That depends on what you define as special characters, but try replaceAll(...):
String result = yourString.replaceAll("[-+.^:,]","");
Note that the ^ character must not be the first one in the list, since you'd then either have to escape it or it would mean "any but these characters".
Another note: the - character needs to be the first or last one on the list, otherwise you'd have to escape it or it would define a range ( e.g. :-, would mean "all characters in the range : to ,).
So, in order to keep consistency and not depend on character positioning, you might want to escape all those characters that have a special meaning in regular expressions (the following list is not complete, so be aware of other characters like (, {, $ etc.):
String result = yourString.replaceAll("[\\-\\+\\.\\^:,]","");
If you want to get rid of all punctuation and symbols, try this regex: \p{P}\p{S} (keep in mind that in Java strings you'd have to escape back slashes: "\\p{P}\\p{S}").
A third way could be something like this, if you can exactly define what should be left in your string:
String result = yourString.replaceAll("[^\\w\\s]","");
This means: replace everything that is not a word character (a-z in any case, 0-9 or _) or whitespace.
Edit: please note that there are a couple of other patterns that might prove helpful. However, I can't explain them all, so have a look at the reference section of regular-expressions.info.
Here's less restrictive alternative to the "define allowed characters" approach, as suggested by Ray:
String result = yourString.replaceAll("[^\\p{L}\\p{Z}]","");
The regex matches everything that is not a letter in any language and not a separator (whitespace, linebreak etc.). Note that you can't use [\P{L}\P{Z}] (upper case P means not having that property), since that would mean "everything that is not a letter or not whitespace", which almost matches everything, since letters are not whitespace and vice versa.
Additional information on Unicode
Some unicode characters seem to cause problems due to different possible ways to encode them (as a single code point or a combination of code points). Please refer to regular-expressions.info for more information.
Answer from Thomas on Stack OverflowThat depends on what you define as special characters, but try replaceAll(...):
String result = yourString.replaceAll("[-+.^:,]","");
Note that the ^ character must not be the first one in the list, since you'd then either have to escape it or it would mean "any but these characters".
Another note: the - character needs to be the first or last one on the list, otherwise you'd have to escape it or it would define a range ( e.g. :-, would mean "all characters in the range : to ,).
So, in order to keep consistency and not depend on character positioning, you might want to escape all those characters that have a special meaning in regular expressions (the following list is not complete, so be aware of other characters like (, {, $ etc.):
String result = yourString.replaceAll("[\\-\\+\\.\\^:,]","");
If you want to get rid of all punctuation and symbols, try this regex: \p{P}\p{S} (keep in mind that in Java strings you'd have to escape back slashes: "\\p{P}\\p{S}").
A third way could be something like this, if you can exactly define what should be left in your string:
String result = yourString.replaceAll("[^\\w\\s]","");
This means: replace everything that is not a word character (a-z in any case, 0-9 or _) or whitespace.
Edit: please note that there are a couple of other patterns that might prove helpful. However, I can't explain them all, so have a look at the reference section of regular-expressions.info.
Here's less restrictive alternative to the "define allowed characters" approach, as suggested by Ray:
String result = yourString.replaceAll("[^\\p{L}\\p{Z}]","");
The regex matches everything that is not a letter in any language and not a separator (whitespace, linebreak etc.). Note that you can't use [\P{L}\P{Z}] (upper case P means not having that property), since that would mean "everything that is not a letter or not whitespace", which almost matches everything, since letters are not whitespace and vice versa.
Additional information on Unicode
Some unicode characters seem to cause problems due to different possible ways to encode them (as a single code point or a combination of code points). Please refer to regular-expressions.info for more information.
This will replace all the characters except alphanumeric
replaceAll("[^A-Za-z0-9]","");
regex - remove all special characters in java - Stack Overflow
java - How to replace special characters in a string? - Stack Overflow
regex - how to remove special characters from string in a file using java - Stack Overflow
regex - Remove special characters in the string in java? - Stack Overflow
use [\\W+] or "[^a-zA-Z0-9]" as regex to match any special characters and also use String.replaceAll(regex, String) to replace the spl charecter with an empty string. remember as the first arg of String.replaceAll is a regex you have to escape it with a backslash to treat em as a literal charcter.
String c= "hjdg$h&jk8^i0ssh6";
Pattern pt = Pattern.compile("[^a-zA-Z0-9]");
Matcher match= pt.matcher(c);
while(match.find())
{
String s= match.group();
c=c.replaceAll("\\"+s, "");
}
System.out.println(c);
You can read the lines and replace all special characters safely this way.
Keep in mind that if you use \\W you will not replace underscores.
Scanner scan = new Scanner(System.in);
while(scan.hasNextLine()){
System.out.println(scan.nextLine().replaceAll("[^a-zA-Z0-9]", ""));
}
That depends on what you mean. If you just want to get rid of them, do this:
(Update: Apparently you want to keep digits as well, use the second lines in that case)
String alphaOnly = input.replaceAll("[^a-zA-Z]+","");
String alphaAndDigits = input.replaceAll("[^a-zA-Z0-9]+","");
or the equivalent:
String alphaOnly = input.replaceAll("[^\\p{Alpha}]+","");
String alphaAndDigits = input.replaceAll("[^\\p{Alpha}\\p{Digit}]+","");
(All of these can be significantly improved by precompiling the regex pattern and storing it in a constant)
Or, with Guava:
private static final CharMatcher ALNUM =
CharMatcher.inRange('a', 'z').or(CharMatcher.inRange('A', 'Z'))
.or(CharMatcher.inRange('0', '9')).precomputed();
// ...
String alphaAndDigits = ALNUM.retainFrom(input);
But if you want to turn accented characters into something sensible that's still ascii, look at these questions:
- Converting Java String to ASCII
- Java change áéőűú to aeouu
- ń ǹ ň ñ ṅ ņ ṇ ṋ ṉ ̈ ɲ ƞ ᶇ ɳ ȵ --> n or Remove diacritical marks from unicode chars
I am using this.
s = s.replaceAll("\\W", "");
It replace all special characters from string.
Here
\w : A word character, short for [a-zA-Z_0-9]
\W : A non-word character
This should work, "if you're looking to retain only ASCII (0-127) characters in your string":
String str = "This is sample CCNA program. it contains CCNP™";
str = str.replaceAll("[^\\x00-\\x7f]+", "");
Do you want to remove all special characters from your strings? If so:
String alphaOnly = input.replaceAll("[^a-zA-Z]+","");
String alphaAndDigits = input.replaceAll("[^a-zA-Z0-9]+","");
Please see Sean Patrick Floyd's answer to a possible duplicate question.
Use replaceAll("[^\\w\\s\\-_]", "");
What I did was add the underscore and hyphen to the regular expression. I added a \\ before the hyphen because it also serves for specifying ranges: a-z means all letters between a and z. Escaping it with \\ makes sure it is treated as an hyphen.
This might help:
replaceAll("[^a-zA-Z0-9_-]", "");
Replace any sequence of non-letters with a single white space:
str.replaceAll("[^a-zA-Z]+", " ")
You also might want to apply trim() after the replace.
If you want to support languages other than English, use "[^\\p{IsAlphabetic}]+" or "[^\\p{IsLetter}]+". See this question about the differences.
The OR operator (|) should work:
System.out.println(str.replaceAll("([^a-zA-Z]|\\s)+", " "));
Actually, the space doesn't have to be there at all:
System.out.println(str.replaceAll("[^a-zA-Z]+", " "));