提取字在bash所有匹配的字符前

Question

如何:之前提取所有的字符串，我只得到了第一个如下

$ 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

短的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，讲述只打印匹配的子

(?=:) - 正预测先行断言，保证所需的子后面:

输出：

cc_dd 
aa_bb 
ee_ff 

---

性能比较：

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来救援！

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