Shell Programming and Scripting

Hi,

My problem is that I have to search a changing pattern and replace it with the wild card char "*"

Code:

 
i/p: 99_*_YYYYMMDD_SRC.txt.tar.gz
o/p: 99_*_*_SRC.txt.tar.gz

The problem is that YYYYMMDD pattern is not static. It could be YYYYMMDDHHMI or could be YYYYMMDDHHMISS.

Can someone please help me here?

-dips

Code:

echo '99_*_YYYYMMDD_SRC.txt.tar.gz' | nawk -F_ '$3="*"' OFS=_

Hi vgersh,

Thanks!! for your quick solution it worked with this kind of pattern.........(had to replace nawk by awk....it seems nawk is not supported on my linux)

But it slipped my mind to mention that I have to first search for YYYYMMDD like pattern meaning it is not certain that it'll be at the 3rd position only. It could be anywhere in the string like

Code:

 
instead of 99_*_YYYYMMDD_SRC.txt.tar.gz
it could be 99_*_SRC_YYYYMMDD.txt.tar.gz
or YYYYMMDD_99_*_SRC.txt.tar.gz

I mean to say it could be anywhere in the string.....sorry for the confusion

-dips

assuming the minimum number of digits in the pattern to be 'blanked out' is 4:

Code:

nawk '{sub("[0-9][0-9][0-9][0-9][0-9]*","*")}1' myFile
OR
sed 's#[0-9]\{4,\}#*#' myFile

Hi vgersh,

I tried your code

Code:

 
echo 45_*_YYYYMMDD_SRC.txt.tar.gz | sed 's#[0-9]\{4,\}#*#'

but it returned the same string

Code:

 
45_*_YYYYMMDD_SRC.txt.tar.gz

-dips

I assumed the 'YYYYMMDD' was actually a mnemonic for the date/time numeric spec.
If you want to take 'YYYYMMDD' literally - that's even easier...:

Code:

sed 's#YYYY[^_.]*#*#' myFile

Hi Vgersh,

Thanks for the solution !!

Can you please explain this sed command ?

-dips