在 Unix shell 中提取子字符串的最简单方法？

Question

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？

简单的意思：

• 功能较少
• 更少的选择
• 少学习

更新

我意识到正则表达式本身与简单性相冲突，因此我选择了最简单的一种cut作为所选答案。我很抱歉含糊不清的问题。我更改了标题以更准确地表示此 QA 的当前状态。

Answer 1

cut 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 内置程序对此也有好处，这是一个示例脚本：

#!/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)

根据上面的 Gnudif，当事情变得非常艰难时，总会有 sed/awk/perl。