Bash Regex Matching digits

Question

I am try to match only the digits from a text file that is in this format:

> 1234

I'm running thru the file with a loop. Every line store in $i

$i | grep "\d{4}"

Output looks like this:

Why is it still outputing the >? I want to remove those.

$i at the start of the line will try to run a command with its value, e.g. a command called 1234. I'm also confused about "what you are doing wrong." Your output doesn't look wrong to me... — Explosion Pills
– Explosion Pills, Commented Feb 26, 2013 at 15:11
Sorry I wasn't clear. I mean why is it outputing the >. I'm trying to get rid of those. — ExceptionLimeCat
– ExceptionLimeCat, Commented Feb 26, 2013 at 15:33
Did you expect grep to extract only the number? Grep will display the entire line containing a match. Maybe you want to use sed instead? (I'm deleting my answer since it didn't address the question you meant to ask.) — William
– William, Commented Feb 26, 2013 at 15:37
@William "Grep will display the entire line containing a match" not necessarily — user000001
– user000001, Commented Feb 26, 2013 at 15:40

user000001 · Accepted Answer · 2013-02-26 15:55:13Z

2

From man grep

   -o, --only-matching
          Print only the matched (non-empty) parts of a matching line, with each such part on a
          separate output line.

Try grep -o ....

Test:

i="> 1234"
$ echo "$i"
> 1234
$ echo "$i" | grep -oP "\d{4}"
1234

edited Feb 26, 2013 at 15:55

answered Feb 26, 2013 at 15:39

user000001

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1 Comment

William Over a year ago

Note to self, check manpages... Obviously things have changed since Solaris 7's grep :-)

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