在Unix
shell(使用正则表达式)上提取子字符串的最简单方法是什么？
简单意味着：
>减少功能
>减少选择
>少学习
更新
我意识到正则表达式本身与简单性相冲突,我选择了最简单的一个作为选择的答案.对于模糊的问题我很抱歉.我更改了标题以更准确地表示此QA的当前状态.
切割可能有用：
$echo hello | cut -c1,3 hl $echo hello | cut -c1-3 hel $echo hello | cut -c1-4 hell $echo hello | cut -c4-5 lo
Shell Builtins对此也有好处,下面是一个示例脚本：
#!/bin/bash
# Demonstrates shells built in ability to split stuff. Saves on
# using sed and awk in shell scripts. Can help performance.
shopt -o nounset
declare -rx FILENAME=payroll_2007-06-12.txt
# Splits
declare -rx NAME_PORTION=${FILENAME%.*} # Left of .
declare -rx EXTENSION=${FILENAME#*.} # Right of .
declare -rx NAME=${NAME_PORTION%_*} # Left of _
declare -rx DATE=${NAME_PORTION#*_} # Right of _
declare -rx YEAR_MONTH=${DATE%-*} # Left of _
declare -rx YEAR=${YEAR_MONTH%-*} # Left of _
declare -rx MONTH=${YEAR_MONTH#*-} # Left of _
declare -rx DAY=${DATE##*-} # Left of _
clear
echo " Variable: (${FILENAME})"
echo " Filename: (${NAME_PORTION})"
echo " Extension: (${EXTENSION})"
echo " Name: (${NAME})"
echo " Date: (${DATE})"
echo "Year/Month: (${YEAR_MONTH})"
echo " Year: (${YEAR})"
echo " Month: (${MONTH})"
echo " Day: (${DAY})"
那输出：
Variable: (payroll_2007-06-12.txt)
Filename: (payroll_2007-06-12)
Extension: (txt)
Name: (payroll)
Date: (2007-06-12)
Year/Month: (2007-06)
Year: (2007)
Month: (06)
Day: (12)
并且根据上面的Gnudif,总是有sed / awk / perl,因为什么时候变得非常艰难.
