run.bat file

I am having weird error. I am passing two args when I run this program.
args[1] is:
However, when I put this url in run.bat file like this:

java <program>  <path>

args[1] comes into main without "%7", and that is why my program doesn't work.
Another word, System.out.println("args[1] " + args[1] ); just print
Of course, run.bat file is excutable file.

Why is that?

  public static void main(String[] args) throws CrawlerException {

    //final String LINE_SEPARATOR = System.getProperty("line.separator");
    //StringBuffer buffy = new StringBuffer("");

      System.out.println("args[1] " + args[1] );
      URL url = new URL(args[1]);
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could u try to use escape before the special charactor.
What platform are you running? For Win it works fine without escaping...
heh, I'm half asleep... What other platform could require *.bat files... :)
Really I don't know, I tried your code for one argument (arg[0]), it works allright without escaping '%' character. Maybe shivsa is right and you should escape it because your Windows parses it as a start of environment variable reference. But again on the WinXP of mine it worked alright.

Good luck!
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In batch files, when you use a percent sign followed by a number, the percent and number are replaced with that argument that was passed to the batch file.  So if in a batch called hello.bat  I were to write:
echo %1
and then to execute it i typed:
hello.bat HelloWorld
then the program would echo HelloWorld

Your batch file is replacing %7 with the 7th argument that you passed the batch file.  Because you didn't pass it seven arguments, it's replacing it with nothing.

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To escape the percent in a batch file, you type %%.  Just the percent character twice.
and thats why i said to escape since windows should not take it as a its own operator and treat it as a special character.
Ah, yeah, well done Breadstick! I launched it directly from command line without batch-file...

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