Need help with regex to remove square brackets and anything between

Question

How can I remove text from between square brackets and the brackets themselves?

For example, I need hello [quote="im sneaky"] world

to become

hello world

Here's what I'm trying to use, but it's not doing the trick

preg_replace("/[\[(.)\]]/", '', $str);

I just end up with hello quote="im sneaky" world

Thanks

Answer 1

I think you actually want parens for your outer brackets since it's a group. square brackets are a range of expressions. Not sure how to type it in SO.

/(\[.*\])/

Answer 2

[ and ] are special characters in a regex. They are used to list characters of a match. [a-z] matches any lowercase letter between "a" and "z". [03b] matches a "0", "3", or "b". To match the characters "[" and "]", you have to escape them with a preceding \.

Your code currently says "replace any character of [](). with an empty string" (reordered from the order in which you typed them for clarity).

---

Greedy match:

preg_replace('/\[.*\]/', '', $str); // Replace from one [ to the last ]

A greedy match could match multiple [s and ]s. That expression would take an example [of "sneaky"] text [with more "sneaky"] here and turn it into an example here.

---

Perl has a syntax for a non-greedy match (you most likely don't want to be greedy):

preg_replace('/\[.*?\]/', '', $str);

Non-greedy matches try to catch as few characters as possible. Using the same example: an example [of "sneaky"] text [with more "sneaky"] here becomes an example text here.

---

Only up to the first following ]:

preg_replace('/\[[^\]]*\]/', '', $str); // Find a [, look for non-] characters, and then a ]

This is more explicit, but harder to read. Using the same example text, you'd get the output of the non-greedy expression.

---

Note that none of these deal explicitly with white space. The spaces on either side of [ and ] will remain.

Also note that all of these can fail for malformed input. Multiple [s and ]s without matches could cause a surprising result.