I must find in a file this string:
200 https://www.example.example
The value 200 is randomic, I must find every HTTP return code (ex 200, 301, 404 etc...etc)
How can I grep only this string with return code variable (I don't want specify every return code in grep command)?
cat file.txt | grep "*** http*"
But it doesn't work.
So you want to match any line starting with a three digit number, followed by "http"?
grep -E '^[0-9]{3} http' file.txt
More accurate as suggested by fedorqui (thanks) would be this:
grep -E '^[1-5][0-9]{2} http' file.txt
This matches numbers in the range 100-599, which is closer to the range of HTTP status codes.
^[1-5][0-9]{2} is more accurate, since it is a HTTP status code.First, there's no need for cat with grep. It can take a file argument itself.
You can do this:
grep '^[1-5][0-9]\{2\}[[:blank:]]\+http' file.txt
This will get you every line that matches with your criteria.
If you want only the return codes, you can run another grep or use other commands, for example with cut:
grep '^[1-5][0-9]\{2\}[[:blank:]]\+http' file.txt | cut -d ' ' -f1
and with grep,
grep '^[1-5][0-9]\{2\}[[:blank:]]\+http' file.txt | grep -o '^[0-9]\+'
grep "*** http*"is not close to the regexp you want. You should find some kind of regexp tutorial if you're going to be trying to use any tools that use regexps (grep, sed, awk, perl, ruby, etc....).