如何在awk中使用if语句，以在Linux中对具有日期和其他值以分隔格式的文件使用

Question

I have a linux file with the below format, my requirement is the date (value) which are coming in the file in the columns (5,7) should be cut as per my substr command.

Moreover, the 5th and 7th columns are those columns which may or may not contain the date(values.), so I need a if statement to correct the date values if present.

Kindly correct me with the proper script.

SAMPLE FILE:

*# cat zx
7b540eda6b89136432213fbe09815c12,50281271950,,,20160524 08:14:26+0400,,20160524 08:14:26+0400,,7b540eda6b89136432213fbe09815c12,,,
7b540eda6b89136432213fbe09815c12,50281271950,,,,,20160524 08:14:26+0400,,7b540eda6b89136432213fbe09815c12,,,

My Script:

awk -F "," '{print $1,$2,"",'BEGIN{if($5==""){print substr($5,1,4)"-"substr($5,5,2)"-"substr($5,7,11)}}',$6,'BEGIN{if($7==""){print substr($7,1,4)"-"substr($7,5,2)"-"substr($7,7,11)}}',$8,$9,$10,$11,$12}' OFS=, zx

ERROR THROWN:

-bash: syntax error near unexpected token `('

Answer 1

awk -F, 'BEGIN{OFS=","} length($5)>0{$5=substr($5,1,4)"-"substr($5,5,2)"-"substr($5,7,11)} length($7)>0{$7=substr($7,1,4)"-"substr($7,5,2)"-"substr($7,7,11)} 1' file

Output:

7b540eda6b89136432213fbe09815c12,50281271950,,,2016-05-24 08:14:26,,2016-05-24 08:14:26,,7b540eda6b89136432213fbe09815c12,,,
7b540eda6b89136432213fbe09815c12,50281271950,,,,,2016-05-24 08:14:26,,7b540eda6b89136432213fbe09815c12,,,

Answer 2

This worked for me on my Solaris machine:

/usr/xpg4/bin/awk -F',' '{printf "%s,%s,%s,", $1,$2,""} length($5)>0{printf "%s,", $5=substr($5,1,4)"-"substr($5,5,2)"-"substr($5,7,11)} {printf "%s,", $6} length($7)>0{printf "%s,", $7=substr($7,1,4)"-"substr($7,5,2)"-"substr($7,7,11)} {printf "%s,%s,%s,%s,%s\n", $8,$9,$10,$11,$12}' zx

Output:

7b540eda6b89136432213fbe09815c12,50281271950,,2016-05-24 08:14:26,,2016-05-24 08:14:26,,7b540eda6b89136432213fbe09815c12,,,
7b540eda6b89136432213fbe09815c12,50281271950,,,2016-05-24 08:14:26,,7b540eda6b89136432213fbe09815c12,,,