I Can't Include ' Symbol to Regular Expressions

I can't include ' symbol to Regular Expressions

I guess that what you really want is this:

"['0-9a-zA-Z]+"

Note that I have removed the ^ (text start) and $ (text end) characters because then your whole text would have to match.

I have merged the groups because otherwise you would not match the text as a whole word. You would get separate apostrophe and then the word.

I have changed the ‘ character into the proper ' character. The automatic conversion from the simple apostrophe is caused by iOS 11 Smart Punctuation. You can turn it off on an input using:

input.smartQuotesType = .no

See https://developer.apple.com/documentation/uikit/uitextinputtraits/2865931-smartquotestype

Regular expression to match a line that doesn't contain a word

The notion that regex doesn't support inverse matching is not entirely true. You can mimic this behavior by using negative look-arounds:

^((?!hede).)*$

Non-capturing variant:

^(?:(?!:hede).)*$

The regex above will match any string, or line without a line break, not containing the (sub)string 'hede'. As mentioned, this is not something regex is "good" at (or should do), but still, it is possible.

And if you need to match line break chars as well, use the DOT-ALL modifier (the trailing s in the following pattern):

/^((?!hede).)*$/s

or use it inline:

/(?s)^((?!hede).)*$/

(where the /.../ are the regex delimiters, i.e., not part of the pattern)

If the DOT-ALL modifier is not available, you can mimic the same behavior with the character class [\s\S]:

/^((?!hede)[\s\S])*$/

Explanation

A string is just a list of n characters. Before, and after each character, there's an empty string. So a list of n characters will have n+1 empty strings. Consider the string "ABhedeCD":

    ┌──┬───┬──┬───┬──┬───┬──┬───┬──┬───┬──┬───┬──┬───┬──┬───┬──┐
S = │e1│ A │e2│ B │e3│ h │e4│ e │e5│ d │e6│ e │e7│ C │e8│ D │e9│
    └──┴───┴──┴───┴──┴───┴──┴───┴──┴───┴──┴───┴──┴───┴──┴───┴──┘
    
index    0      1      2      3      4      5      6      7

where the e's are the empty strings. The regex (?!hede). looks ahead to see if there's no substring "hede" to be seen, and if that is the case (so something else is seen), then the . (dot) will match any character except a line break. Look-arounds are also called zero-width-assertions because they don't consume any characters. They only assert/validate something.

So, in my example, every empty string is first validated to see if there's no "hede" up ahead, before a character is consumed by the . (dot). The regex (?!hede). will do that only once, so it is wrapped in a group, and repeated zero or more times: ((?!hede).)*. Finally, the start- and end-of-input are anchored to make sure the entire input is consumed: ^((?!hede).)*$

As you can see, the input "ABhedeCD" will fail because on e3, the regex (?!hede) fails (there is "hede" up ahead!).

Regular expression to filter special characters not working

I want a function that returns true if a given string does NOT match.

You can negate the character set. (Note the ^ symbol within the square brackets). This will return true for strings that don't contain any of these special characters.

^[^`~!@#$%^&*()_+={}\[\]|\\:;“’<,>.?๐฿]*$

https://regex101.com/r/CqtqoK/1

Regex pattern including all special characters

Please don't do that... little Unicode BABY ANGELs like this one are dying! ◕◡◕ (← these are not images) (nor is the arrow!)

☺

And you are killing 20 years of DOS :-) (the last smiley is called WHITE SMILING FACE... Now it's at 263A... But in ancient times it was ALT-1)

and his friend

☻

BLACK SMILING FACE... Now it's at 263B... But in ancient times it was ALT-2

Try a negative match:

Pattern regex = Pattern.compile("[^A-Za-z0-9]");

(this will ok only A-Z "standard" letters and "standard" 0-9 digits.)

Regex - Does not contain certain Characters

^[^<>]+$

The caret in the character class ([^) means match anything but, so this means, beginning of string, then one or more of anything except < and >, then the end of the string.

Which regular expression operator means 'Don't' match this character?

You can use negated character classes to exclude certain characters: for example [^abcde] will match anything but a,b,c,d,e characters.

Instead of specifying all the characters literally, you can use shorthands inside character classes: [\w] (lowercase) will match any "word character" (letter, numbers and underscore), [\W] (uppercase) will match anything but word characters; similarly, [\d] will match the 0-9 digits while [\D] matches anything but the 0-9 digits, and so on.

If you use PHP you can take a look at the regex character classes documentation.

Regular Expressions and negating a whole character group

Use negative lookahead:

^(?!.*ab).*$

UPDATE: In the comments below, I stated that this approach is slower than the one given in Peter's answer. I've run some tests since then, and found that it's really slightly faster. However, the reason to prefer this technique over the other is not speed, but simplicity.

The other technique, described here as a tempered greedy token, is suitable for more complex problems, like matching delimited text where the delimiters consist of multiple characters (like HTML, as Luke commented below). For the problem described in the question, it's overkill.

For anyone who's interested, I tested with a large chunk of Lorem Ipsum text, counting the number of lines that don't contain the word "quo". These are the regexes I used:

(?m)^(?!.*\bquo\b).+$

(?m)^(?:(?!\bquo\b).)+$

Whether I search for matches in the whole text, or break it up into lines and match them individually, the anchored lookahead consistently outperforms the floating one.

---

Related Topics