line 17 which is the same line of InputStream response yada yada ... :P
@Dan: Your URL is still missing its protocol part. See my answer.
@Bakkal (nice name btw): What do you mean? I'm lost. :(
@Bakkal: SO ate the protocol part. @Dan: why are you trying to connect it anyway inside a catch block? If an MalformedURLException has been thrown, you can't reasonably expect that the URL is correctly well formed (and thus valid).
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