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!");
}

Answer 1

Please don't do that... little Unicode BABY ANGEL s 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 AZ "standard" letters and "standard" 0-9 digits.)

Answer 2

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:

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

Answer 3

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:

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

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

which contains all unicode punctuation and symbols.

Answer 4

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_]");

Answer 5

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;)

Answer 6

Here is my regex variant of a special character:

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

(Java code)

Answer 7

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.

Answer 8

Try:

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

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

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

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

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

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

Answer 9

使用此正则表达式模式 ("^[a-zA-Z0-9]*$") 。它验证不包括特殊字符的字母数字字符串

Answer 10

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.

Answer 11

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 ).

Answer 12

尝试将其用于相同的事情 - StringUtils.isAlphanumeric(value)

Answer 13

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();

Answer 14

这是我的正则表达式，用于从任何字符串中删除所有特殊字符：

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

Answer 15

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("");

Answer 16

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

Answer 17

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)

Answer 18

(^\\W$)

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

Answer 19

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}]+"

Answer 20

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.

Answer 21

我使用下面的 reg 在字符串中查找特殊字符

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

Answer 22

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

An example:

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

Answer 23

You have to escape some symbols

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

OR

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