How do I fetch only numbers in grep?

Question

I have file like this :

 other lines . . .    
 blah blah blah (:34)

I wish to find the occurrence of numbers in the above file. I came up with:

grep [0-9] filename

But that is printing the whole:

blah blah blah (:34)

Rather I want only 34. Is there any way to do so?

Solution 1

You can use grep -E to access the extended regular expression syntax( Same as egrep)

I have created a testfile with below contents:

>cat testfile
this is some text
with some random lines

again some text
ok now going for numbers (:32)
ok now going for numbers (:12)
ok now going for numbers (:132)
ok now going for numbers (:1324)

Now to grep the numbers alone from the text you can use

>grep -Eo '[0-9]{1,4}' testfile
32
12
132
1324

will be output.

Here "-o" is used to only output the matching segment of the line, rather than the full contents of the line.

The squiggly brackets (e.g. { and }) indicate the number of instances of the match. {1,4} requires that the previous character or character class must occur at least once, but no more than four times.

Hope this helps

Solution 2

You can use RE bracket expression [:digit:] specified by section 9.3.5 of POSIX standard , in combination with -o flag to print only matching "words"

$ grep -o '[[:digit:]]*' <<< $'No number in this line\nbut 123 here'                                                     
123

Solution 3

You can use Perl style regular expressions as well

grep -Po "\\d+" filename

-P Interpret PATTERNS as Perl-compatible regular expressions (PCREs).

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

Solution 4

grep -o will print only the matching part of the line. Otherwise grep will print any lines with the pattern.

Solution 5

I would use curl to access your file locally or remotely, then I would grep lines with numbers wrapped in (: ) then cut those pieces off and write to file

the accepted answer ignores there could be numbers in the previous lines of the file, it does work for the example data, but what if the file was remote?

Local

curl file:///home/$USER/Public/input.txt  | grep -o '(:.*)' | cut -d ":" -f 2 | cut -d ")" -f 1 > output.txt

In this example output.txt in your current folder will be written over, we are accessing input.txt from your Public folder.

Remote

curl https://yoursite.com/Public/input.txt  | grep -o '(:.*)' | cut -d ":" -f 2 | cut -d ")" -f 1 > output.txt

In this example output.txt in your current folder will be written over, we are accessing input.txt from https://yoursite.com/Public/.

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Comments

[Admin]

In the future, also check out the man page for grep (or any other program). The man page details the options required for many common uses of the program. e.g. man grep

[FractalSpace]

Nice. Also, to match more than 4 or arbitrary number of digits, use grep -Eo '[0-9]{1,}' testfile

[PerlDuck]

Why use curl when a simple cat would do?

[Stef]

The question doesn't clearly state that the file is local. this answer allows for both.

[alec]

deserves way more than 2 upvotes :)

[Karl Xu]

@FractalSpace when have the constrain of repeating at least one time in [0-9]{1,}, the -o option might not necessary, since repeating at least one time, by definition remove the empty output?

[FractalSpace]

It's been a while, so I have no recollection of what I wrote and why :)

[Jacob]

@Stef it's inconsequential where the file is or if the input is even a file. This answer has a lot of cruft that just adds confusion.

[spuder]

Note: You may not have -P option if running BSD grep (e.g. OSX. )