正则表达式可以在字符串的倒数第二次出现之前进行匹配

Question

If I have the following string:

hello/everyone/good/bye/world

I want to match everything up through the second to last forward slash:

hello/everyone/good/

But the number of slashes may vary.

What would the regex be to do this? Been googling around to no avail.

Answer 1

Instead of a regular expression, use split and take advantage of "slicing".

t = "hello/everyone/good/bye/world"
ts = t.split("/")
print "/".join( ts[0:-2] ) + "/"

This prints

hello/everyone/good/

This answer is for python2 (you did not specify which language), there are equivalent commands in some other languages. For example, java has a String.split() method (to complicate things, it requires a regular expression). Consult the documentation for the language you are using.

Answer 2

Maybe try this https://regex101.com/r/bH2vI0/1 :

((\w+\/)+(?!\w+\/\w+))

Not sure if sublime allows lookaheads though, but basically just match every word + / except for last two. [Az] might be better if you don't want digits and _