I would like to replace all ONLY single equal signs.
var mystr = 'ONE == TWO ... THREE==FOUR ... FIVE = SIX ... SEVEN=EIGHT' ... NINE := TEN;
return mystr.replace(/(?=\=)([=]{1})(?!\=)/gm, '==');
I get the following:
ONE === TWO ... THREE===FOUR ... FIVE == SIX ... SEVEN==EIGHT ... NINE :== TEN
Numbers 5-6, 7-8, are ok. But, I would like this:
ONE == TWO ... THREE==FOUR ... FIVE == SIX ... SEVEN==EIGHT ... NINE := TEN
Whats wrong with my regex?
Since Javascript doesn't support lookbehind assertions you can't check if the char before is an equal char or not. But you could match it an insert it again.
return mystr.replace(/([^=:])=(?!=)/g, '$1==');
See it here on Regexr.
([^=:]) is a negated character class, that matches any char, but = and :
. This char is reinserted in the replacement string by the $1.
This would not work, when the first char in the string is a single "=", since the start of the string would not be matched by [^=:].
You could perhaps use this regex:
([^=:])=(?!=)
So this will match something that starts with anything not being = or : and then an = sign and no = after.
Then use a replace of $1==.
Tested here
And if you want a picture too...
:. So, how could it possibly know not to duplicate the last=character? (Hint: since JavaScript doesn't have lookbehind assertions this is pretty tricky... the easiest way would be a second pass to replace:==with:=.)([=]{1})is the same as([=])is the same as(=)(parenthesis are not even needed here). Don't make it overly complicated.