Regex.prototype.exec returns null on second iteration of the search [duplicate]

Answer 1

When calling exec on a global regex, it remembers where was the last match and starts the next one from there.

var rx = /a/g, matches = rx.exec('aa');
rx.lastIndex; //1

When you modify your input string, between exec calls, the lastIndex becomes desychronised.

While there's a few way to fix this, I would rather perform the replace in a single statement.

'QUESTION1 + QUESTION2 + QUESTION3'.replace(/QUESTION\d+/g, function ($1) { 
    return 'VALUE_FOR_' + $1; 
});

//"VALUE_FOR_QUESTION1 + VALUE_FOR_QUESTION2 + VALUE_FOR_QUESTION3"

Answer 2

Since you're using global flag in exec call regex engine remembers lastIndex which is is a read/write integer property of regular expressions that specifies the index at which to start the next match.

Place this to reset it:

re.lastIndex=0;

Just before next call to RegExp.exec

re.lastIndex = 0 allows you to reset the search index to perform a second search starting from the beginning. Without it, the search begins at the index of the last match, which would yield null in this case.

Read this Mozilla doc for more info on this. As per this manual:

This property is set only if the regular expression used the "g" flag to indicate a global search.