Menu
In Java, suppose I have a String variable S, and I want to search for it inside of another String T, like so:
if (T.matches(S)) ...
(note: the above line was T.contains() until a few posts pointed out that that method does not use regexes. My bad.)
But now suppose S may have unsavory characters in it. For instance, let S = "[hi". The left square bracket is going to cause the regex to fail. Is there a function I can call to escape S so that this doesn't happen? In this particular case, I would like it to be transformed to "\[hi".
299
The backslash in a regular expression precedes a literal character. You also escape certain letters that represent common character classes, such as \w for a word character or \s for a space.
Characters can be escaped in Java Regex in two ways which are listed as follows which we will be discussing upto depth: Using \Q and \E for escaping. Using backslash(\\) for escaping.
However, we know that the backslash character is an escape character in Java String literals as well. Therefore, we need to double the backslash character when using it to precede any character (including the \ character itself).
In order to use a literal ^ at the start or a literal $ at the end of a regex, the character must be escaped. Some flavors only use ^ and $ as metacharacters when they are at the start or end of the regex respectively. In those flavors, no additional escaping is necessary. It's usually just best to escape them anyway.
String.contains does not use regex, so there isn't a problem in this case.
Where a regex is required, rather rejecting strings with regex special characters, use java.util.regex.Pattern.quote to escape them.
176
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
