grep capture regex

Question

I am trying to use grep to capture data below:


"\\.xy$"
"\\.ab$"
"\\.ef\\.hi$"

I have

grep  -Eo "((\\\\\.[a-zA-Z]+)){1,2}\\$" file

two problems:

Try using the *extended grep* `egrep` with better support for regular expressions. — pavium, Jun 13 '11 at 02:09
@pavium: I just tried with egrep -o, exactly the same results — user775187, Jun 13 '11 at 02:11
@pavium: Egrep is the same as `grep -E`, which the OP is using. — Johnsyweb, Jun 13 '11 at 02:12

score 2 · Accepted Answer · edited Sep 25 '11 at 03:44

2

Precede the dollar with a single backslash:

% grep -Eo '"(\\\\\.[[:alpha:]]+){1,2}\$"' input
"\\.xy$"
"\\.ab$"
"\\.ef\\.hi$"

Or put the special characters into square brackets, which I find more readable:

% grep -Eo '"([\]{2}[.][[:alpha:]]+)+"' input 
"\\.xy$"
"\\.ab$"
"\\.ef\\.hi$"

edited Sep 25 '11 at 03:44

hobs

answered Jun 13 '11 at 02:11

Johnsyweb

almost works, I like the second approach, one problem: the $ still shows up in the results, but it's not part of the capture group, why? – user775187 Jun 13 '11 at 02:35
`-o` means "show only matching part of line", not the capture group `'(...)'`. What do you mean "almost works?" – Johnsyweb Jun 13 '11 at 02:48
@user775187: You can't with grep, but this will work: `grep '[$]"$' input | grep -Eo '"([\]{2}\.[[:alpha:]]+){1,2}'` – Johnsyweb Jun 13 '11 at 08:45
1

you can eliminate one more backslash in your readability version: `grep -Eo '"([\]{2}[.][[:alpha:]]+)+[$]"' input` – hobs Sep 25 '11 at 03:05

score 0 · Answer 2 · answered Jun 13 '11 at 02:12

0

You've double-escaped the $ - try this:

grep  -Eo '"((\\\\\.[a-zA-Z]+)){1,2}\$"' file

answered Jun 13 '11 at 02:12

Bohemian

2 Answers2