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The following will replace ASCII control characters (shorthand for [\x00-\x1F\x7F]):

my_string.replaceAll("\\p{Cntrl}", "?"); 

The following will replace all ASCII non-printable characters (shorthand for [\p{Graph}\x20]), including accented characters:

my_string.replaceAll("[^\\p{Print}]", "?"); 

However, neither works for Unicode strings. Does anyone has a good way to remove non-printable characters from a unicode string?

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  • 2
    Just as an addendum: the list of Unicode General Categories can be found in UAX #44 Commented Jun 1, 2011 at 10:32
  • 1
    Possible duplicate of Fastest way to strip all non-printable characters from a Java String Commented Oct 14, 2016 at 17:34
  • 1
    @Stewart: hi, have you looked at the question/answers besides the title?!? Commented Oct 14, 2016 at 18:09
  • 1
    @Stewart: that other question covers only the ascii subset of non-printable characters!!! Commented Oct 14, 2016 at 18:12

9 Answers 9

169
my_string.replaceAll("\\p{C}", "?"); 

See more about Unicode regex. java.util.regexPattern/String.replaceAll supports them.

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8 Comments

In java 1.6 at least, there is no support for them. download.oracle.com/javase/6/docs/api/java/util/regex/… ...I also tried your line out, and besides of missing a backslash, it plainly simply doesn't work.
This works: char c = 0xFFFA; String.valueOf(c).replaceAll("\\p{C}", "?"); also in the javadoc for pattern look in the Unicode support section, says it supports the categories
There are also invisible whitespace characters (like 0x0200B), which are part of \p{Zs} group. Unfortunately, this one includes also normal whitespaces. For those who are trying to filter an input string that shouldn't contain any spaces, the string s.replaceAll("[\\p{C}\\p{Z}]", "") will do the charm
This is what I was looking for, I was trying replaceAll("[^\\u0000-\\uFFFF]", "") but had no success
Attention: this solution presented here (with 150 upvotes) also removes line breaks, which you might don't want to be replaced.
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73

Op De Cirkel is mostly right. His suggestion will work in most cases:

myString.replaceAll("\\p{C}", "?"); 

But if myString might contain non-BMP codepoints then it's more complicated. \p{C} contains the surrogate codepoints of \p{Cs}. The replacement method above will corrupt non-BMP codepoints by sometimes replacing only half of the surrogate pair. It's possible this is a Java bug rather than intended behavior.

Using the other constituent categories is an option:

myString.replaceAll("[\\p{Cc}\\p{Cf}\\p{Co}\\p{Cn}]", "?"); 

However, solitary surrogate characters not part of a pair (each surrogate character has an assigned codepoint) will not be removed. A non-regex approach is the only way I know to properly handle \p{C}:

StringBuilder newString = new StringBuilder(myString.length()); for (int offset = 0; offset < myString.length();) { int codePoint = myString.codePointAt(offset); offset += Character.charCount(codePoint); // Replace invisible control characters and unused code points switch (Character.getType(codePoint)) { case Character.CONTROL: // \p{Cc} case Character.FORMAT: // \p{Cf} case Character.PRIVATE_USE: // \p{Co} case Character.SURROGATE: // \p{Cs} case Character.UNASSIGNED: // \p{Cn} newString.append('?'); break; default: newString.append(Character.toChars(codePoint)); break; } } 

Comments

12

methods below for your goal

public static String removeNonAscii(String str) { return str.replaceAll("[^\\x00-\\x7F]", ""); } public static String removeNonPrintable(String str) // All Control Char { return str.replaceAll("[\\p{C}]", ""); } public static String removeSomeControlChar(String str) // Some Control Char { return str.replaceAll("[\\p{Cntrl}\\p{Cc}\\p{Cf}\\p{Co}\\p{Cn}]", ""); } public static String removeFullControlChar(String str) { return removeNonPrintable(str).replaceAll("[\\r\\n\\t]", ""); } 

Comments

8

You may be interested in the Unicode categories "Other, Control" and possibly "Other, Format" (unfortunately the latter seems to contain both unprintable and printable characters).

In Java regular expressions you can check for them using \p{Cc} and \p{Cf} respectively.

1 Comment

Well, too bad java expressions don't have them, but at least I got the list right now... better than nothing. thanks
0

I have used this simple function for this:

private static Pattern pattern = Pattern.compile("[^ -~]"); private static String cleanTheText(String text) { Matcher matcher = pattern.matcher(text); if ( matcher.find() ) { text = text.replace(matcher.group(0), ""); } return text; } 

Hope this is useful.

1 Comment

What does that regular expression do?
0

Based on the answers by Op De Cirkel and noackjr, the following is what I do for general string cleaning: 1. trimming leading or trailing whitespaces, 2. dos2unix, 3. mac2unix, 4. removing all "invisible Unicode characters" except whitespaces:

myString.trim.replaceAll("\r\n", "\n").replaceAll("\r", "\n").replaceAll("[\\p{Cc}\\p{Cf}\\p{Co}\\p{Cn}&&[^\\s]]", "")

Tested with Scala REPL.

Comments

0

I propose it remove the non printable characters like below instead of replacing it

private String removeNonBMPCharacters(final String input) { StringBuilder strBuilder = new StringBuilder(); input.codePoints().forEach((i) -> { if (Character.isSupplementaryCodePoint(i)) { strBuilder.append("?"); } else { strBuilder.append(Character.toChars(i)); } }); return strBuilder.toString(); } 

Comments

0

Supported multilanguage

public static String cleanUnprintableChars(String text, boolean multilanguage) { String regex = multilanguage ? "[^\\x00-\\xFF]" : "[^\\x00-\\x7F]"; // strips off all non-ASCII characters text = text.replaceAll(regex, ""); // erases all the ASCII control characters text = text.replaceAll("[\\p{Cntrl}&&[^\r\n\t]]", ""); // removes non-printable characters from Unicode text = text.replaceAll("\\p{C}", ""); return text.trim(); } 

Comments

-4

I have redesigned the code for phone numbers +9 (987) 124124 Extract digits from a string in Java

 public static String stripNonDigitsV2( CharSequence input ) { if (input == null) return null; if ( input.length() == 0 ) return ""; char[] result = new char[input.length()]; int cursor = 0; CharBuffer buffer = CharBuffer.wrap( input ); int i=0; while ( i< buffer.length() ) { //buffer.hasRemaining() char chr = buffer.get(i); if (chr=='u'){ i=i+5; chr=buffer.get(i); } if ( chr > 39 && chr < 58 ) result[cursor++] = chr; i=i+1; } return new String( result, 0, cursor ); } 

Comments

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