Parameter problem in my script

Laibro · #1 (**permalink**) 06-01-2008

Hi all

I hope that any one help me with my small problem

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

xxmenxx · #2 (**permalink**) 06-01-2008

Try this

Code:

x=example.tar.z
y=`echo $x |cut -f 1 -d . `
echo $y

danmero · #3 (**permalink**) 06-01-2008

Useless use of echo and cut

Code:

x=example.tar.z
y=${x%.tar.z}
echo $y

xxmenxx · #4 (**permalink**) 06-01-2008

danmero, can you explain your code please
what "%" means here??
thanks in advance

danmero · #5 (**permalink**) 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,

mschwage · #6 (**permalink**) 06-01-2008

Two things to note:

The ${var%pattern} construct is a ksh (and perhaps bash) shell thing, not bourne shell or csh.
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.

danmero · #7 (**permalink**) 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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