Java：将字符串“\uFFFF”转换为字符
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Java: Convert String "\uFFFF" into char
提问by Dima
Is there a standard method to convert a string like "\uFFFF" into character meaning that the string of six character contains a presentation of one unicode character?
是否有一种标准方法可以将像“\uFFFF”这样的字符串转换为字符，这意味着六个字符的字符串包含一个 unicode 字符的表示？
采纳答案by Bozho
char c = "\uFFFF".toCharArray()[0];
The value is directly interpreted as the desired string, and the whole sequence is realized as a single character.
该值被直接解释为所需的字符串，整个序列被实现为单个字符。
Another way, if you are going to hard-code the value:
另一种方式，如果您要对值进行硬编码：
char c = '\uFFFF';
Note that \uFFFFdoesn't seem to be a proper unicode character, but try with \u041ffor example.
请注意，这\uFFFF似乎不是一个合适的 unicode 字符，但请尝试使用\u041f例如。
回答by SyntaxT3rr0r
The backslash is escaped here (so you see two of them but the sString is really only 6 characters long). If you're surethat you have exactly "\u" at the beginning of your string, simply skip them and converter the hexadecimal value:
反斜杠在此处转义（因此您可以看到其中两个，但s字符串实际上只有 6 个字符长）。如果您确定字符串的开头正好有“\u”，只需跳过它们并转换十六进制值：
String s = "\u20ac";
char c = (char) Integer.parseInt( s.substring(2), 16 );
After that cshall contain the euro symbol as expected.
之后c应包含预期的欧元符号。
回答by Yoni
String charInUnicode = "\u0041"; // ascii code 65, the letter 'A'
Integer code = Integer.parseInt(charInUnicode.substring(2), 16); // the integer 65 in base 10
char ch = Character.toChars(code)[0]; // the letter 'A'
回答by stoivane
If you are parsing input with Java style escaped characters you might want to have a look at StringEscapeUtils.unescapeJava. It handles Unicode escapes as well as newlines, tabs etc.
如果您使用 Java 风格的转义字符解析输入，您可能需要查看StringEscapeUtils.unescapeJava。它处理 Unicode 转义以及换行符、制表符等。
String s = StringEscapeUtils.unescapeJava("\u20ac\n"); // s contains the euro symbol followed by newline
