怎样从最后一列往前取

t069064449

123456]
SDFDFF]
WWW111]

我想以]为分隔符，从后面取3列，结果为
456
DFF
111

请问应该怎么做

moperyblue

1. echo '123456]
   
2. SDFDFF]
   
3. WWW111]'|sed -r 's/.*(.{3})].*/\1/'
   
4. 456
   
5. DFF
   
6. 111

复制代码

t069064449

sed -r 貌似我的机器没有这个选项
回复 2# moperyblue


   

moperyblue

1. echo '123456]
   
2. SDFDFF]
   
3. WWW111]'|sed  's/.*\(.\{3\}\)].*/\1/'
   
4. 456
   
5. DFF
   
6. 111

复制代码

1. awk '{$0=substr($0,length($0)-3,3)}1'

复制代码

回复 3# t069064449


   

jason680

回复 1# t069064449

$ grep -oP "...(?=])" FILE

$ awk 'match($0,"(...)]",a){print a[1]}' FILE

$ perl -lane '{print $1 if(m/(...)]/)}' FILE

hz_oracle

grep -Po ".{3,3}]" file | sed 's/]//'

sunzhiguolu

1. perl -nle 'print $& if(/\w{3}\b/)' f

复制代码

456
DFF
111

ll104567

sed 's/^...\(...\)]/\1/'

awk 'BEGIN{FS=""}{print $4FS$5FS$6}'

t069064449

谢谢大家，我都试试

疯子影子

awk -F"]" '{print substr($1,length($1)-2,3)}' str1.txt