How to extract text from a string using sed?
bash, regex, sed
Solution
The pattern `\d` might not be supported by your `sed`. Try `[0-9]` or `[[:digit:]]` instead.
To only print the actual match (not the entire matching line), use a substitution.
sed -n 's/.*\([0-9][0-9]*G[0-9][0-9]*\).*/\1/p'
The parentheses capture the text they match into a back reference. Here, the first (and only) parentheses capture the string we want to keep, and we replace the entire line with just the captured string `\1`, and print the resulting line. (The `p` option says to print the resulting line after performing a successful substitution, and the `-n` option prevents `sed` from performing its normal printing of every other line.)
Problem
My example string is as follows: ``` This is 02G05 a test string 20-Jul-2012 ``` Now from the above string I want to extract `02G05`. For that I tried the following regex with sed ``` $ echo "This is 02G05 a test string 20-Jul-2012" | sed -n '/\d+G\d+/p' ``` But the above command prints nothing and the reason I believe is it is not able to match anything against the pattern I supplied to sed. So, my question is what am I doing wrong here and how to correct it. When I try the above string and pattern with python I get my result ``` >>> re.findall(r'\d+G\d+',st) ['02G05'] >>> ```