How to specify only one instead of multiple of a character in a row in grep regex?

Question

I am trying to use a single grep command to search for all instances of $ until a single =

I have the following command so far grep -E '\$.*=' which returns all lines with a $ and then a = but it also returns lines with == which I dont want. I want it to match the lines below

$string = 'content';    (match)
$_array['key']=12;      (match)

if ($str == 'pass')     (no match)
return $var1==$var2;    (no match)

I get my intended results with this command grep -E '\$.*=' | grep -v '==' but I want to make it work with one grep command and no piping

grep -E '^[^=]*(=[^=]+)*\$[^=]*=([^=]+(=[^=]+)*)?$' Demo I'd explain it but your question was closed for some totaly unknown / bizzarro / obscure reason.. Voting to Reopen ! — sln
– sln, Commented Nov 3 at 19:42

Ed Morton · Accepted Answer · 2025-11-06 17:54:10Z

1

$ grep -E '\$(.*[^=])?=([^=]|$)' file
$string = 'content';    (match)
$_array['key']=12;      (match)

It looks for the $ followed by either any chars then not-= (.*[^=]) or no chars (?), then =, then not-= or the end of the line ([^=]|$).

edited Nov 6 at 17:54

answered Nov 4 at 17:12

Ed Morton

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2 Comments

stevesliva Nov 12 at 3:23

grep -E '\$[^=]+=([^=]|$)' is a few chars more succinct but not different enough for its own answer.

Ed Morton Nov 12 at 13:18

@stevesliva funny, I thought I'd made sure mine didn't match $a = ("v"==$b); as yours an @Cyrus's would, but now I find mine also matches it. Ah well, it wasnt clear if that was desirable or not and the OP seems to have left the building anyway

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