在Unix shell中提取子串的最简单方法是什么?

前端之家收集整理的这篇文章主要介绍了在Unix shell中提取子串的最简单方法是什么?前端之家小编觉得挺不错的,现在分享给大家,也给大家做个参考。
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})"

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.

原文链接:https://www.f2er.com/bash/385708.html

猜你在找的Bash相关文章