Comments on: Messing up with command-line arguments in Bash: $*, $@, "$*", "$@",… http://tadek.pietraszek.org/blog/2006/01/05/messing-up-with-command-line-arguments-in-bash/ Some random notes about computers, security, cool links and others. Thu, 15 Jul 2010 15:44:43 +0000 http://wordpress.org/?v=2.9.2 hourly 1 By: lowrykun http://tadek.pietraszek.org/blog/2006/01/05/messing-up-with-command-line-arguments-in-bash/comment-page-1/#comment-37322 lowrykun Fri, 28 Aug 2009 22:13:55 +0000 http://tadek.pietraszek.org/blog/?p=230#comment-37322 <p>Dude - it's 2009 and this page is still dolling out love!</p> <p>Thanks for the tip!</p> Dude – it’s 2009 and this page is still dolling out love!

Thanks for the tip!

]]> By: Mac http://tadek.pietraszek.org/blog/2006/01/05/messing-up-with-command-line-arguments-in-bash/comment-page-1/#comment-36616 Mac Mon, 28 Jul 2008 11:22:04 +0000 http://tadek.pietraszek.org/blog/?p=230#comment-36616 <p>I found the problem and the solution after about an hour as well, then discarded the solution because it didn't look like it was working. I was using 'echo' display the expanded java command and parameters, and it seems that echo strips off quotes, so the output looked the same and I assumed it wasn't going to work.</p> <p>In this example, the command is:</p> <p>test.sh -p "a b c"</p> <p>echo $JAVA_HOME/bin/java com.example.MyClass "$@" ---? /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c</p> <p>echo $JAVA_HOME/bin/java com.example.MyClass $@ ---> /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c</p> <p>No difference, I thought, so that back to the drawing board.</p> <p>BUT, after reading this blog, I tried again and executed with sh -x, I see:</p> <ul> <li>echo /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p 'a b c' /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c</li> <li>echo /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c</li> </ul> <p>doh!</p> <p>Thanks!</p> I found the problem and the solution after about an hour as well, then discarded the solution because it didn’t look like it was working. I was using ‘echo’ display the expanded java command and parameters, and it seems that echo strips off quotes, so the output looked the same and I assumed it wasn’t going to work.

In this example, the command is:

test.sh -p “a b c”

echo $JAVA_HOME/bin/java com.example.MyClass “$@” —? /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c

echo $JAVA_HOME/bin/java com.example.MyClass $@ —> /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c

No difference, I thought, so that back to the drawing board.

BUT, after reading this blog, I tried again and executed with sh -x, I see:

  • echo /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p ‘a b c’ /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c
  • echo /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c /usr/java/jdk1.5.0_12/bin/java com.example.MyClass -p a b c

doh!

Thanks!

]]> By: Roman http://tadek.pietraszek.org/blog/2006/01/05/messing-up-with-command-line-arguments-in-bash/comment-page-1/#comment-30125 Roman Thu, 17 Jan 2008 20:41:13 +0000 http://tadek.pietraszek.org/blog/?p=230#comment-30125 <p>Excellent. This was very useful. Thanks!</p> Excellent. This was very useful. Thanks!

]]> By: David http://tadek.pietraszek.org/blog/2006/01/05/messing-up-with-command-line-arguments-in-bash/comment-page-1/#comment-14425 David Sun, 22 Apr 2007 00:21:17 +0000 http://tadek.pietraszek.org/blog/?p=230#comment-14425 <p>Nice one. If found the problem with $@ quickly but not the solution (actually I first thought there might be a problem with the parsing library). Of course, in hindsight, a peek at the man page would have helped:</p> <p>man bash</p> <p>... @ Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a sepa- rate word. That is, "$@" is equivalent to "$1" "$2" ... When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed). ...</p> Nice one. If found the problem with $@ quickly but not the solution (actually I first thought there might be a problem with the parsing library). Of course, in hindsight, a peek at the man page would have helped:

man bash

… @ Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a sepa- rate word. That is, “$@” is equivalent to “$1″ “$2″ … When there are no positional parameters, “$@” and $@ expand to nothing (i.e., they are removed). …

]]> By: Rogan Creswick http://tadek.pietraszek.org/blog/2006/01/05/messing-up-with-command-line-arguments-in-bash/comment-page-1/#comment-3697 Rogan Creswick Mon, 21 Aug 2006 23:06:34 +0000 http://tadek.pietraszek.org/blog/?p=230#comment-3697 <p>ah, you just saved me an hour or more :) Thanks for sharing!</p> ah, you just saved me an hour or more :) Thanks for sharing!

]]> By: Dan White http://tadek.pietraszek.org/blog/2006/01/05/messing-up-with-command-line-arguments-in-bash/comment-page-1/#comment-2201 Dan White Tue, 20 Jun 2006 19:33:42 +0000 http://tadek.pietraszek.org/blog/?p=230#comment-2201 <p>I just spent an hour trying to figure this out. My problem was that I was using $* instead of $@. Watch out for that!</p> I just spent an hour trying to figure this out. My problem was that I was using $* instead of $@. Watch out for that!

]]>