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

Question

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 *

Answer 1

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".