Cutting certain string from a longer string

Question

I have strings:

fvvDataFolders/DDB/DDB2018-02-21oM]
fbbDataFolders/DDB/DDB2018-02-22oM]

I want to strip everything that starts with Data and ends in what looks like a date:

DataFolders/DDB/DDB2018-02-21
DataFolders/DDC/DDB2018-02-22

How I can do it?

Please be more specific. Do you simply want to strip the first three and last three characters? Or everyhing that starts with Data and ends in what looks like a date? Or just the characters 4-32? That would be cut -c 4-32. — PerlDuck, Feb 25 '18 at 12:43
Would also be nice to know if these are in a file or variable — Sergiy Kolodyazhnyy, May 22 '18 at 21:04

pa4080 · Answer 1 · 2018-05-22T20:43:52.940

You can use the command grep in this way:

grep -oP 'Data.*[0-9]{4}-[0-9]{2}-[0-9]{2}' input-file > output-file

-o, --only-matching - show only the part of a line matching PATTERN.
-P, --perl-regexp - PATTERN is a Perl regular expression; or in this case could be used also the option -E, --extended-regexp - PATTERN is an extended regular expression (ERE).
the regexp 'Data.*[0-9]{4}-[0-9]{2}-[0-9]{2}' matches to your requirements. It begin with the string Data, followed by unknown number * of any characters ., and ends with the date format: 4 digits from 0 to 9 dash 2 digits from 0 to 9 dash 2 digits from 0 to 9.

Here is also a sed solution:

sed -r 's/^.*(Data.*[0-9]{4}-[0-9]{2}-[0-9]{2}).*$/\1/' /tmp/input-file

redirect the output to a new file > output-file or use the option -i.bak to make the changes in their places and create a backup file.
-r, --regexp-extended - use extended regular expressions in the script.
the command s means substitute: /<string-or-regexp>/<replacement>/.
^.* will match to the beginning ^ of the line, followed by unknown number of any characters.
.*$ will match to the end $ of the line, precede by unknown number of any characters.
within the the , the capture group (...), will be treated as the variable \1. So the whole line ^.*$ will be substituted by the part that matces to what is in the brackets.

@PerlDuck, I was inspired by your perl example and added also a sed solution. — pa4080, Feb 25 '18 at 15:09
Nice. I also tried sed but as I'm not particular familiar with it I didn't get it working. The -r switch is crucial and without it my attempts all failed. /// If I were the asker I'd accept your answer because it also explains things. Plus: you answered 17 seconds quicker. :-) — PerlDuck, Feb 25 '18 at 15:18
@PerlDuck, I really like topics as these, in which a simple question receives large number of answers with different possible solutions. Here is my favourite: How to remove particular words from lines of a text file? — pa4080, Feb 25 '18 at 15:26

score 4 · Accepted Answer · answered Feb 25 '18 at 14:38

Either

grep -P -o 'Data.+?\d\d\d\d-\d\d-\d\d'

or

perl -pe 's/^.+(Data.+?\d\d\d\d-\d\d-\d\d).+$/$1/'

will do. They both print the minimal string that starts with Data and ends in what looks like a date (YYYY-MM-DD).

echo "fvvDataFolders/DDB/DDB2018-02-21oM]" > input.txt
echo "fbbDataFolders/DDB/DDB2018-02-22oM]" >> input.txt
grep -P -o 'Data.+?\d\d\d\d-\d\d-\d\d' input.txt

# output:
DataFolders/DDB/DDB2018-02-21
DataFolders/DDB/DDB2018-02-22

perl -pe 's/^.+(Data.+?\d\d\d\d-\d\d-\d\d).+$/$1/' input.txt

# output:
DataFolders/DDB/DDB2018-02-21
DataFolders/DDB/DDB2018-02-22

Cutting certain string from a longer string

2 Answers2