How to use sed to grab regular expression

Question

I'd like to grab the digits in a string like so :

 "sample_2341-43-11.txt"   to   2341-43-11

And so I tried the following command:

echo "sample_2341-43-11.txt" | sed -n -r 's|[0-9]{4}\-[0-9]{2}\-[0-9]{2}|\1|p'

I saw this answer, which is where I got the idea. Use sed to grab a string, but it doesn't work on my machine:

it gives an error "illegal option -r".
it doesn't like the \1, either.

I'm using sed on MacOSX yosemite.

Is this the easiest way to extract that information from the file name?

you do not need the \1 for the expression you are looking for — Dalorzo, Sep 18 '15 at 01:24
-r is only for GNU sed (the one that comes with bash) isn't it? — Jerry Jeremiah, Sep 18 '15 at 01:47

score 2 · Answer 1 · answered Sep 18 '15 at 01:33

2

You need to set your grouping and match the rest of the line to remove it with the group. Also the - does not need to be escaped. And the -n will inhibit the output (It just returns exit level for script conditionals).

echo "sample_2341-43-11.txt" | sed -r 's/^.*([0-9]{4}-[0-9]{2}-[0-9]{2}).*$/\1/'

answered Sep 18 '15 at 01:33

Zen Momentum

133
1
6

It will not work on MacOSX yosemite! see: http://stackoverflow.com/questions/12178924/os-x-sed-e-doesnt-accept-extended-regular-expressions – Nir Alfasi Sep 18 '15 at 01:51

score 2 · Answer 2 · edited May 23 '17 at 11:44

2

Enhanced regular expressions are not supported in the Mac version of sed.

You can use grep instead:

echo "sample_2341-43-11.txt" | grep -Eo "((\d+|-)+)"

OUTPUT

2341-43-11

edited May 23 '17 at 11:44

Community

1
1

answered Sep 18 '15 at 01:50

Nir Alfasi

53,191
11
86
129

score 0 · Answer 3 · answered Sep 18 '15 at 06:07

echo "one1sample_2341-43-11.txt" \
 | sed 's/[^[:digit:]-]\{1,\}/ /g;s/ \{1,\}/ /g;s/^ //;s/ $//'

1 2341-43-11

Extract all numbers(digit) completed with - (thus allow here --12 but can be easily treated)
posix compliant
all number of the line are on same line (if several) separate by a space character (could be changed to new line if wanted)

score 0 · Answer 4 · answered Sep 19 '15 at 05:17

You can try this ways also

sed 's/[^_]\+_\([^.]\+\).*/\1/' <<< sample_2341-43-11.txt

OutPut:

2341-43-11

Explanation:

[^_]\+       - Match the content untile _ ( sample_)
\([^.]\+\)   - Match the content until . and capture the pattern (2341-43-11)
.*           - Discard remaining character (.txt)

james jelo4kul · Answer 5 · 2015-09-18T01:43:36.277

-1

You can go with what the poster above said. Well, making use of this

pattern "\d+-\d+-\d+" would match what you are looking for. See demo here

https://regex101.com/r/kO2cZ1/3

edited Sep 18 '15 at 01:43

answered Sep 18 '15 at 01:31

james jelo4kul

839
4
17

I believe the \1 was a reference to his matching group. – David H. Bennett Sep 18 '15 at 01:35

How to use sed to grab regular expression

5 Answers5