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Parameter problem in my script

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# 1  
Old 06-01-2008
Parameter problem in my script
Hi all


I hope that any one help me with my small problem Smilie


My problem is:

inside my script their is a parameter x=example.tar.Z

I want to write a script to let y=example (without .tar.Z)


Thank you in advanced Smilie
# 2  
Old 06-01-2008
Try this
Code:
x=example.tar.z
y=`echo $x |cut -f 1 -d . `
echo $y

# 3  
Old 06-01-2008
Useless use of echo and cut Smilie
Code:
x=example.tar.z
y=${x%.tar.z}
echo $y

# 4  
Old 06-01-2008
danmero, can you explain your code please
what "%" means here??
thanks in advance
# 5  
Old 06-01-2008
You can retrieve (expand) variables value by enclosing the variable name in curly braces and preceding the left curly brace with a dollar sign, like that you eliminate the use of echo and backticks.
% will expand the variable removing the regexp(.tar.z) from the end of the variable value.

Regards,
# 6  
Old 06-01-2008
Two things to note:
  1. The ${var%pattern} construct is a ksh (and perhaps bash) shell thing, not bourne shell or csh.
  2. The thing on the righthandside of the % is a pattern, not a regexp.
See SH(1) USER COMMANDS SH(1), do a search in the page for a percent sign. It's only about the 2nd or third entry down; you should find information about using this construct fairly readily.
# 7  
Old 06-01-2008
Quote:
Originally Posted by mschwage
The ${var%pattern} construct is a ksh (and perhaps bash) shell thing, not bourne shell or csh.
I use Bourne shell(/vin/sh)sh
Quote:
Originally Posted by mschwage
[*]The thing on the righthandside of the % is a pattern, not a regexp.
Yes, however you can use regexp.
Code:
y=${x%.*.?}
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