java – Regex pattern including all special characters-ThrowExceptions

Exception or error:

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!");
}
How to solve:

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” digits.

Answer:

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:

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.

Answer:

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:

Here is my regex variant of a special character:

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

(Java code)

Answer:

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:

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.

Answer:

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

Answer:

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:

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:

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

Answer:

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

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

Answer:

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:

(^\W$)

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

Answer:

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:

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)

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