Most simple way of extracting substring in Unix shell

regexshellunix

What's the most simple way to extract substring on Unix shell (with regex)?

Simple means:

  • less feature
  • less options
  • less study

Update

I realized regex itself is conflicting with simplicity, and I chose the simplest one cut as the chosen answer. I am sorry for vague question. I changed title to represent current state of this QA more precisely.

Best Answer

cut might be useful:

$ 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 are good for this too, here is a sample script:

#!/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})"

That outputs:

  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)

And as per Gnudif above, there are always sed/awk/perl for when the going gets really tough.

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