Regular expression and grep not working

Question

I have the following regular expression:

([:digit:]{4})-([:digit:]{1,2})-([:digit:]{1,2})

It should get dates in this format:

2010-12-19

And I am using it on filenames that look like this:

2010-12-19-xxx-xxx-xxx.markdown

And, when I use it with grep like this:

echo $POST | grep -oE "([:digit:]{4})-([:digit:]{1,2})-([:digit:]{1,2})" # $POST is the filename

It doesn't work, I just get emptiness.

Answer 1

Try this:-

echo $POST | grep -oE "[0-9]{4}-[0-9]{1,2}-[0-9]{1,2}"

If I try it here, I get:-

[andys@daedalus ~]$ echo "2010-12-19-aaa-bbb-ccc-ddd.markdown" | grep -oE "[0-9]{4}-[0-9]{1,2}-[0-9]{1,2}"
2010-12-19

Hope that's what you're looking for.

Answer 2

Andy's answer is fine, but if you want something closer to your original syntax, you could try:

echo $POST | egrep -oE "([[:digit:]]{4})-([[:digit:]]{1,2})-([[:digit:]]{1,2})"

You need egrep here for extended regular expressions, and the double brackets for character classes.

Answer 3

You don't need the parentheses, but you do need more square brackets. Character classes have the same characteristics as individual characters. Just as you might search for vowels like this: [aeiou], or digits like this: [0123456789] or this: [0-9], you need to enclose a class such as [:digit:] or [:upper:] in a bracket expression as well: [[:xdigit:]] (hex digits).

grep -oE "[[:digit:]]{4}-[[:digit:]]{1,2}-[[:digit:]]{1,2}"