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awk script to extract a column, replace one of the header and replace year(from ddmmyy to yyyy)

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Top Forums UNIX for Beginners Questions & Answers awk script to extract a column, replace one of the header and replace year(from ddmmyy to yyyy)
# 8  
Old 05-10-2019
Code:
$ cat testk3.csv
"PERIOD_END_DATE" "PERIOD_END_DATE"
"17-FEB-19" "17-FEB-19"
"24-FEB-19" "24-FEB-19"
"24-FEB-19" "24-FEB-19"
$ awk -F, -v yr='"YEAR"' ' { $0=$0 OFS (NR == 1 ? yr : "20" substr($0,length($0)-2,3)) } 1 ' OFS="," testk3.csv
"PERIOD_END_DATE" "PERIOD_END_DATE","YEAR"
"17-FEB-19" "17-FEB-19",2019"
"24-FEB-19" "24-FEB-19",2019"
"24-FEB-19" "24-FEB-19",2019"

# 9  
Old 05-10-2019
Strange, when i run the following command:
Code:
awk -F, -v yr='"YEAR"' ' { $0=$0 OFS (NR == 1 ? yr : "20" substr($0,length($0)-2,3)) } 1 ' OFS="," testk3.csv > testk5.csv


it shows me output as before:


Code:
$ head testk3.csv
"PERIOD_END_DATE" "PERIOD_END_DATE"
"17-FEB-19" "17-FEB-19"
"24-FEB-19" "24-FEB-19"
"24-FEB-19" "24-FEB-19"
"17-FEB-19" "17-FEB-19"
"03-MAR-19" "03-MAR-19"
"10-MAR-19" "10-MAR-19"
"17-FEB-19" "17-FEB-19"
"03-MAR-19" "03-MAR-19"
"24-FEB-19" "24-FEB-19"



$ head testk5.csv
,"YEAR"_END_DATE" "PERIOD_END_DATE"
,209"EB-19" "17-FEB-19"
,209"EB-19" "24-FEB-19"
,209"EB-19" "24-FEB-19"
,209"EB-19" "17-FEB-19"
,209"AR-19" "03-MAR-19"
,209"AR-19" "10-MAR-19"
,209"EB-19" "17-FEB-19"
,209"AR-19" "03-MAR-19"
,209"EB-19" "24-FEB-19"

# 10  
Old 05-10-2019
You seem to have DOS line terminators (<CR> = ^M = 0x0D = \r) in your file - how /where did you produce it? Remove them like
Code:
$ awk -F, -v yr='"YEAR"' ' { sub (/\r$/, ""); $0=$0 OFS (NR == 1 ? yr : "20" substr($0,length($0)-2,2)) } 1 ' OFS="," 
"17-FEB-19" "17-FEB-19",2019
"24-FEB-19" "24-FEB-19",2019
"24-FEB-19" "24-FEB-19",2019

# 11  
Old 05-10-2019
If we go back to post #1 in this thread, you might notice that Kunalcurious used the awk script:
Code:
awk -F ',' '{print $11,$11"\r"}' test1.csv > test2.csv

so there no reason to wonder where the <carriage-return>s came from...
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