Remove Special Characters For A Pattern Java

Question

I want to remove that characters from a String:

+ - ! ( ) { } [ ] ^ ~ : \

also I want to remove them:

/*
*/
&&
||

I mean that I will not remove & or | I will remove them if the second character follows the first one (/* */ && ||)

How can I do that efficiently and fast at Java?

Example:

a:b+c1|x||c*(?)

will be:

abc1|xc*?

Best advice? Try it and see what works. Don't worry about efficiency or speed. It isn't likely to make any difference; worry about it only when it does. — duffymo, Oct 08 '13 at 11:49
What did you try? You should take a look at `String#replaceAll()`. — Arnaud Denoyelle, Oct 08 '13 at 11:50
possible duplicate of [how to remove special characters from an string?](http://stackoverflow.com/questions/7552253/how-to-remove-special-characters-from-an-string) — Gilles 'SO- stop being evil', Jun 20 '14 at 21:45

score 1 · Answer 1 · answered Oct 08 '13 at 11:54

1

This can be done via a long, but actually very simple regex.

String aString = "a:b+c1|x||c*(?)";
String sanitizedString = aString.replaceAll("[+\\-!(){}\\[\\]^~:\\\\]|/\\*|\\*/|&&|\\|\\|", "");
System.out.println(sanitizedString);

answered Oct 08 '13 at 11:54

Alex Shesterov

26,085
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score 1 · Answer 2 · answered Oct 08 '13 at 11:57

1

I think that the java.lang.String.replaceAll(String regex, String replacement) is all you need:

http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#replaceAll(java.lang.String, java.lang.String).

answered Oct 08 '13 at 11:57

Albert Sadowski

642
5
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score 0 · Answer 3 · edited May 23 '17 at 12:29

Thomas wrote on How to remove special characters from a string?:

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]","");
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.

score 0 · Answer 4 · answered Oct 08 '13 at 12:20

there is two way to do that :

1)

ArrayList<String> arrayList = new ArrayList<String>();
    arrayList.add("+");
    arrayList.add("-");
    arrayList.add("||");
    arrayList.add("&&");
    arrayList.add("(");
    arrayList.add(")");
    arrayList.add("{");
    arrayList.add("}");
    arrayList.add("[");
    arrayList.add("]");
    arrayList.add("~");
    arrayList.add("^");
    arrayList.add(":");
    arrayList.add("/");
    arrayList.add("/*");
    arrayList.add("*/");
    String string = "a:b+c1|x||c*(?)";
    for (int i = 0; i < arrayList.size(); i++) {
        if (string.contains(arrayList.get(i)));
            string=string.replace(arrayList.get(i), "");
    }
    System.out.println(string);

2)

    String string = "a:b+c1|x||c*(?)";
    string = string.replaceAll("[+\\-!(){}\\[\\]^~:\\\\]|/\\*|\\*/|&&|\\|\\|", "");
    System.out.println(string);

Remove Special Characters For A Pattern Java

4 Answers4