我正在使用Java处理一些Java源代码。我正在提取字符串文字并将它们提供给一个带字符串的函数。问题是我需要将未转义的String版本传递给函数（即这意味着将\n转换为换行符，将\\转换为单个\等。< / p>
Java API中是否有一个函数可以执行此操作？如果没有，我可以从某个库获得这样的功能吗？显然，Java编译器必须进行这种转换。
如果有人想知道，我正试图在反编译的混淆Java文件中取消混淆字符串文字。
答案 0 :(得分：94)
这里给出的org.apache.commons.lang.StringEscapeUtils.unescapeJava()作为另一个答案实际上是没什么帮助的。
\0为空。java.util.regex.Pattern.compile()及其使用的所有转义，包括\a，\e，尤其是\cX。 charAt接口而不是codePoint接口，从而揭示了Java char的妄想。保证保持Unicode字符。不是。他们只是逃避这一点，因为没有UTF-16代理人会找他们正在寻找的任何东西。 我写了一个字符串unescaper，解决了OP的问题，没有Apache代码的所有烦恼。
/*
*
* unescape_perl_string()
*
* Tom Christiansen <tchrist@perl.com>
* Sun Nov 28 12:55:24 MST 2010
*
* It's completely ridiculous that there's no standard
* unescape_java_string function. Since I have to do the
* damn thing myself, I might as well make it halfway useful
* by supporting things Java was too stupid to consider in
* strings:
*
* => "?" items are additions to Java string escapes
* but normal in Java regexes
*
* => "!" items are also additions to Java regex escapes
*
* Standard singletons: ?\a ?\e \f \n \r \t
*
* NB: \b is unsupported as backspace so it can pass-through
* to the regex translator untouched; I refuse to make anyone
* doublebackslash it as doublebackslashing is a Java idiocy
* I desperately wish would die out. There are plenty of
* other ways to write it:
*
* \cH, \12, \012, \x08 \x{8}, \u0008, \U00000008
*
* Octal escapes: \0 \0N \0NN \N \NN \NNN
* Can range up to !\777 not \377
*
* TODO: add !\o{NNNNN}
* last Unicode is 4177777
* maxint is 37777777777
*
* Control chars: ?\cX
* Means: ord(X) ^ ord('@')
*
* Old hex escapes: \xXX
* unbraced must be 2 xdigits
*
* Perl hex escapes: !\x{XXX} braced may be 1-8 xdigits
* NB: proper Unicode never needs more than 6, as highest
* valid codepoint is 0x10FFFF, not maxint 0xFFFFFFFF
*
* Lame Java escape: \[IDIOT JAVA PREPROCESSOR]uXXXX must be
* exactly 4 xdigits;
*
* I can't write XXXX in this comment where it belongs
* because the damned Java Preprocessor can't mind its
* own business. Idiots!
*
* Lame Python escape: !\UXXXXXXXX must be exactly 8 xdigits
*
* TODO: Perl translation escapes: \Q \U \L \E \[IDIOT JAVA PREPROCESSOR]u \l
* These are not so important to cover if you're passing the
* result to Pattern.compile(), since it handles them for you
* further downstream. Hm, what about \[IDIOT JAVA PREPROCESSOR]u?
*
*/
public final static
String unescape_perl_string(String oldstr) {
/*
* In contrast to fixing Java's broken regex charclasses,
* this one need be no bigger, as unescaping shrinks the string
* here, where in the other one, it grows it.
*/
StringBuffer newstr = new StringBuffer(oldstr.length());
boolean saw_backslash = false;
for (int i = 0; i < oldstr.length(); i++) {
int cp = oldstr.codePointAt(i);
if (oldstr.codePointAt(i) > Character.MAX_VALUE) {
i++; /****WE HATES UTF-16! WE HATES IT FOREVERSES!!!****/
}
if (!saw_backslash) {
if (cp == '\\') {
saw_backslash = true;
} else {
newstr.append(Character.toChars(cp));
}
continue; /* switch */
}
if (cp == '\\') {
saw_backslash = false;
newstr.append('\\');
newstr.append('\\');
continue; /* switch */
}
switch (cp) {
case 'r': newstr.append('\r');
break; /* switch */
case 'n': newstr.append('\n');
break; /* switch */
case 'f': newstr.append('\f');
break; /* switch */
/* PASS a \b THROUGH!! */
case 'b': newstr.append("\\b");
break; /* switch */
case 't': newstr.append('\t');
break; /* switch */
case 'a': newstr.append('\007');
break; /* switch */
case 'e': newstr.append('\033');
break; /* switch */
/*
* A "control" character is what you get when you xor its
* codepoint with '@'==64. This only makes sense for ASCII,
* and may not yield a "control" character after all.
*
* Strange but true: "\c{" is ";", "\c}" is "=", etc.
*/
case 'c': {
if (++i == oldstr.length()) { die("trailing \\c"); }
cp = oldstr.codePointAt(i);
/*
* don't need to grok surrogates, as next line blows them up
*/
if (cp > 0x7f) { die("expected ASCII after \\c"); }
newstr.append(Character.toChars(cp ^ 64));
break; /* switch */
}
case '8':
case '9': die("illegal octal digit");
/* NOTREACHED */
/*
* may be 0 to 2 octal digits following this one
* so back up one for fallthrough to next case;
* unread this digit and fall through to next case.
*/
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7': --i;
/* FALLTHROUGH */
/*
* Can have 0, 1, or 2 octal digits following a 0
* this permits larger values than octal 377, up to
* octal 777.
*/
case '0': {
if (i+1 == oldstr.length()) {
/* found \0 at end of string */
newstr.append(Character.toChars(0));
break; /* switch */
}
i++;
int digits = 0;
int j;
for (j = 0; j <= 2; j++) {
if (i+j == oldstr.length()) {
break; /* for */
}
/* safe because will unread surrogate */
int ch = oldstr.charAt(i+j);
if (ch < '0' || ch > '7') {
break; /* for */
}
digits++;
}
if (digits == 0) {
--i;
newstr.append('\0');
break; /* switch */
}
int value = 0;
try {
value = Integer.parseInt(
oldstr.substring(i, i+digits), 8);
} catch (NumberFormatException nfe) {
die("invalid octal value for \\0 escape");
}
newstr.append(Character.toChars(value));
i += digits-1;
break; /* switch */
} /* end case '0' */
case 'x': {
if (i+2 > oldstr.length()) {
die("string too short for \\x escape");
}
i++;
boolean saw_brace = false;
if (oldstr.charAt(i) == '{') {
/* ^^^^^^ ok to ignore surrogates here */
i++;
saw_brace = true;
}
int j;
for (j = 0; j < 8; j++) {
if (!saw_brace && j == 2) {
break; /* for */
}
/*
* ASCII test also catches surrogates
*/
int ch = oldstr.charAt(i+j);
if (ch > 127) {
die("illegal non-ASCII hex digit in \\x escape");
}
if (saw_brace && ch == '}') { break; /* for */ }
if (! ( (ch >= '0' && ch <= '9')
||
(ch >= 'a' && ch <= 'f')
||
(ch >= 'A' && ch <= 'F')
)
)
{
die(String.format(
"illegal hex digit #%d '%c' in \\x", ch, ch));
}
}
if (j == 0) { die("empty braces in \\x{} escape"); }
int value = 0;
try {
value = Integer.parseInt(oldstr.substring(i, i+j), 16);
} catch (NumberFormatException nfe) {
die("invalid hex value for \\x escape");
}
newstr.append(Character.toChars(value));
if (saw_brace) { j++; }
i += j-1;
break; /* switch */
}
case 'u': {
if (i+4 > oldstr.length()) {
die("string too short for \\u escape");
}
i++;
int j;
for (j = 0; j < 4; j++) {
/* this also handles the surrogate issue */
if (oldstr.charAt(i+j) > 127) {
die("illegal non-ASCII hex digit in \\u escape");
}
}
int value = 0;
try {
value = Integer.parseInt( oldstr.substring(i, i+j), 16);
} catch (NumberFormatException nfe) {
die("invalid hex value for \\u escape");
}
newstr.append(Character.toChars(value));
i += j-1;
break; /* switch */
}
case 'U': {
if (i+8 > oldstr.length()) {
die("string too short for \\U escape");
}
i++;
int j;
for (j = 0; j < 8; j++) {
/* this also handles the surrogate issue */
if (oldstr.charAt(i+j) > 127) {
die("illegal non-ASCII hex digit in \\U escape");
}
}
int value = 0;
try {
value = Integer.parseInt(oldstr.substring(i, i+j), 16);
} catch (NumberFormatException nfe) {
die("invalid hex value for \\U escape");
}
newstr.append(Character.toChars(value));
i += j-1;
break; /* switch */
}
default: newstr.append('\\');
newstr.append(Character.toChars(cp));
/*
* say(String.format(
* "DEFAULT unrecognized escape %c passed through",
* cp));
*/
break; /* switch */
}
saw_backslash = false;
}
/* weird to leave one at the end */
if (saw_backslash) {
newstr.append('\\');
}
return newstr.toString();
}
/*
* Return a string "U+XX.XXX.XXXX" etc, where each XX set is the
* xdigits of the logical Unicode code point. No bloody brain-damaged
* UTF-16 surrogate crap, just true logical characters.
*/
public final static
String uniplus(String s) {
if (s.length() == 0) {
return "";
}
/* This is just the minimum; sb will grow as needed. */
StringBuffer sb = new StringBuffer(2 + 3 * s.length());
sb.append("U+");
for (int i = 0; i < s.length(); i++) {
sb.append(String.format("%X", s.codePointAt(i)));
if (s.codePointAt(i) > Character.MAX_VALUE) {
i++; /****WE HATES UTF-16! WE HATES IT FOREVERSES!!!****/
}
if (i+1 < s.length()) {
sb.append(".");
}
}
return sb.toString();
}
private static final
void die(String foa) {
throw new IllegalArgumentException(foa);
}
private static final
void say(String what) {
System.out.println(what);
}
如果它对其他人有帮助，那么欢迎你这样做 - 没有任何附加条件。如果你改进它，我希望你能把你的增强功能邮寄给我，但你当然不需要。
答案 1 :(得分：46)
您可以使用String unescapeJava(String)中的StringEscapeUtils Apache Commons Lang方法。
以下是一个示例代码段：
String in = "a\\tb\\n\\\"c\\\"";
System.out.println(in);
// a\tb\n\"c\"
String out = StringEscapeUtils.unescapeJava(in);
System.out.println(out);
// a b
// "c"
实用程序类具有对Java，Java Script，HTML，XML和SQL进行转义和转义字符串的方法。它还具有直接写入java.io.Writer的重载。
看起来StringEscapeUtils处理带有一个u的Unicode转义，但不处理八进制转义，或带有无关u s的Unicode转义。
/* Unicode escape test #1: PASS */
System.out.println(
"\u0030"
); // 0
System.out.println(
StringEscapeUtils.unescapeJava("\\u0030")
); // 0
System.out.println(
"\u0030".equals(StringEscapeUtils.unescapeJava("\\u0030"))
); // true
/* Octal escape test: FAIL */
System.out.println(
"\45"
); // %
System.out.println(
StringEscapeUtils.unescapeJava("\\45")
); // 45
System.out.println(
"\45".equals(StringEscapeUtils.unescapeJava("\\45"))
); // false
/* Unicode escape test #2: FAIL */
System.out.println(
"\uu0030"
); // 0
System.out.println(
StringEscapeUtils.unescapeJava("\\uu0030")
); // throws NestableRuntimeException:
// Unable to parse unicode value: u003
JLS引用：
提供八进制转义以与C兼容，但只能表达
\u0000到\u00FF的Unicode值，因此通常首选Unicode转义。
如果您的字符串可以包含八进制转义符，您可能希望首先将它们转换为Unicode转义符，或者使用其他方法。
无关的u也记录如下：
Java编程语言指定了一种将用Unicode编写的程序转换为ASCII的标准方法，该程序将程序更改为可由基于ASCII的工具处理的形式。转换涉及通过添加额外的
u将程序源文本中的任何Unicode转义转换为ASCII - 例如，\uxxxx变为\uuxxxx- 同时转换非ASCII字符源文本到Unicode转义，每个包含一个u。这个转换版本对于Java编程语言的编译器同样可以接受，并且代表完全相同的程序。稍后可以通过将存在多个
u的每个转义序列转换为一个较少u的Unicode字符序列，从而转换每个转义序列，从而可以从此ASCII格式恢复确切的Unicode源。单个u到相应的单个Unicode字符。
如果您的字符串可以包含带有无关u的Unicode转义符，那么您可能还需要在使用StringEscapeUtils之前对其进行预处理。
或者，您可以尝试从头开始编写自己的Java字符串文字unescaper，确保遵循确切的JLS规范。
答案 2 :(得分：11)
遇到类似的问题，对提出的解决方案并不满意，并自己实施了这个。
也可作为Github上的要点：
/**
* Unescapes a string that contains standard Java escape sequences.
* <ul>
* <li><strong>\b \f \n \r \t \" \'</strong> :
* BS, FF, NL, CR, TAB, double and single quote.</li>
* <li><strong>\X \XX \XXX</strong> : Octal character
* specification (0 - 377, 0x00 - 0xFF).</li>
* <li><strong>\uXXXX</strong> : Hexadecimal based Unicode character.</li>
* </ul>
*
* @param st
* A string optionally containing standard java escape sequences.
* @return The translated string.
*/
public String unescapeJavaString(String st) {
StringBuilder sb = new StringBuilder(st.length());
for (int i = 0; i < st.length(); i++) {
char ch = st.charAt(i);
if (ch == '\\') {
char nextChar = (i == st.length() - 1) ? '\\' : st
.charAt(i + 1);
// Octal escape?
if (nextChar >= '0' && nextChar <= '7') {
String code = "" + nextChar;
i++;
if ((i < st.length() - 1) && st.charAt(i + 1) >= '0'
&& st.charAt(i + 1) <= '7') {
code += st.charAt(i + 1);
i++;
if ((i < st.length() - 1) && st.charAt(i + 1) >= '0'
&& st.charAt(i + 1) <= '7') {
code += st.charAt(i + 1);
i++;
}
}
sb.append((char) Integer.parseInt(code, 8));
continue;
}
switch (nextChar) {
case '\\':
ch = '\\';
break;
case 'b':
ch = '\b';
break;
case 'f':
ch = '\f';
break;
case 'n':
ch = '\n';
break;
case 'r':
ch = '\r';
break;
case 't':
ch = '\t';
break;
case '\"':
ch = '\"';
break;
case '\'':
ch = '\'';
break;
// Hex Unicode: u????
case 'u':
if (i >= st.length() - 5) {
ch = 'u';
break;
}
int code = Integer.parseInt(
"" + st.charAt(i + 2) + st.charAt(i + 3)
+ st.charAt(i + 4) + st.charAt(i + 5), 16);
sb.append(Character.toChars(code));
i += 5;
continue;
}
i++;
}
sb.append(ch);
}
return sb.toString();
}
答案 3 :(得分：10)
答案 4 :(得分：8)
我知道这个问题很老，但是我想要一个不涉及JRE6以外的库的解决方案（即Apache Commons是不可接受的），我想出了一个使用内置{{3的简单解决方案的解决方案}}：
import java.io.*;
// ...
String literal = "\"Has \\\"\\\\\\\t\\\" & isn\\\'t \\\r\\\n on 1 line.\"";
StreamTokenizer parser = new StreamTokenizer(new StringReader(literal));
String result;
try {
parser.nextToken();
if (parser.ttype == '"') {
result = parser.sval;
}
else {
result = "ERROR!";
}
}
catch (IOException e) {
result = e.toString();
}
System.out.println(result);
输出：
Has "\ " & isn't
on 1 line.
答案 5 :(得分：6)
我有点迟到了，但我想我会提供我的解决方案，因为我需要相同的功能。我决定使用Java Compiler API，这会让它变慢，但会使结果准确。基本上我活着创建一个类然后返回结果。这是方法：
public static String[] unescapeJavaStrings(String... escaped) {
//class name
final String className = "Temp" + System.currentTimeMillis();
//build the source
final StringBuilder source = new StringBuilder(100 + escaped.length * 20).
append("public class ").append(className).append("{\n").
append("\tpublic static String[] getStrings() {\n").
append("\t\treturn new String[] {\n");
for (String string : escaped) {
source.append("\t\t\t\"");
//we escape non-escaped quotes here to be safe
// (but something like \\" will fail, oh well for now)
for (int i = 0; i < string.length(); i++) {
char chr = string.charAt(i);
if (chr == '"' && i > 0 && string.charAt(i - 1) != '\\') {
source.append('\\');
}
source.append(chr);
}
source.append("\",\n");
}
source.append("\t\t};\n\t}\n}\n");
//obtain compiler
final JavaCompiler compiler = ToolProvider.getSystemJavaCompiler();
//local stream for output
final ByteArrayOutputStream out = new ByteArrayOutputStream();
//local stream for error
ByteArrayOutputStream err = new ByteArrayOutputStream();
//source file
JavaFileObject sourceFile = new SimpleJavaFileObject(
URI.create("string:///" + className + Kind.SOURCE.extension), Kind.SOURCE) {
@Override
public CharSequence getCharContent(boolean ignoreEncodingErrors) throws IOException {
return source;
}
};
//target file
final JavaFileObject targetFile = new SimpleJavaFileObject(
URI.create("string:///" + className + Kind.CLASS.extension), Kind.CLASS) {
@Override
public OutputStream openOutputStream() throws IOException {
return out;
}
};
//file manager proxy, with most parts delegated to the standard one
JavaFileManager fileManagerProxy = (JavaFileManager) Proxy.newProxyInstance(
StringUtils.class.getClassLoader(), new Class[] { JavaFileManager.class },
new InvocationHandler() {
//standard file manager to delegate to
private final JavaFileManager standard =
compiler.getStandardFileManager(null, null, null);
@Override
public Object invoke(Object proxy, Method method, Object[] args) throws Throwable {
if ("getJavaFileForOutput".equals(method.getName())) {
//return the target file when it's asking for output
return targetFile;
} else {
return method.invoke(standard, args);
}
}
});
//create the task
CompilationTask task = compiler.getTask(new OutputStreamWriter(err),
fileManagerProxy, null, null, null, Collections.singleton(sourceFile));
//call it
if (!task.call()) {
throw new RuntimeException("Compilation failed, output:\n" +
new String(err.toByteArray()));
}
//get the result
final byte[] bytes = out.toByteArray();
//load class
Class<?> clazz;
try {
//custom class loader for garbage collection
clazz = new ClassLoader() {
protected Class<?> findClass(String name) throws ClassNotFoundException {
if (name.equals(className)) {
return defineClass(className, bytes, 0, bytes.length);
} else {
return super.findClass(name);
}
}
}.loadClass(className);
} catch (ClassNotFoundException e) {
throw new RuntimeException(e);
}
//reflectively call method
try {
return (String[]) clazz.getDeclaredMethod("getStrings").invoke(null);
} catch (Exception e) {
throw new RuntimeException(e);
}
}
它需要一个数组，因此您可以批量转发。因此，以下简单测试成功：
public static void main(String[] meh) {
if ("1\02\03\n".equals(unescapeJavaStrings("1\\02\\03\\n")[0])) {
System.out.println("Success");
} else {
System.out.println("Failure");
}
}
答案 6 :(得分：3)
我遇到了同样的问题，但我并不喜欢我在这里找到的任何解决方案。所以，我写了一个使用匹配器迭代字符串的字符来查找和替换转义序列。此解决方案假定输入格式正确。也就是说，它很乐意跳过无意义的转义，它解码换行和回车的Unicode转义（否则不能出现在字符文字或字符串文字中，因为这些文字的定义和Java的转换阶段顺序资源）。道歉，代码有点简洁。
import java.util.Arrays;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Decoder {
// The encoded character of each character escape.
// This array functions as the keys of a sorted map, from encoded characters to decoded characters.
static final char[] ENCODED_ESCAPES = { '\"', '\'', '\\', 'b', 'f', 'n', 'r', 't' };
// The decoded character of each character escape.
// This array functions as the values of a sorted map, from encoded characters to decoded characters.
static final char[] DECODED_ESCAPES = { '\"', '\'', '\\', '\b', '\f', '\n', '\r', '\t' };
// A pattern that matches an escape.
// What follows the escape indicator is captured by group 1=character 2=octal 3=Unicode.
static final Pattern PATTERN = Pattern.compile("\\\\(?:(b|t|n|f|r|\\\"|\\\'|\\\\)|((?:[0-3]?[0-7])?[0-7])|u+(\\p{XDigit}{4}))");
public static CharSequence decodeString(CharSequence encodedString) {
Matcher matcher = PATTERN.matcher(encodedString);
StringBuffer decodedString = new StringBuffer();
// Find each escape of the encoded string in succession.
while (matcher.find()) {
char ch;
if (matcher.start(1) >= 0) {
// Decode a character escape.
ch = DECODED_ESCAPES[Arrays.binarySearch(ENCODED_ESCAPES, matcher.group(1).charAt(0))];
} else if (matcher.start(2) >= 0) {
// Decode an octal escape.
ch = (char)(Integer.parseInt(matcher.group(2), 8));
} else /* if (matcher.start(3) >= 0) */ {
// Decode a Unicode escape.
ch = (char)(Integer.parseInt(matcher.group(3), 16));
}
// Replace the escape with the decoded character.
matcher.appendReplacement(decodedString, Matcher.quoteReplacement(String.valueOf(ch)));
}
// Append the remainder of the encoded string to the decoded string.
// The remainder is the longest suffix of the encoded string such that the suffix contains no escapes.
matcher.appendTail(decodedString);
return decodedString;
}
public static void main(String... args) {
System.out.println(decodeString(args[0]));
}
}
我应该注意到Apache Commons Lang3似乎没有遭受已接受的解决方案中指出的弱点。也就是说，StringEscapeUtils似乎处理八进制转义和Unicode转义的多个u字符。这意味着除非你有一些迫切的理由要避免使用Apache Commons，否则你应该使用它而不是我的解决方案（或其他任何解决方案）。
答案 7 :(得分：3)
Java 13添加了执行此操作的方法：String#translateEscapes。
它是Java 13和14中的预览功能，但在Java 15中被提升为完整功能。
答案 8 :(得分：2)
对于记录，如果您使用Scala，则可以执行以下操作：
StringContext.treatEscapes(escaped)
答案 9 :(得分：1)
org.apache.commons.lang3.StringEscapeUtils现已标记为已弃用。您可以改用org.apache.commons.text.StringEscapeUtils#unescapeJava(String)。它需要额外的Maven dependency：
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-text</artifactId>
<version>1.4</version>
</dependency>
并且似乎处理了一些更特殊的情况，例如反转义：
\\b，\\n，\\t，\\f，\\r 答案 10 :(得分：0)
如果你正在从文件中读取unicode转义字符，那么你将很难做到这一点，因为字符串将被逐字地读取以及反斜杠的转义：
my_file.txt
Blah blah...
Column delimiter=;
Word delimiter=\u0020 #This is just unicode for whitespace
.. more stuff
这里，当您从文件中读取第3行时，字符串/行将具有：
"Word delimiter=\u0020 #This is just unicode for whitespace"
并且字符串中的char []将显示：
{...., '=', '\\', 'u', '0', '0', '2', '0', ' ', '#', 't', 'h', ...}
Commons StringUnescape不会为你取消这个（我试过unescapeXml（））。您必须手动执行described here。
因此，子字符串“\ u0020”应该成为1个单个字符'\ u0020'
但是如果你使用这个“\ u0020”做String.split("... ..... ..", columnDelimiterReadFromFile)内部真正使用正则表达式，它将直接工作，因为从文件中读取的字符串是转义的，并且非常适合在正则表达式模式中使用！ （困惑？）