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Join Date: Aug 2007
Posts: 4
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Newbie Poster
I hve code like
echo $1
echo $2
and am executing the code as ./script.sh /home/*.txt /home/*.xsd
The corresponding .txt files and .xsd files of the directory are getting substitued as the command line argument .
and it prints smething like
data.txt
value.xsd
Apart frm using * if i use anything else it works fine . But my requirement is to pass it in this format in the command line ...
Am completely new to shell scripting ....Pls Help me ...
echo $1
echo $2
and am executing the code as ./script.sh /home/*.txt /home/*.xsd
The corresponding .txt files and .xsd files of the directory are getting substitued as the command line argument .
and it prints smething like
data.txt
value.xsd
Apart frm using * if i use anything else it works fine . But my requirement is to pass it in this format in the command line ...
Am completely new to shell scripting ....Pls Help me ...
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Join Date: Feb 2006
Posts: 1,509
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Rep Power: 10
Solved Threads: 136
Because the shell will automatically expand those filename expansion characters before passing the arguments to the script. You might try using \* instead of *, but then, don't expect that the * will work as designed later (it may, then again it may not, I have not tried to "mask" the file expansion during initialisation then using it later).
Java Programmer and Sun Systems Administrator
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Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan
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Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan
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