How to match “anything up until this sequence of characters” in a regular expression?

Posted By: Anonymous

Take this regular expression: /^[^abc]/. This will match any single character at the beginning of a string, except a, b, or c.

If you add a * after it – /^[^abc]*/ – the regular expression will continue to add each subsequent character to the result, until it meets either an a, or b, or c.

For example, with the source string "qwerty qwerty whatever abc hello", the expression will match up to "qwerty qwerty wh".

But what if I wanted the matching string to be "qwerty qwerty whatever "

…In other words, how can I match everything up to (but not including) the exact sequence "abc"?

Solution

You didn’t specify which flavor of regex you’re using, but this will
work in any of the most popular ones that can be considered “complete”.

/.+?(?=abc)/

How it works

The .+? part is the un-greedy version of .+ (one or more of
anything). When we use .+, the engine will basically match everything.
Then, if there is something else in the regex it will go back in steps
trying to match the following part. This is the greedy behavior,
meaning as much as possible to satisfy.

When using .+?, instead of matching all at once and going back for
other conditions (if any), the engine will match the next characters by
step until the subsequent part of the regex is matched (again if any).
This is the un-greedy, meaning match the fewest possible to
satisfy.

/.+X/  ~ "abcXabcXabcX"        /.+/  ~ "abcXabcXabcX"
          ^^^^^^^^^^^^                  ^^^^^^^^^^^^

/.+?X/ ~ "abcXabcXabcX"        /.+?/ ~ "abcXabcXabcX"
          ^^^^                          ^

Following that we have (?={contents}), a zero width
assertion, a look around. This grouped construction matches its
contents, but does not count as characters matched (zero width). It
only returns if it is a match or not (assertion).

Thus, in other terms the regex /.+?(?=abc)/ means:

Match any characters as few as possible until a “abc” is found,
without counting the “abc”.

Answered By: Anonymous

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