如何从字符串中删除非 ASCII 字符？

我有字符串 "A função"、 "Ãugent"，其中需要用空字符串替换诸如 ç、 ã和 Ã之类的字符。

如何从字符串中删除这些非 ASCII 字符？

我试图使用下面的函数来实现它，但是它不能正常工作。一个问题是不需要的字符被空格字符替换。

public static String matchAndReplaceNonEnglishChar(String tmpsrcdta) {
String newsrcdta = null;
char array[] = Arrays.stringToCharArray(tmpsrcdta);
if (array == null)
return newsrcdta;


for (int i = 0; i < array.length; i++) {
int nVal = (int) array[i];
boolean bISO =
// Is character ISO control
Character.isISOControl(array[i]);
boolean bIgnorable =
// Is Ignorable identifier
Character.isIdentifierIgnorable(array[i]);
// Remove tab and other unwanted characters..
if (nVal == 9 || bISO || bIgnorable)
array[i] = ' ';
else if (nVal > 255)
array[i] = ' ';
}
newsrcdta = Arrays.charArrayToString(array);


return newsrcdta;
}
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最佳答案

This will search and replace all non ASCII letters:

String resultString = subjectString.replaceAll("[^\\x00-\\x7F]", "");
小开

This would be the Unicode solution

String s = "A função, Ãugent";
String r = s.replaceAll("\\P{InBasic_Latin}", "");

\p{InBasic_Latin} is the Unicode block that contains all letters in the Unicode range U+0000..U+007F (see regular-expression.info)

\P{InBasic_Latin} is the negated \p{InBasic_Latin}

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You can try something like this. Special Characters range for alphabets starts from 192, so you can avoid such characters in the result.

String name = "A função";


StringBuilder result = new StringBuilder();
for(char val : name.toCharArray()) {
if(val < 192) result.append(val);
}
System.out.println("Result "+result.toString());
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FailedDev's answer is good, but can be improved. If you want to preserve the ascii equivalents, you need to normalize first:

String subjectString = "öäü";
subjectString = Normalizer.normalize(subjectString, Normalizer.Form.NFD);
String resultString = subjectString.replaceAll("[^\\x00-\\x7F]", "");


=> will produce "oau"

That way, characters like "öäü" will be mapped to "oau", which at least preserves some information. Without normalization, the resulting String will be blank.

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Or you can use the function below for removing non-ascii character from the string. You will get know internal working.

private static String removeNonASCIIChar(String str) {
StringBuffer buff = new StringBuffer();
char chars[] = str.toCharArray();


for (int i = 0; i < chars.length; i++) {
if (0 < chars[i] && chars[i] < 127) {
buff.append(chars[i]);
}
}
return buff.toString();
}
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[Updated solution]

can be used with "Normalize" (Canonical decomposition) and "replaceAll", to replace it with the appropriate characters.

import java.text.Normalizer;
import java.text.Normalizer.Form;
import java.util.regex.Pattern;


public final class NormalizeUtils {


public static String normalizeASCII(final String string) {
final String normalize = Normalizer.normalize(string, Form.NFD);


return Pattern.compile("\\p{InCombiningDiacriticalMarks}+")
.matcher(normalize)
.replaceAll("");
} ...
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CharMatcher.retainFrom can be used, if you're using the Google Guava library:

String s = "A função";
String stripped = CharMatcher.ascii().retainFrom(s);
System.out.println(stripped); // Prints "A funo"
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The ASCII table contains 128 codes, with a total of 95 printable characters, of which only 52 characters are letters:

  • [0-127] ASCII codes
    • [32-126] printable characters
      • [48-57] digits [0-9]
      • [65-90] uppercase letters [A-Z]
      • [97-122] lowercase letters [a-z]

You can use String.codePoints method to get a stream over int values of characters of this string and filter out non-ASCII characters:

String str1 = "A função, Ãugent";


String str2 = str1.codePoints()
.filter(ch -> ch < 128)
.mapToObj(Character::toString)
.collect(Collectors.joining());


System.out.println(str2); // A funo, ugent

Or you can explicitly specify character ranges. For example filter out everything except letters:

String str3 = str1.codePoints()
.filter(ch -> ch >= 'A' && ch <= 'Z'
|| ch >= 'a' && ch <= 'z')
.mapToObj(Character::toString)
.collect(Collectors.joining());


System.out.println(str3); // Afunougent

See also: How do I not take Special Characters in my Password Validation (without Regex)?

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String s = "A função";
String stripped = s.replaceAll("\\P{ASCII}", "");
System.out.println(stripped); // Prints "A funo"

or

private static final Pattern NON_ASCII_PATTERN = Pattern.compile("\\P{ASCII}");


public static String matchAndReplaceNonEnglishChar(String tmpsrcdta) {
return NON_ASCII_PATTERN.matcher(s).replaceAll("");
}


public static void main(String[] args) {
matchAndReplaceNonEnglishChar("A função"); // Prints "A funo"
}

Explanation

The method String.replaceAll(String regex, String replacement) replaces all instances of a given regular expression (regex) with a given replacement string.

Replaces each substring of this string that matches the given regular expression with the given replacement.

Java has the "\p{ASCII}" regular expression construct which matches any ASCII character, and its inverse, "\P{ASCII}", which matches any non-ASCII character. The matched characters can then be replaced with the empty string, effectively removing them from the resulting string.

String s = "A função";
String stripped = s.replaceAll("\\P{ASCII}", "");
System.out.println(stripped); // Prints "A funo"

The full list of valid regex constructs is documented in the Pattern class.

Note: If you are going to be calling this pattern multiple times within a run, it will be more efficient to use a compiled Pattern directly, rather than String.replaceAll. This way the pattern is compiled only once and reused, rather than each time replaceAll is called:

public class AsciiStripper {
private static final Pattern NON_ASCII_PATTERN = Pattern.compile("\\P{ASCII}");
    

public static String stripNonAscii(String s) {
return NON_ASCII_PATTERN.matcher(s).replaceAll("");
}
}
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An easily-readable, ascii-printable, streams solution:

String result = str.chars()
.filter(c -> isAsciiPrintable((char) c))
.mapToObj(c -> String.valueOf((char) c))
.collect(Collectors.joining());


private static boolean isAsciiPrintable(char ch) {
return ch >= 32 && ch < 127;
}

To convert to "_": .map(c -> isAsciiPrintable((char) c) ? c : '_')

32 to 127 is equivalent to the regex [^\\x20-\\x7E] (from comment on the regex solution)

Source for isAsciiPrintable: http://www.java2s.com/Code/Java/Data-Type/ChecksifthestringcontainsonlyASCIIprintablecharacters.htm


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