awk 替换文本内容 学习的感觉很好
[root@localhost test]# cat awk 1a 9,100.34 1b 1,999.00 1c 5,656.55 [root@localhost test]# awk '{sub(/1/,"test")}{print "\n",$1,$2}' awk testa 9,100.34 testb 1,999.00 testc 5,656.55 [root@localhost test]# awk '{gsub(/1/,"test")}{print "\n",$1,$2}' awk testa 9,test00.34 testb test,999.00 testc 5,656.55[root@localhost test]# awk '{sub(/[0-9]+/,"")}{print "\n",$1,$2}' awk a 9,100.34 b 1,999.00 c 5,656.55
打印出$1只包含4个字符的 awk '$1~/^....$/{print $1}' file
http://bbs.linuxtone.org/thread-17620-1-1.html 看到的学习一下记录一下 效果是有了 但时间和我系统时间对不上
[root@localhost test]# cat awk 1a 9,100.34 dkjfjkdkjf 45 lopo 1b 1,999.00 dgfg 456 ll 1c 5,656.55 fghgf 465 df
[root@localhost test]# awk '{$2=strftime("%F %T",$2);print $1,$2,$3 >"bbb.txt";print $1,$2,$4 >"ccc.txt"}' awk [root@localhost test]# cat bbb.txt 1a 1969-12-31 16:00:09 dkjfjkdkjf 1b 1969-12-31 16:00:01 dgfg 1c 1969-12-31 16:00:05 fghgf [root@localhost test]# date Wed Dec 14 22:49:28 PST 2011 [root@localhost test]# cat ccc.txt 1a 1969-12-31 16:00:09 45 1b 1969-12-31 16:00:01 456 1c 1969-12-31 16:00:05 465
[root@localhost test]# dateWed Dec 14 23:07:09 PST 2011
问题已解决 把{$2=strftime("%F %T",$2)中的$2去掉就可得到正确的格式了 见下图一个文件，列数是不一样的，如果有5列，就取前4列，如果有6列，就取前5列
当第一列大于2的时候 打印