Grep to extract the word matching the given pattern

shell, unix

Solution

Try following:

$ grep  -Eio '[a-z]+:[0-9]+' log.txt 
abc:12
abcd:12
def:134

- `-i` to ignore case.

- `-o` to print only matched part.

UPDATE

To match only `abc` / `def`:

$ grep  -Eio '\b(abc|def):[0-9]+\b' log.txt 
abc:12
def:134

- `(abc|def):`: match `abc` or(|) `def` followed by `:`.

- `[0-9]+`: matched numbers.

- `\b`: match word boundary

Problem

I have a log file with entries like ``` INFO 2013-08-16 13:46:48,660 Index=abc:12 insertTotal=11 INFO 2013-08-16 13:46:48,660 Index=abcd:12 insertTotal=11 INFO 2013-08-16 13:46:48,660 Index=def:134 insertTotal=11 INFO 2013-08-16 13:46:48,660 Index=abkfe insertTotal=11 INFO 2013-08-16 13:46:48,660 Index=lmkfe insertTotal=11 INFO 2013-08-16 13:46:48,660 Index=lmkfe insertTotal=11 ``` I would like to grep and extract the words that match my pattern which is `abc:<some_number>` and `def:<some_number>`. ``` $ cat "log.txt" | grep -w "abc" -w "def" >> "failed_values.txt"; ``` So in this case, my `failed_values.txt` should only have ``` abc:12 def:134 ``` The key to note is that my pattern ends with a `:` followed by a number and then a space e. g. `abc:122`.

Original source