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Sub string after the last occurence of "."

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# 1  
Old 07-30-2008
Bug Sub string after the last occurence of "."
Hi,
Suppose there is a string
x.y.z , where x, y & z could be any integer from 0-999.

How do I extract the value of z?
Also how to extract "x.y" ?

I'm OK with using sh, sed, awk grep or cut.
Please help me figure this out.

Thanks
# 2  
Old 07-30-2008
What is your sh? This is normally a link to e.g. bash, ksh or tcsh. Given bash:
Code:
$ a=x.y.z
$ echo ${a//*.}
z

# 3  
Old 07-30-2008
Quote:
Originally Posted by fabtagon
What is your sh? This is normally a link to e.g. bash, ksh or tcsh. Given bash:
Code:
$ a=x.y.z
$ echo ${a//*.}
z

.. and the answer for OP's second question, if the shell support Parameter Expansion
Code:
echo ${a##*.}
echo ${a%.*}

.. if not you can use cut(beside other)
Code:
echo x.y.z | cut -d '.' -f3
echo x.y.z | cut -d '.' -f1-2

# 4  
Old 07-30-2008
if you have distinct field delimiters like that, ie ".", its easier to use fields than parameter expansion or substitution.
Code:
# a=x.y.z
# IFS="."
# set -- $a
# echo $1
x
# echo $2
y
# echo $3
z

# 5  
Old 08-04-2008
Code:
cat filename | cut -d"." -f3

# 6  
Old 08-04-2008
Thanks a lot guys....
I appreciate your help.Smilie
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