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

我有一个具有以下格式的linux文件，我的要求是文件（5,7）中即将出现的日期（值）应按照我的substr命令进行剪切。

此外，第5列和第7列是那些可能包含也可能不包含date（values）的列，因此我需要一个if语句来更正日期值（如果存在）。

请使用正确的脚本纠正我。

样本文件：

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

我的剧本：

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

错误抛出：

-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

输出：

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

这在我的Solaris机器上对我有用：

/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

输出：

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