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So I'm trying to match a year number sequence with grep and this should be easy. I'm just a bit stumped that my simplistic regex isn't working.

What I'm doing is running a tool which archives some files but it needs to check for the date of the file to put it in the correct directory. I already have properly formatted input which comes to me as:

<span class='t-d'>1994-Oct-28</span>

This is just one example, when I have this I want to grab just the 1994 part of it and use this to continue archiving to the correct year. I was assuming something like this would be sufficient:

grep -o '[0-9]{4}'

But this doesn't seem to match on anything. When I try something like:

grep -o '[0-9]'

it matches all the separate numbers, so 1 9 9 4 2 and 8.

So my syntax is wrong but as for as my knowledge goes this matches a number of 0 to 9 4 times, the {} specifying length either in a range or exact range. If someone could help me with this simple syntax it would be highly appreciated.

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up vote 4 down vote accepted

The {4} is an extended regular expression. grep will not understand it unless you specify the -E option:

-E, --extended-regexp
       Interpret PATTERN as an extended regular expression (ERE, see below).  (-E is specified by POSIX.)

try with

grep -E '[0-9]{4}'


$ echo abcd1234abcd | grep -o -E '[0-9]{4}'

you can also use the [:digit:] character class to avoid problems with locales where the order of the symbols could be different:

$ echo abcd1234abcd | grep -o -E '[[:digit]]{4}'

if for any reason you don't want to use extended regular expressions you can use

grep -o '[0-9][0-9][0-9][0-9]'
share|improve this answer
You're a life saver, thanks! Something so simple. – Yonathan Klijnsma Jul 22 '12 at 13:00

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