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Replace multiple file by passing parameter value

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# 1  
Old 12-13-2017
Replace multiple file by passing parameter value
Hello All,
I want to change date part in file name to yesterday date in the file name.

example file name
Code:
file-12122017-06-30-41.dat

want
Code:
file-12112017-06-30-41.dat

I am doing like below. Below it is not changing the filename. Actually it is not parsing the $today and $yesterday value in replacing part.
Code:
yesterday=`date +%m%d%Y -d "1 day ago"`
today=`date +%m%d%Y`
find . -name "*$today-*" -exec bash -c 'mv $0 ${0/"$today"/"$yesterday"}' {} \;

Kindly help on this.
Last edited by looney; 12-13-2017 at 03:03 PM.. Reason: output example
# 2  
Old 12-13-2017
Single quotes prevent variables from being expanded - try double quotes, and leave the inner ones out. Be careful - in that context, $0 will not yield the desired file name but the name of the running shell. Escape the $ , then.

Why, btw, that find overkill? How about a simple
Code:
for FN in *$today-*; do echo mv $FN ${FN/$today/$yesterday}; done

remove echo when happy with the result presented.
Last edited by RudiC; 12-13-2017 at 03:37 PM..
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