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

Simple means:

  • less feature
  • less options
  • less study
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It would help if you could describe what you want to extract from where. Even with complex tools like grep and sed, simple things tend to be simple. –  SvW Sep 4 '10 at 16:14
Your question is vague and too broad. –  Dennis Williamson Sep 4 '10 at 17:30
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3 Answers

up vote 4 down vote accepted

cut might be useful:

$ echo hello | cut -c1,3
$ echo hello | cut -c1-3
$ echo hello | cut -c1-4
$ echo hello | cut -c4-5

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

# 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 _


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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While you might sometimes need to declare a variable read-only, it's rare in a context such as this one to need to export a variable. It would be much simpler to just do an assignment: var=value without using declare at all. –  Dennis Williamson Sep 4 '10 at 17:30
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Unix shells do not traditionally have regex support built-in. Bash and Zsh both do, so if you use the =~ operator to compare a string to a regex, then:

You can get the substrings from the $BASH_REMATCH array in bash.

In Zsh, if the BASH_REMATCH shell option is set, the value is in the $BASH_REMATCH array, else it's in the $MATCH/$match tied pair of variables (one scalar, the other an array). If the RE_MATCH_PCRE option is set, then the PCRE engine is used, else the system regexp libraries, for an extended regexp syntax match, as per bash.

So, most simply: if you're using bash:

if [[ "$variable" =~ unquoted.*regex ]]; then

If you're not using Bash or Zsh, it gets more complicated as you need to use external commands.

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grep and sed are probably the tools you want, depending on the structure of text.

sed should do the trick, if you do not know what the substring is, but know some pattern that is around it.

for example, if you want to find a substring of digits that starts with a "#" sign, you could write something like:

sed 's/^.*#([0-9]+)/\1/g' yourfile

grep could do something similar, but the question is what you need to do with the substring and whether we are talking normal line-end delimited text or not.

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