CODEWITHSUNDEEP
regex
keldar
keldar

Reputation: 6252

Regex - match up to first literal

I have some lines of code I am trying to remove some leading text from which appears like so:

Line 1: myApp.name;
Line 2: myApp.version
Line 3: myApp.defaults, myApp.numbers;

I am trying and trying to find a regex that will remove anything up to (but excluding) myApp.

I have tried various regular expressions, but they all seem to fail when it comes to line 3 (because myApp appears twice).

The closest I have come so far is:

.*?myApp

Pretty simple - but that matches both instances of myApp occurrences in Line 3 - whereas I'd like it to match only the first.

There's a few hundred lines - otherwise I'd have deleted them all manually by now.

Can somebody help me? Thanks.

Upvotes: 0

Views: 1331

Answers (2)

Avinash Raj
Avinash Raj

Reputation: 174796

You need to add an anchor ^ which matches the starting point of a line ,

^.*?(myApp)

DEMO

Use the above regex and replace the matched characters with $1 or \1. So that you could get the string myApp in the final result after replacement.

Pattern explanation:

  • ^ Start of a line.
  • .*?(myApp) Shortest possible match upto the first myApp. The string myApp was captured and stored into a group.(group 1)
  • All matched characters are replaced with the chars present inside the group 1.

Upvotes: 2

Stephen Ostermiller
Stephen Ostermiller

Reputation: 25535

Your regular expression works in Perl if you add the ^ to ensure that you only match the beginnings of lines:

cat /tmp/test.txt  | perl -pe 's/^.*?myApp/myApp/g'
myApp.name;
myApp.version
myApp.defaults, myApp.numbers;

If you wanted to get fancy, you could put the "myApp" into a group that doesn't get captured as part of the expression using (?=) syntax. That way it doesn't have to be replaced back in.

cat /tmp/test.txt  | perl -pe 's/^.*?(?=myApp)//g'
myApp.name;
myApp.version
myApp.defaults, myApp.numbers;

Upvotes: 1

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