Pulling out a certain value from a string via regex

Question

I've see a bunch of examples, but I just can't seem to make this work. Can grep output only specified groupings that match? for example seems like it should work, but I get either errors or no output at all.

I want to do:

pathname="/a/long/path/of/mine/2x02 - bar.mp4"

All of the examples will be the long path, one or two digits, x and then 2 digits followed by a space, a - and a filename.

I want to parse this for the 02 value: https://regex101.com/ shows that \d{1,2}x(\d\d) should have match 1 = 02 in this example.

What I can't figure out is if I have

echo "$pathname" | sed -n 's/.*\d{1,2}x\(\d\d\)/\1/p'

or

echo $pathname | grep -oP '\d{1,2}x(\d\d)'

I get nothing. I can do:

echo $pathname | grep -oP '(\d\d)'

but there could be cases where there are other 2 digit values in a row, like if I had

/a/long/path/of/mine/12x02 - bar.mp4

in which case I don't think the above will specify the second match, so I prefer the more specific regex... IF I can use matching groups or something. I'm trying to do this in bash on Scientific Linux 7.1.

Answer 1

As you were using grep with PCRE (-P), you can use this Regex pattern :

grep -Po '\d{1,2}x\K\d{2}(?= )' <<<"$pathname"

• \d{1,2}x will match one or two digits followed by x, then \K will discard the match

• \d{2} will match exactly two digits, the zero width positive lookahead pattern (?= ) ensures that we have a space after the two digits.

So this should fulfill your requirements.

Example :

$ grep -Po '\d{1,2}x\K\d{2}(?= )' <<<'/a/long/path/of/mine/2x02 - bar.mp4'
02

$ grep -Po '\d{1,2}x\K\d{2}(?= )' <<<'/a/long/path/of/mine/34x12 - bar.mp4'
12

$ grep -Po '\d{1,2}x\K\d{2}(?= )' <<<'/a/long/path/of/mine/0x1 - bar.mp4'
## No match

$ grep -Po '\d{1,2}x\K\d{2}(?= )' <<<'/a/long/path/of/mine/00x1 - bar.mp4'
## No match

Answer 2

Using sed

With sed in basic mode, the braces need to be escaped:

$ echo "$pathname" | sed -n 's/.*[[:digit:]]\{1,2\}x\([[:digit:]][[:digit:]]\).*/\1/p'
02

For greater portability, I used [[:digit:]] in place of \d. I also added .* to the end so as to remove the trailing text.

Using grep -P

grep -P supports a look-behind feature but the look-behind text has to be of fixed length. So, we can look for a single digit preceding the x preceding the two digits that we want to display:

$ echo "$pathname" | grep -oP '(?<=\dx)(\d\d)'
02

Alternate path

Both of the above also work with the alternate path:

$ echo '/a/long/path/of/mine/12x02 - bar.mp4' | grep -oP '(?<=\dx)(\d\d)'
02
$ echo '/a/long/path/of/mine/12x02 - bar.mp4' | sed -n 's/.*[[:digit:]]\{1,2\}x\([[:digit:]][[:digit:]]\).*/\1/p'
02

Answer 3

Using just a posix shell

p=$pathname
p=${p##*/}
p=${p#*x}
p=${p%% *}
echo "$p"

#or on one line
p=${pathname##*/};p=${p#*x};p=${p%% *};echo "$p"