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how to get the last part of a string followed by a pattern

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# 1  
Old 09-08-2007
how to get the last part of a string followed by a pattern
assuming "cat" is the pattern, string (regardless length)
asdadfcat4

I need to get 4

for eirtrjkkkcat678- I'd get 678

(in b-shell)
Thanks in advance!!!
# 2  
Old 09-08-2007
Code:
echo "1test23string456" | sed "s/^.*[^0-9]\([0-9][0-9]*\)$/\1/"

# 3  
Old 09-08-2007
Code:
# var=eirtrjkkkcat678
# echo ${var##*cat}
678

# 4  
Old 09-08-2007
robotronic:

I can't get the digits followed "string", instead, I get "Variable syntax".
Why? Would you explain each syntax in sed- I'm a newbie & confused. It looks to me it tries to get at least two digits out of from the end of the string...
Thanks for your help!

ghostdog:
I assume that wasn't b-shell script? I got "bad substitution" running it.
Thanks
# 5  
Old 09-08-2007
Hi gentle peopel:

I think I got it all right:

echo "beginingPattern3568" | sed 's/^.*Pattern\([0-9]*\)$/\1/'

Thanks for helping, again.

bluemoon
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