Regular expression syntax for "match nothing"?

Question

Hello.

I have a python template engine that heavily uses regexp. It's uses concatenation like

re.compile( regexp1 + "|" + regexp2 + "*|" + regexp3 + "+" )

I can modify individual substrings (regexp1, regexp2 etc). Is it any small and light expression that match nothing so i can use it inside a template where i don't want any matches? Unfortunately, it's sometimes '+' or '*' after regexp atom so i can't use empty string - "nothing to repeat" error will rise.

Answer 1

A null will find nothing, so "re.compile()" should theoretically work.

Correspondingly in Perl or VIM, // as "/" is the regex delimiter marker.

Answer 2

This shouldn't match anything:

re.compile('$^')

So if you replace regexp1, regexp2 and regexp3 with '$^' it will be impossible to find a match. Unless you are using the multi line mode.

---

After some tests I found a better solution

re.compile('a^')

It is impossible to match and will fail earlier than the previous solution. You can replace a with any other character and it will always be impossible to match

Answer 3

Maybe '.{0}'?

Answer 4

"()"

matches nothing and nothing only.

Answer 5

To match an empty string - even in multiline mode - you can use \A\Z, so:

re.compile('\A\Z|\A\Z*|\A\Z+')

The difference is that \A and \Z are start and end of string, whilst ^ and $ these can match start/end of lines, so $^|$^*|$^+ could potentially match a string containing newlines (if the flag is enabled).

And to fail to match anything (even an empty string), simply attempt to find content before the start of the string, e.g:

re.compile('.\A|.\A*|.\A+')

Since no characters can come before \A (by definition), this will always fail to match.

Answer 6

You could use
\z..
This is the absolute end of string, followed by two of anything

If + or * is tacked on the end this still works refusing to match anything

Answer 7

Or, use some list comprehension to remove the useless regexp entries and join to put them all together. Something like:

re.compile('|'.join([x for x in [regexp1, regexp2, ...] if x != None]))

Be sure to add some comments next to that line of code though :-)

Answer 8

(?!) should always fail to match. It is the zero-width negative look-ahead. If what is in the parentheses matches then the whole match fails. Given that it has nothing in it, it will fail the match for anything (including nothing).