System.out.println(Normalizer.normalize("àèé", Normalizer.Form.NFD).replaceAll("\\p{InCombiningDiacriticalMarks}+", ""));

为我工作。上面代码段的输出给出了“aee”，这是我想要的，但是

System.out.println(Normalizer.normalize("àèé", Normalizer.Form.NFD).replaceAll("[^\\p{ASCII}]", ""));

没有做任何替换。

@David Conrad solution is the fastest I tried using the Normalizer, but it does have a bug. It basically strips characters which are not accents, for example Chinese characters and other letters like æ, are all stripped. The characters that we want to strip are non spacing marks, characters which don't take up extra width in the final string. These zero width characters basically end up combined in some other character. If you can see them isolated as a character, for example like this `, my guess is that it's combined with the space character.

public static String flattenToAscii(String string) {
    char[] out = new char[string.length()];
    String norm = Normalizer.normalize(string, Normalizer.Form.NFD);

    int j = 0;
    for (int i = 0, n = norm.length(); i < n; ++i) {
        char c = norm.charAt(i);
        int type = Character.getType(c);

        //Log.d(TAG,""+c);
        //by Ricardo, modified the character check for accents, ref: http://stackoverflow.com/a/5697575/689223
        if (type != Character.NON_SPACING_MARK){
            out[j] = c;
            j++;
        }
    }
    //Log.d(TAG,"normalized string:"+norm+"/"+new String(out));
    return new String(out);
}

@virgo47的解决方案非常快，但很接近。接受的答案使用Normalizer和正则表达式。我想知道Normalizer和正则表达式占用了多少时间，因为删除所有非ascii字符可以在没有正则表达式的情况下完成:

import java.text.Normalizer;

public class Strip {
    public static String flattenToAscii(String string) {
        StringBuilder sb = new StringBuilder(string.length());
        string = Normalizer.normalize(string, Normalizer.Form.NFD);
        for (char c : string.toCharArray()) {
            if (c <= '\u007F') sb.append(c);
        }
        return sb.toString();
    }
}

小的额外加速可以通过写入char[]而不调用toCharArray()来获得，尽管我不确定代码清晰度的降低是否值得这样做:

public static String flattenToAscii(String string) {
    char[] out = new char[string.length()];
    string = Normalizer.normalize(string, Normalizer.Form.NFD);
    int j = 0;
    for (int i = 0, n = string.length(); i < n; ++i) {
        char c = string.charAt(i);
        if (c <= '\u007F') out[j++] = c;
    }
    return new String(out);
}

这种变化具有使用Normalizer的正确性和使用表的一些速度方面的优点。在我的机器上，这个答案比公认的答案快4倍，比@virgo47的答案慢6.6倍到7倍(公认的答案比我机器上的@virgo47的答案慢26倍)。

如果你没有库，使用regex和Normalizer的最好方法之一是:

    public String flattenToAscii(String s) {
                if(s == null || s.trim().length() == 0)
                        return "";
                return Normalizer.normalize(s, Normalizer.Form.NFD).replaceAll("[\u0300-\u036F]", "");
}

这比replaceAll("[^\p{ASCII}]"， ""))更有效，而且如果你不需要变音符符(就像你的例子一样)。

否则，您必须使用p{ASCII}模式。

的问候。

编辑:如果你不困于Java <6，速度不是关键，/或翻译表太有限，请使用David的回答。重点是使用Normalizer(在Java 6中引入)，而不是在循环中使用转换表。

虽然这不是“完美”的解决方案，但当你知道范围(在我们的例子中是latin1,2)时，它工作得很好，在Java 6之前工作(虽然不是一个真正的问题)，并且比大多数建议的版本快得多(可能是也可能不是一个问题):

    /**
 * Mirror of the unicode table from 00c0 to 017f without diacritics.
 */
private static final String tab00c0 = "AAAAAAACEEEEIIII" +
    "DNOOOOO\u00d7\u00d8UUUUYI\u00df" +
    "aaaaaaaceeeeiiii" +
    "\u00f0nooooo\u00f7\u00f8uuuuy\u00fey" +
    "AaAaAaCcCcCcCcDd" +
    "DdEeEeEeEeEeGgGg" +
    "GgGgHhHhIiIiIiIi" +
    "IiJjJjKkkLlLlLlL" +
    "lLlNnNnNnnNnOoOo" +
    "OoOoRrRrRrSsSsSs" +
    "SsTtTtTtUuUuUuUu" +
    "UuUuWwYyYZzZzZzF";

/**
 * Returns string without diacritics - 7 bit approximation.
 *
 * @param source string to convert
 * @return corresponding string without diacritics
 */
public static String removeDiacritic(String source) {
    char[] vysl = new char[source.length()];
    char one;
    for (int i = 0; i < source.length(); i++) {
        one = source.charAt(i);
        if (one >= '\u00c0' && one <= '\u017f') {
            one = tab00c0.charAt((int) one - '\u00c0');
        }
        vysl[i] = one;
    }
    return new String(vysl);
}

在我使用32位JDK的HW上进行的测试表明，这在~100ms内执行了从àèéľšťč89FDČ到aeelstc89FDC的100万次转换，而Normalizer方式使其在3.7s(慢37倍)。如果您的需求与性能有关，并且您知道输入范围，那么这可能适合您。

喜欢:-)

面对同样的问题，这里是使用Kotlin扩展的解决方案

   val String.stripAccents: String
    get() = Regex("\\p{InCombiningDiacriticalMarks}+")
        .replace(
            Normalizer.normalize(this, Normalizer.Form.NFD),
            ""
        )

使用

val textWithoutAccents = "some accented string".stripAccents