Regex pattern including all special characters

Question

I want to write a simple regular expression to check if in given string exist any special character. My regex works but I don't know why it also includes all numbers, so when I put some number it returns an error.

My code:

//pattern to find if there is any special character in string
Pattern regex = Pattern.compile("[$&+,:;=?@#|'<>.-^*()%!]");
//matcher to find if there is any special character in string
Matcher matcher = regex.matcher(searchQuery.getSearchFor());

if(matcher.find())
{
    errors.rejectValue("searchFor", "wrong_pattern.SearchQuery.searchForSpecialCharacters","Special characters are not allowed!");
}

the dash in `[]` should be escaped, it has special meaning there. — MightyPork, Aug 05 '13 at 12:21
Exactly. It would be better to define all "non-special" charactes and make that negative. — NeplatnyUdaj, Aug 05 '13 at 12:22
yes maybe it would be wiser to assert the use of only those characters you want to allow. — d'alar'cop, Aug 05 '13 at 12:23
Can you please provide the solution String.replace("\"", """) — Lova Chittumuri, Mar 06 '19 at 15:36
To easily do live tests of your regex patterns, I would suggest this very useful tool : https://regexr.com/ — Grégory C, Dec 03 '20 at 13:41

xanatos · Accepted Answer · 2021-01-15T21:33:47.337

264

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.)

edited Jan 15 '21 at 21:33

answered Aug 05 '13 at 12:23

xanatos

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@AbdullahShoaib Clearly not :) You'll need to do a full list of what you consider "special" and/or what you consider "good". – xanatos Mar 09 '15 at 09:58
I notice many people use `[A-Za-z0-9]` to represent any letter or digit, both lowercase and upper, but is it not better to just do `[0-z]`? – Abraham Murciano Benzadon Jun 25 '17 at 16:09
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@AbrahamMurcianoBenzadon: The decimal digits, the upper case roman letters, and the lower case roman letters occupy three _disjoint_ ranges of character code space. – Solomon Slow Jun 25 '17 at 17:31
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@AbrahamMurcianoBenzadon You can see what James wrote in the handy screenshot of Character Map posted by Sina in another response: your regex would accept *:;<=>?@[\]^_`* (other than 0-9, a-z, A-Z) – xanatos Jun 26 '17 at 06:15
@AbdullahShoaib to handle non-english letters correctly, it's better to use `[^\p{Alnum}]` – arekolek Aug 07 '17 at 11:09
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Lets assume that we use [A-Za-z0-9] , but if we need to take care about Cyrillic or a few more alphabets, than how to do the regex? – Kaloyan Stamatov Oct 18 '18 at 15:14
@KaloyanStamatov See https://stackoverflow.com/questions/6256825/cyrillic-alphabet-validation – xanatos Oct 19 '18 at 06:46
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there are more languages than English ... – PeiSong Dec 15 '21 at 01:51

score 49 · Answer 2 · answered Aug 05 '13 at 12:22

49

You have a dash in the middle of the character class, which will mean a character range. Put the dash at the end of the class like so:

[$&+,:;=?@#|'<>.^*()%!-]

answered Aug 05 '13 at 12:22

Jerry

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Can you please provide the solution String.replace("\"", """); – Lova Chittumuri Mar 06 '19 at 15:36
@LovaChittumuri Please state your problem clearly. What are you trying to achieve, input, and desired output. – Jerry Mar 06 '19 at 16:17

score 33 · Answer 3 · answered Aug 05 '13 at 12:31

That's because your pattern contains a .-^ which is all characters between and including . and ^, which included digits and several other characters as shown below:

enter image description here

If by special characters, you mean punctuation and symbols use:

[\p{P}\p{S}]

which contains all unicode punctuation and symbols.

score 30 · Answer 4 · answered Aug 05 '13 at 12:27

SInce you don't have white-space and underscore in your character class I think following regex will be better for you:

Pattern regex = Pattern.compile("[^\w\s]");

Which means match everything other than [A-Za-z0-9\s_]

Unicode version:

Pattern regex = Pattern.compile("[^\p{L}\d\s_]");

score 17 · Answer 5 · answered May 14 '19 at 13:42

For people (like me) looking for an answer for special characters like Ä etc. just use this pattern:

Only text (or a space): "[A-Za-zÀ-ȕ ]"
Text and numbers: "[A-Za-zÀ-ȕ0-9 ]"
Text, numbers and some special chars: "[A-Za-zÀ-ȕ0-9(),-_., ]"

Regex just starts at the ascii index and checks if a character of the string is in within both indexes [startindex-endindex].

So you can add any range.

Eventually you can play around with a handy tool: https://regexr.com/

Good luck;)

Just the stuff I looked for. Thanks for your explanation. – azurecorn Aug 09 '22 at 22:15 — azurecorn, Aug 09 '22 at 22:15

score 13 · Answer 6 · answered Mar 02 '21 at 08:13

Use this to catch the common special characters excluding .-_.

/[!"`'#%&,:;<>=@{}~\$\(\)\*\+\/\\\?\[\]\^\|]+/

If you want to include .-_ as well, then use this:

/[-._!"`'#%&,:;<>=@{}~\$\(\)\*\+\/\\\?\[\]\^\|]+/

If you want to filter strings that are URL friendly and do not contain any special characters or spaces, then use this:

/^[^ !"`'#%&,:;<>=@{}~\$\(\)\*\+\/\\\?\[\]\^\|]+$/

When you use patterns like /[^A-Za-z0-9]/, then you will start catching special alphabets like those of other languages and some European accented alphabets (like é, í ).

score 11 · Answer 7 · answered Oct 17 '22 at 10:32

Shout out to Mohamed Yusuff 's solution!

We can match all 32 special characters using range.

[!-\/:-@[-`{-~]

1st Group

[!-\/]

Match ASCII code from 33 to 47:
!"#$%&'()*+,-./

-- 15 out of 32 characters matched

2nd Group

[:-@]

Match ASCII code from 58 to 64:
:;<=>?@

-- 7 out of 32 characters matched

3rd Group

[[-`]

Match ASCII code from 91 to 96:
[\]^_`

-- 6 out of 32 characters matched

4th Group

[{-~]

Match ASCII code from 123 to 126:
{|}~

-- 4 out of 32 characters matched

In total matched back all 32 chars (15+7+6+4)

Reference

Special Character table_Arranged

Extended ASCII table

score 9 · Answer 8 · answered Jun 14 '19 at 07:58

I have defined one pattern to look for any of the ASCII Special Characters ranging between 032 to 126 except the alpha-numeric. You may use something like the one below:

To find any Special Character:
[ -\/:-@\[-\`{-~]
To find minimum of 1 and maximum of any count:
(?=.*[ -\/:-@\[-\`{-~]{1,})

These patterns have Special Characters ranging between 032 to 047, 058 to 064, 091 to 096, and 123 to 126.

score 8 · Answer 9 · answered Mar 20 '17 at 14:52

8

Here is my regex variant of a special character:

String regExp = "^[^<>{}\"/|;:.,~!?@#$%^=&*\\]\\\\()\\[¿§«»ω⊙¤°℃℉€¥£¢¡®©0-9_+]*$";

(Java code)

answered Mar 20 '17 at 14:52

VKostenc

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score 6 · Answer 10 · answered Jul 04 '16 at 06:41

6

Use this regular expression pattern ("^[a-zA-Z0-9]*$") .It validates alphanumeric string excluding the special characters

answered Jul 04 '16 at 06:41

sam

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Serguei Fedorov · Answer 11 · 2016-02-17T18:40:52.993

5

If you only rely on ASCII characters, you can rely on using the hex ranges on the ASCII table. Here is a regex that will grab all special characters in the range of 33-47, 58-64, 91-96, 123-126

[\x21-\x2F\x3A-\x40\x5B-\x60\x7B-\x7E]

However you can think of special characters as not normal characters. If we take that approach, you can simply do this

^[A-Za-z0-9\s]+

Hower this will not catch _ ^ and probably others.

edited Feb 17 '16 at 18:40

answered Feb 03 '16 at 22:06

Serguei Fedorov

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I finally used `(?i)^([[a-z][^a-z0-9\\s\$\$\\[\\]\\{\\}\\\\^\\$\\|\\?\\*\\+\\.\\<\\>\\-\\=\\!\\_]]*)$` to match any character. – cdaiga Feb 17 '16 at 11:25
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**Never use `[A-z]` in a regex.** It matches all uppercase and lowercase ASCII letters as you would expect. but it also matches several punctuation characters whose code points lie between `Z` and `a`. Use `[A-Za-z]` instead, or `[a-z]` in case-insensitive mode. – Alan Moore Feb 17 '16 at 18:37
@AlanMoore, good to know! I'll make the change to the answer. – Serguei Fedorov Feb 17 '16 at 18:39
how about '.' dot character . It supporsed to match any character except new line. In python re.DOTALL matches all including newline. Check out the regular expression faq in the python tutorial https://docs.python.org/2/howto/regex.html – Dr Deo Sep 06 '18 at 15:01

score 4 · Answer 12 · edited Feb 17 '16 at 18:42

4

Try:

(?i)^([[a-z][^a-z0-9\\s\\(\\)\\[\\]\\{\\}\\\\^\\$\\|\\?\\*\\+\\.\\<\\>\\-\\=\\!\\_]]*)$

(?i)^(A)$: indicates that the regular expression A is case insensitive.

[a-z]: represents any alphabetic character from a to z.

[^a-z0-9\\s\$\$\\[\\]\\{\\}\\\\^\\$\\|\\?\\*\\+\\.\\<\\>\\-\\=\\!\\_]: represents any alphabetic character except a to z, digits, and special characters i.e. accented characters.

[[a-z][^a-z0-9\\s\$\$\\[\\]\\{\\}\\\\^\\$\\|\\?\\*\\+\\.\\<\\>\\-\\=\\!\\_]]: represents any alphabetic(accented or unaccented) character only characters.

*: one or more occurrence of the regex that precedes it.

edited Feb 17 '16 at 18:42

Alan Moore

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answered Feb 17 '16 at 11:26

cdaiga

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Inside a character class, none of those characters need to be escaped except `\` and `-`. Many of them never need to be escaped at all. "Better safe than sorry" is a fine philosophy, but readability is important, too. – Alan Moore Feb 17 '16 at 18:46
@AlanMoore (If you're the comic book author, extra credit), the "-" I've found can be left unescaped if left as the trailing character. `[a-z_=-]` matches a-z, _, =, or -. I place readability above all else in anything of the form "regex", but yeah, using the shortcuts can lead to woes eventually. – alife Sep 04 '22 at 19:35

score 3 · Answer 13 · edited Dec 07 '16 at 23:54

3

Try using this for the same things - StringUtils.isAlphanumeric(value)

edited Dec 07 '16 at 23:54

Buddy

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answered Dec 07 '16 at 23:52

Ash

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space/blank is also a special char if you use this method. Better to replace the space and tabs chars before calling this method. – Deepu Sahni Mar 21 '18 at 01:14

score 3 · Answer 14 · answered Sep 20 '18 at 12:18

3

We can achieve this using Pattern and Matcher as follows:

Pattern pattern = Pattern.compile("[^A-Za-z0-9 ]");
Matcher matcher = pattern.matcher(trString);
boolean hasSpecialChars = matcher.find();

answered Sep 20 '18 at 12:18

KayV

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score 3 · Answer 15 · edited Nov 12 '18 at 12:05

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Here is my regular expression, that I used for removing all the special characters from any string :

String regex = ("[ \\\\s@  [\\\"]\\\\[\\\\]\\\\\\\0-9|^{#%'*/<()>}:`;,!& .?_$+-]+")

edited Nov 12 '18 at 12:05

Jan Černý

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answered Nov 12 '18 at 10:56

Suchismita Barik

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Perfectly worked for me but small change is that to escape '\' (backslash) we should use "\\\\\\\\" – Naga Srinu Kapusetti Jul 09 '20 at 08:35

score 3 · Answer 16 · edited Jan 21 '19 at 02:56

3

Please use this.. it is simplest.

\p{Punct} Punctuation: One of !"#$%&'()*+,-./:;<=>?@[]^_`{|}~

https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html

    StringBuilder builder = new StringBuilder(checkstring);
    String regex = "\\p{Punct}"; //Special character : `~!@#$%^&*()-_+=\|}{]["';:/?.,><
    //change your all special characters to "" 
    Pattern  pattern = Pattern.compile(regex);
    Matcher matcher = pattern.matcher(builder.toString());
    checkstring=matcher.replaceAll("");

edited Jan 21 '19 at 02:56

Nick

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answered Jan 21 '19 at 02:36

Energy

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this answer matches the common list of special characters used by (american?) login systems. https://support.okta.com/help/s/article/What-special-characters-are-accepted-by-the-Okta-password?language=en_US – john k Feb 15 '23 at 20:42

score 2 · Answer 17 · edited Jan 08 '20 at 06:46

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You can use a negative match:

Pattern regex = Pattern.compile("([a-zA-Z0-9])*"); (For zero or more characters)

or

Pattern regex = Pattern.compile("([a-zA-Z0-9])+"); (For one or more characters)

edited Jan 08 '20 at 06:46

Tek Nath Acharya

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answered Jan 08 '20 at 05:50

Anirudh Sharma

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Question is not about allowing only roman numerals and english alphabets, what if user wanted to except japanese text, your solution is not going to work. – mightyWOZ Jan 08 '20 at 06:19

score 1 · Answer 18 · answered Apr 20 '20 at 15:37

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Try this. It works on C# it should work on java also. If you want to exclude spaces just add \s in there @"[^\p{L}\p{Nd}]+"

answered Apr 20 '20 at 15:37

danvasiloiu

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score 1 · Answer 19 · edited Oct 05 '20 at 12:07

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To find any number of special characters use the following regex pattern: ([^(A-Za-z0-9 )]{1,})

[^(A-Za-z0-9 )] this means any character except the alphabets, numbers, and space. {1,0} this means one or more characters of the previous block.

edited Oct 05 '20 at 12:07

Dharman

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answered Oct 05 '20 at 12:01

mlnbhargav

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It won't find `(` and `)`. – Wiktor Stribiżew Oct 19 '20 at 12:45
The `(` and `)` are problematic here. `[^A-Za-z_=]`, for example, allow for anything other than A-Z or a-z or _ or = to trigger. `[^[:alnum:][:punct:]]` similarly triggers on any character not alphanumeric nor punctuation. – alife Sep 04 '22 at 19:31

score 0 · Answer 20 · answered Aug 26 '16 at 09:18

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(^\W$)

^ - start of the string, \W - match any non-word character [^a-zA-Z0-9_], $ - end of the string

answered Aug 26 '16 at 09:18

Sareesh Krishnan

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score 0 · Answer 21 · edited Oct 05 '22 at 11:14

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A small addition to include all special characters like: ū and Ā:

An example:

Pattern regex = Pattern.compile("[A-Za-zÀ-ÖØ-öø-ū]");

edited Oct 05 '22 at 11:14

Alessandro Candeloro

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answered Sep 30 '22 at 09:32

Andries van Weeren

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score 0 · Answer 22 · answered Dec 26 '22 at 02:06

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You have to escape some symbols

/([!`\-\_.\"\'#%,:;<>=@{}~\$\(\)\*\+\/\\\?\[\]\^\|]+)/

OR

/([\!\"\#\$\%\&\'\(\)\*\+\,\-\.\/\:\;\<\>\=\?\@\[\]\{\}\\\\\^\_\`\~]+$)/

answered Dec 26 '22 at 02:06

Dexter

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score 0 · Answer 23 · answered Mar 08 '23 at 21:21

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To match the common Ascii special characters you can simply use this [!-\/].

So, it will be Pattern regex = Pattern.compile("[!-\/]");

answered Mar 08 '23 at 21:21

Mostafa Wael

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score 0 · Answer 24 · answered Apr 26 '23 at 11:59

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what about [ -~] This will match all ASCII characters from the space to tilde

answered Apr 26 '23 at 11:59

Wasim A.

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score -1 · Answer 25 · answered Jul 08 '22 at 02:30

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I use reg below for find special character in string

var reg = new RegExp("[`~!@#$%^&*()\\]\\[+={}/|:;\"\'<>,.?-_]");

answered Jul 08 '22 at 02:30

Nguyễn Huỳnh Hữu Tài

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