如何检查字符串是否是有效的类标识符

Question

有没有一些好的方法可以检查某个字符串是否表示有效的（完全限定的）Java类名？例如，org.comPAny.ClassName甚至mYcRAZYcLASSnAME是有效的类名，但像org..package.MyClass或org.ClassName.之类的东西不是。我想在不加载该类的情况下检查类名的有效性。

Java 中是否有一些方便的方法来检查这一点？或者，您能否提供涵盖所有情况的正则表达式？

编辑：请不要建议第三方库。

Answer 1

如果你真的想要一个防弹的解决方案，按"\."分割字符串并首先使用

Character.isJavaIdentifierStart

和循环

Character.isJavaIdentifierPart

以检查部件是否为有效的标识符名称。

 

编辑：#split(String)将字符串解释为正则表达式，因此请确保按"\."拆分字符串，而不是像您期望的最小意外原则那样"."。多么糟糕的 API 设计，哎呀！这就是你从你的语言中没有正则表达式文字中得到的......

Answer 2

正确的方法是使用 javax.lang.model.SourceVersion.isName 方法（Java 标准库的一部分）。

Answer 3

不同版本的 Java 平台支持不同的 Unicode 版本。

因此，不同的版本将支持不同的有效标识符。

下面说明了 BMP 外部的标识符和 Java 保留字。

import java.util.Arrays;
public enum PackageName {
  SIMPLE, QUALIFIED, INVALID;
    public static final PackageName check(String name) {
    PackageName ret = PackageName.INVALID;
    int[] codePoint;
    int index = 0, dotex = -1;
    boolean needStart = true;
    escape: {
      if(name == null || name.isEmpty()) break escape;
      if(name.codePointAt(0) == '.') break escape;
      codePoint = name.codePoints().toArray();
      while (index <= codePoint.length) {
        if(index == codePoint.length) {
          if(codePoint[index - 1] == '.'){
            ret = PackageName.INVALID; break escape;}
          int start = dotex + 1;
          int end = index;
          start = name.offsetByCodePoints(0, start);
          end = name.offsetByCodePoints(0, end);
          String test = name.substring(start, end);
          if(!(Arrays.binarySearch(reserved, test) < 0)){
            ret = PackageName.INVALID; break escape;}
          if(!(ret == PackageName.QUALIFIED)) ret = PackageName.SIMPLE;
          break escape;
        }
        if(codePoint[index] == '.') {
          if(codePoint[index - 1] == '.'){
            ret = PackageName.INVALID;  break escape;}
          else {
            needStart = true;
            int start = dotex + 1;
            int end = index;
            start = name.offsetByCodePoints(0, start);
            end = name.offsetByCodePoints(0, end);
            String test = name.substring(start, end);
            if(!(Arrays.binarySearch(reserved, test) < 0)) break escape;
            dotex = index;
            ret = PackageName.QUALIFIED;
          }
        } else if(Character.isJavaIdentifierStart(codePoint[index])) {
          if(needStart) needStart = false;
        } else if((!Character.isJavaIdentifierPart(codePoint[index]))){
          ret = PackageName.INVALID; break escape;
        } 
        index++;
      }
    }
    return ret;
  }
  private static final String[] reserved;
  static {
    reserved = new String[] { "abstract", "assert", "boolean", "break", "byte",
        "case", "catch", "char", "class", "const", "continue", "default", "do",
        "double", "else", "enum", "extends", "false", "final", "finally",
        "float", "for", "if", "goto", "implements", "import", "instanceof",
        "int", "interface", "long", "native", "new", "null", "package",
        "private", "protected", "public", "return", "short", "static",
        "strictfp", "super", "switch", "synchronized", "this", "throw",
        "throws", "transient", "true", "try", "void", "volatile", "while" };
  }
}

Answer 4

public static boolean isValidJavaIdentifier(String s)
{
 // an empty or null string cannot be a valid identifier
 if (s == null || s.length() == 0)
 {
    return false;
 }
 char[] c = s.toCharArray();
 if (!Character.isJavaIdentifierStart(c[0]))
 {
    return false;
 }
 for (int i = 1; i < c.length; i++)
 {
    if (!Character.isJavaIdentifierPart(c[i]))
    {
       return false;
    }
 }
 return true;
}

参考：http://www.java2s.com/Code/Java/Reflection/CheckwhetherthegivenStringisavalididentifieraccordingtotheJavaLanguagespecifications.htm

Answer 5

您的问题有很好的解决方案：

public static boolean isFullyQualifiedClassname( String classname ) {
      if (classname == null) return false;
      String[] parts = classname.split("[\.]");
      if (parts.length == 0) return false;
      for (String part : parts) {
          CharacterIterator iter = new StringCharacterIterator(part);
          // Check first character (there should at least be one character for each part) ...
          char c = iter.first();
          if (c == CharacterIterator.DONE) return false;
          if (!Character.isJavaIdentifierStart(c) && !Character.isIdentifierIgnorable(c)) return false;
          c = iter.next();
          // Check the remaining characters, if there are any ...
          while (c != CharacterIterator.DONE) {
              if (!Character.isJavaIdentifierPart(c) && !Character.isIdentifierIgnorable(c)) return false;
              c = iter.next();
          }
      }
      return true;
  }

来源： http://www.java2s.com/Code/Java/Reflection/DeterminewhetherthesuppliedstringrepresentsawellformedfullyqualifiedJavaclassname.htm

Answer 6

你见过这个吗？

http://commons.apache.org/sandbox/classscan/apidocs/org/apache/commons/classscan/builtin/ClassNameHelper.html

此处的方法isValidIdentifier *检查提供的名称是否是有效的 java 包名称部分或类标识符。

Answer 7

使用正则表达式使防弹解决方案变得容易：

String NAME = "\p{javaJavaIdentifierStart}\p{javaJavaIdentifierPart}*";
String DOT = "\.";
String DOTTED_NAME = NAME + "(?:" + DOT + NAME + ")*";
Pattern DOTTED_NAME_PATTERN = Pattern.compile(DOTTED_NAME);
boolean isJavaName(String name) {
    Matcher matcher = DOTTED_NAME_PATTERN.matcher(name);
    return matcher.matches();
}