在bash

Question

如何在:之前提取所有字符串，我只能得到第一个

$ echo "cc_dd:qqq www;aa_bb:ll fi;ee_ff:bb dd ee" | sed 's/:.*//g'
$ cc_dd
# want to print cc_dd aa_bb and ee_ff

Answer 1

短 grep 方法：

s="cc_dd:qqq www;aa_bb:ll fi;ee_ff:bb dd ee"
grep -Po '[^;:]+(?=:)' <<< $s

-P选项，允许Perl正则表达式

-o选项，告诉只打印匹配的子字符串

(?=:) - 积极的先行断言，确保所需的子字符串后跟:

输出：

cc_dd
aa_bb
ee_ff

效果比较：

time (for i in {1..1000}; do grep -Po '[^;:]+(?=:)' <<< $s > /dev/null; done;)

real    0m1.936s
user    0m0.036s
sys     0m0.236s

time (for i in {1..1000}; do awk -v RS=\; '{split($0,a,":"); print a[1]}' <<< $s > /dev/null; done;)

real    0m2.633s
user    0m0.056s
sys     0m0.264s

Answer 2

awk救援！

echo "cc_dd:qqq www;aa_bb:ll fi;ee_ff:bb dd ee" | 
awk -v RS=\; '{split($0,a,":"); print a[1]}'

cc_dd
aa_bb
ee_ff

Answer 3

awk -F'[;:]' '{print $1,$3,$5}' OFS='\n' file

cc_dd
aa_bb
ee_ff