__（'…'）的正则表达式

Question

I'm trying to get a regular expression to work and I'm having problems: I want to match everything that starts with

__(' or __("  

and ends with

') or ") 

I tried with

/__\(['"][^']*['"]\)/g  and /__\(['"].*['"]\)/g

but they all have problems with this sample text:

text that should not match
__('all text and html<a href="#">link</a> that should match') text that should not match __('all text and html<a     href="#">link</a>')
__("all text and html<a href="#">link</a>") text that should not match __("all text and html<a     href="#">link</a>")
other text that should not match

who has the winning RegExp?

Answer 1

Using the second example, use *? for a non-greedy match and consider using a backreference for the quotes.

/__\((['"]).*?\1\)/g

Live Demo

Answer 2

I guess that you want non-greedy match, then the best solution is to use:

/__\(['"][^']*?['"]\)/g

Also, escaping the single and double quotes may help:

You can check it at this regex101

http://regex101.com/r/sM8yF9/1