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Get all last 4 characters of files

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# 1  
Old 04-26-2016
Get all last 4 characters of files
Hi All.

Seeking for your assistance to get all the last 4 characters of each file before "." and put it in the variable

Code:
ex.
abcdZWU1501.csv
abcdXYZ1501.csv
abcdEFG1502.csv
abcdHIJ1501.csv

output will be:
Code:
1501
1502

What i did was
Code:
find *.csv | awk -F "." '{print $1}' | tail -c5
my output is 1502 only.

Please advise,

Thanks,
# 2  
Old 04-26-2016
If you use awk anyhow, why not do everything with it?
Code:
find *.csv | awk -F "." '{print substr ($1, length($1)-3)}'
1502
1501
1501
1501

Shell version (recent bash):
Code:
 while read FN; do T=${FN%.*}; echo ${T#${T%????}}; done < <(ls *.csv)
1502
1501
1501
1501

This User Gave Thanks to RudiC For This Post:
znesotomayor
# 3  
Old 04-26-2016
Thanks Sir Rudic. what i did was i put it in the for loop statement.

BR,
# 4  
Old 04-26-2016
I believe the request was to take it one step further:
Code:
find . -name '*.csv'|awk '!((X=substr($0,length($0)-7,4)) in Y){print X;Y[X]}'

producing the output:
Code:
1502
1501

# 5  
Old 04-26-2016
I think this will help..

Code:
find *.csv | cut -d "." -f1 | rev | cut -c1-4 | rev

it will produce output like below

Code:
1501
1501
1502
1501

# 6  
Old 04-26-2016
A tad bit compact Smilie
Code:
find *.csv | awk -F"[^0-9]+" '{print $2}'

# 7  
Old 04-26-2016
try also:
Code:
find *.csv | sed 's/.*\(....\)[.]csv/\1/'

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