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To take file based on date passed

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# 1  
Old 09-21-2017
To take file based on date passed
Hi Guys,

I have certain files in my directory which gets appended with dates something like this

Code:
T1_aug17.txt
T1_Aug17.txt
T1_Sep17.txt
config.txt
T1
T2

my code:
curr_date=`date -d "$date" +%Y-%m-%d`
path=mydir
for file in `cat config.txt`
do
final_file=$(ls  $path/ | grep -i $file | grep .txt )
done

I need to take only below files based on current month i.e. its sep 17 , i need to take previous named file based on curr_date aug 17

Code:
T1_aug17.txt
T1_Aug17.txt

# 2  
Old 09-21-2017
This is far from clear. Do you need today's day-of-month, one month earlier? Is that always the three-letter-abbreviated month name, or other month representation as well? Case doesn't matter, just for the initial, or the entire string as well? What to do in January?
# 3  
Old 09-21-2017
yes i would need todays day as one month earlier because the curr_date will be always monthly

for eg
Code:
2017-09-11
2017-10-11
2017-10-11

based on the above i need to take files for eg 2017-09-11 then take files of previous month which is Aug17.

The fiiles wil be always in this format that is contains aug17 constant

Code:
T1_aug17.txt/T1_Aug17.txt/T1_aug17_1.txt

# 4  
Old 09-21-2017
So 17 is the year, NOT the day?
# 5  
Old 09-21-2017
Exactly its not day its year
# 6  
Old 09-21-2017
Please look back to your post#1. Where did you state THAT?

Howsoever, try
Code:
DT=$(date +"%b%y" -d"-1month")
ls -1 *.txt | while read FN; do [[ "${FN^^}" =~ "${DT^^}" ]] && echo $FN; done
T1_aug17_1.txt
T1_aug17.txt
T1_Aug17.txt

This User Gave Thanks to RudiC For This Post:
rohit_shinez
# 7  
Old 09-21-2017
Thanks,

Can this is be done in my existing code

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