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 ended up with:
hello quote="im sneaky" world
If you want to remove the [ or the ] , use the expression: "\\[|\\]" . The two backslashes escape the square bracket and the pipe is an "or".
In other words, square brackets match exactly one character. (a-z0-9) will match two characters, the first is one of abcdefghijklmnopqrstuvwxyz , the second is one of 0123456789 , just as if the parenthesis weren't there. The () will allow you to read exactly which characters were matched.
Square brackets ( “[ ]” ): Any expression within square brackets [ ] is a character set; if any one of the characters matches the search string, the regex will pass the test return true.
replaceAll() to replace square brackets in a String To replace any character from a String, we generally use replaceAll() method of String.
[
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.
Just in case you are looking for a recursive removal:
$str = preg_replace("/\[([^\[\]]++|(?R))*+\]/", "", $str);
That will convert this:
This [text [more text]] is cool
to this:
This is cool
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