Why is this grep not working?

Question

The following regex with grep doesn't seem to be working: grep "(?=\_\(\").*?(?=\"\))" ./testfile.js

testfile.js is as follows:

asoijf oaisdjf _("string 1") fodijsasf _("string 2")
fasdoij _("string 3");
console.log(_("string 4"));

My aim is to grab all the strings enclosed in _() function calls, without greps -P flag (the option doesn't exist for me). Expected output would be:

string 1
string 2
string 3
string 4

Any idea's?

Update: The -P flag was removed from bash in my version of mac (see grep -P no longer works how can I rewrite my searches)

.*? always gives me problems. Try: _\("([^"]+)"\) if that works I'll write up a nice answer — sniperd, Sep 25 '17 at 12:57
`grep`'s regex syntax is different from and more limited than what you might be used to (Perl syntax). Parentheses are only meta-characters when escaped. It doesn't support lookahead `(?=...)` or non-greediness `.*?`. — John Kugelman, Sep 25 '17 at 13:00

RavinderSingh13 · Accepted Answer · 2017-09-25T13:32:41.493

1

Could you please try following awk and let me know if this helps you.

grep -oE '_\([^\)]*' Input_file | cut -c3-

Output will be as follows.

"string 1"
"string 2"
"string 3"
"string 4"

EDIT: Since OP doesn't have -P option in it's O.S so providing an awk approach here too.

awk '{
  while($0){
    match($0,/_\([^\)]*/);
    st=RSTART;
    le=RLENGTH;
    if(substr($0,RSTART+2,RLENGTH-2)){
      print substr($0,RSTART+2,RLENGTH-2)
};
  $0=substr($0,st+va+3)
}
}
'   Input_file

edited Sep 25 '17 at 13:32

answered Sep 25 '17 at 12:57

RavinderSingh13

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1

That misses string 2 – 123 Sep 25 '17 at 12:59
I thought only 1st should be picked, edited it now, thanks. – RavinderSingh13 Sep 25 '17 at 13:04
the -P flag doesn't work for me, is this something to do with not having perl installed? – Nodeocrat Sep 25 '17 at 13:05
@Nodeocrat, could you please try my edit solution with awk and let me know if this helps you please. – RavinderSingh13 Sep 25 '17 at 13:34
@Nodeocrat, I edited my grep solution too now, let me know on same. – RavinderSingh13 Sep 25 '17 at 13:36
1

The revised answer ```grep -oE '_\([^\)]*' Input_file | cut -c3-``` works just as desired, thanks :) – Nodeocrat Sep 25 '17 at 13:48

P.... · Answer 2 · 2017-09-25T13:17:37.887

0

grep -oP '_\("\K[^"]+' inputfile
string 1
string 2
string 3
string 4

Here, -o will print only the matched result, not the whole line. \K will be used for the look behind, Means matches the string left of \K but do not print it. [^"]+ means anything except " one or more time.

Or without -P option:

grep -oE '_\("[^"]+' inputfile|cut -d'"' -f2
string 1
string 2
string 3
string 4

edited Sep 25 '17 at 13:17

answered Sep 25 '17 at 13:01

P....

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Why is this grep not working?

2 Answers2