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06-25-2008
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string replacement in a sequence of characters
Hi All,
I have a string "TBM630300000000020080506094041000003818".I want to replace the last nine digits with another string stored in a variabe called "count".The variabe is also having nine digits.Could any one please help me on this how to accomplish.I need a detail syntax(not in the short form) since i am new to the unix scripting. I tried with some flavour of "string" commands but those are not supported by my KSH environment. Appreciate your help!!!!!!!!!!!!!!! Thanks Narasimharao. |
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06-25-2008
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Quote:
Code:
echo "your_string" | sed -e "s/\(.*\)........./\1$count/" A . represents any character. A .* means any number of characters. \(.*\) would collect the characters into a buffer. Since we need to replace the last nine numbers, you can represent them as nine consecutive dots or as [0-9]\{9\}. This will get replaced with the contents of $count. |
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06-25-2008
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Using ksh93 Code:
$ str="TBM630300000000020080506094041000003818"
$ print ${str%{9}([[:digit:]])}
TBM630300000000020080506094041
$ count="987654321"
$ print ${str%{9}([[:digit:]])}$count
TBM630300000000020080506094041987654321
$
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07-08-2008
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string replace ment
Hi vino,
First of all Thanks for you.Its working fine for my requirement.Still i need a little bit clarification on the below code.What is the 1 before the "$count" variable signifies? Please explain me.with out 1 its not working fine. Thanks in Advance. Narasimharao. Quote:
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