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Shell Programming and Scripting

To take file based on date passed

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# 8  
Old 09-21-2017
Why don't you try it and present the resulting code here for analysis?
# 9  
Old 09-21-2017
I have added something like this

Code:
DT=$(date +"%b%y" -d"-1month")

final_file=$(ls  $path/ | grep -i $file$DT | grep .txt )

# 10  
Old 09-21-2017
And - does it come close to something like you want?
# 11  
Old 09-21-2017
Quote:
Originally Posted by RudiC
And - does it come close to something like you want?
Almost but it's searching pattern like filenameaug17. Can you help where is the mistake
# 12  
Old 09-21-2017
No, it doesn't:
Code:
touch filenameaug17.
ls -1 *.txt | while read FN; do [[ "${FN^^}" =~ "${DT^^}" ]] && echo $FN; done
T1_aug17_1.txt
T1_aug17.txt
T1_Aug17.txt

With your approach, try to make the ls command or the regex more specific
# 13  
Old 09-22-2017
Hi,

I didn't get how it will change , because grep $file$DT will search for filenameAug17 pattern.



Code:
DT=$(date +"%b%y" -d"-1month")

final_file=$(ls  $path/ | grep -i $file$DT | grep .txt )

Are you suggesting something like this

Code:
DT=$(date +"%b%y" -d"-1month")

final_file=$(ls  $path/ | grep -i $file| grep | $DT| grep .txt )
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