Regular expression pattern for preg_replace the last slash and all the following characters

Question

I can't figure out how to write the right pattern for my preg_replace pattern.
My situation is this: I've got an URL, let's suppose this:

$url = "first/second/third/fourth"

Now, I need to remove only the last "/" and all the characters after that. So my result should become like this:

first/second/third

For this moment I solved this way:

$result = substr($url, 0, (strrpos($url, "/")));

but I know there should be the right patter to be writted into my preg_replace.
Any suggestion?

Ry- · Accepted Answer · 2013-04-20 16:10:28Z

4

Use what you have; it's simple and efficient. Less parentheses could make it look less complicated:

$result = substr($url, 0, strrpos($url, '/'));

The regular expression would look like this:

$result = preg_replace('#/[^/]*$#', '', $url);

Just as long, slightly more confusing.

answered Apr 20, 2013 at 16:10

Ry-♦

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5 Comments

Souleiman Over a year ago

Interesting how they are exactly the same number of characters :) But which would be more efficient?

Ry- Over a year ago

@Soule: I imagine the first one (the regular expression would have to be compiled, and then probably in to something that's less efficient - on the other hand, though, it is PHP), but it shouldn't be important.

Souleiman Over a year ago

"Premature optimizations are the root of all evil" ;)

ilSavo Over a year ago

Yeah! It work! I'm just a little confused about the use of "#"...but it worked! Thanks...

Ry- Over a year ago

@ilSavo: It's a delimiter! Normally, I'd use a slash, but there's a slash in the expression, so I picked something else. Another common delimiter is the tilde.

Thomas · Accepted Answer · 2013-04-20 16:09:50Z

0

Match a slash, followed by zero or more characters that are not slashes, followed by the end of the string:

preg_replace("/[^/]*$", "", $url)

Or, using a non-greedy match (*?), this should also work:

preg_replace("/.*?$", "", $url)

answered Apr 20, 2013 at 16:09

Thomas

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