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https://www.gnu.org/software/sed/manual/sed.html#Command_002dLine-Options

-n
--quiet
--silent
-e script
--expression=script Add the commands in script to the set of commands to be run while processing the input.

https://stackoverflow.com/questions/14593320/insert-field-separator-to-arrange-and-capture-value-in-variable-in-shell-script/14593411#14593411

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echo $'For example:\nThis is counter1 1000\nthis counter2 2000\n
     this counter3 is higher value 3000\ndone.\n' |
  sed -ne 's/^.* \([0-9]\{1,99\}\)/\1/p'

I understand ([0-9]\{1,99\}\) refer to pattern integer.
^refer to beginning of a the string.
.* refer to 1 or more character.

But overall, I am still confused with sed -ne 's/^.* \([0-9]\{1,99\}\)/\1/p'

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  • Not clear on precisely what you're not clear on : ) But here's a useful resource for regular expressions Commented Jan 9, 2022 at 10:50
  • As stated in the answer .* means 0 or more characters. For 1 or more, you need .+ (with -E) or .\+. And ^ matches the beginning of the line Commented Jan 9, 2022 at 11:04

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(First, this is not a bash integer regex pattern. sed is handling the regex and has nothing to do with bash)

So I understand that you are unable to figure out the regex in sed -ne 's/^.* \([0-9]\{1,99\}\)/\1/p'

Let's make it simpler. The equivalent code would be sed -E -ne 's/^.* ([0-9]{1,99})/\1/p'

The -E enables extended (modern) regex and you won't have to escape stuff using \

The s/pattern/replacement/ operator looks for a string matching the regex pattern in the left side and replaces it with the string on the right.

Now, let's focus on /^.* ([0-9]{1,99})/

  • ^ matches at the beginning of the line
  • .* matches any character 0 or more times
  • (space) matches a single space character
  • [0-9]{1,99} matches any digits occurring a minimum of 1 time to a max of 99; so any number with a length of 1 to 99

([0-9]{1,99}) is in parentheses which means sed will "capture" this substring and use it in the replacement part where \1 was used.

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  • Note that it's functionally equivalent to sed -ne 's/^.* \([0-9]\)/\1/p'. The interval operator serves no purpose here as there's nothing that needs to be found after that sequence of digits. [0-9]{1,99} would also match on the first 99 digits of a 300 digit number. It would make a difference in sed -ne 's/^.* \([0-9]\{1,99\}\)$/\1/p' for instance where the $ matches on the end of the subject. Commented Jan 9, 2022 at 9:18
  • I understand s/pattern/replacement/ but what does the last character p refer to? Commented Jan 9, 2022 at 15:36
  • p is a sed directive meaning print whatever was matched. Commented Jan 9, 2022 at 15:46
  • @StéphaneChazelas: But yur expression only matches one digit. Commented Jan 9, 2022 at 18:02

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