使用 awk 更改文件中的日期格式

Question

var1= date -d "19521029 1010" +"%Y-%m-%d %H:%M"  

echo$var1

its working properly but date -d inside awk is not working

file

KOD19|KAD37748|DEL37728|VIDYA|19521029 1010|201209111625
SASI19|NAS38228|DEL37728|KARTHIKA|19521029 0000|201308071912
RADHA94|VAS37748|DEL37728|LALINKA|19521029 0000|201407061815

First method

awk 'BEGIN {FS=OFS="|"} $5=date -d "$5"+"%Y-%m-%d %H:%M" {print}' file

second method

awk -F '|' '{$5=date -d $5+"%Y-%m-%d %H:%M"; {OFS="|"}; print}' file

Desired output

KOD19|KAD37748|DEL37728|VIDYA|1952-10-29 10:10|201209111625
SASI19|NAS38228|DEL37728|KARTHIKA|1952-10-29 10:10|201308071912
RADHA94|VAS37748|DEL37728|LALINKA|1952-10-29 10:10|201407061815

I want to convert the fifth column of "file" to user input date format. Actually column number and date formats are dynamic ie dt="%Y-%m-%d %H:%M" and num=$5 it will very depends on user requirement.

Answer 1

Just use match() to catch the data in the 5th field and print it back:

awk -F"|" -v OFS="|" '{
        match($5, /([0-9]{4})([0-9]{2})([0-9]{2}) ([0-9]{2})([0-9]{2})/, a);
        $5=sprintf("%d-%d-%d %s:%s", a[1], a[2], a[3], a[4], a[5])
     }1' file

It returns:

$ awk -F"|" -v OFS="|" '{match($5, /([0-9]{4})([0-9]{2})([0-9]{2}) ([0-9]{2})([0-9]{2})/, a); $5=sprintf("%d-%d-%d %s:%s", a[1], a[2], a[3], a[4], a[5])}1' file
KOD19|KAD37748|DEL37728|VIDYA|1952-10-29 10:10|201209111625
SASI19|NAS38228|DEL37728|KARTHIKA|1952-10-29 00:00|201308071912
RADHA94|VAS37748|DEL37728|LALINKA|1952-10-29 00:00|201407061815

See a full description of this technique in How to filter logs easily with awk? .