Antlr 解析没有显式endtag的文本

解析词典条目的问题（参见下面的示例）基于 没有明确的开始和结束标记，但：

• 一个元素的结束标记已经是下一个元素的开始标记
• 或者：开始标记不是一个语法元素，但它是当前的解析状态（因此它取决于您在inputstream中已经“看到”的内容）

示例1，简单输入：

wordWithoutSpace [phonetic information]
definition as everything until colon: example sentence until EOF

示例2，多定义条目：

wordWithoutSpace [phonetic information]
1. first definition until colon: example sentence until second definition
2. second definition until colon: example sentence until EOF

正如我用文字或伪代码所说：

dictionary-entry : 
     word = .+ ' ' // catch everything as word until you see a space
     phon = '[' .+ ']' // then follows phonetic, which is everything in brackets
     (MultipleMeaning | UniqueMeaning)

MultipleMeaning : Int '.' definition= .+ ':' // a MultipleMeaning has a number 
                                             // before the definition

UniqueMeaning : definition= .+ ':'

import org.antlr.runtime.ANTLRStringStream;
import org.antlr.runtime.CharStream;
import org.antlr.runtime.Token;

public class TestLexer {


    public static void main(String[] args) {


        String str = "Word [phon]1.definition:";
        CharStream input = new ANTLRStringStream(str);
        DudenLexer lexer = new DudenLexer(input);
        Token token;
        while ((token = lexer.nextToken())!=Token.EOF_TOKEN) {
          System.out.println("Token: "+token);
        }
    }
}

我已经试用过带门的Lexer（antlr版本：3.2）

TestLexer:

dictionary-entry : 
     word = .+ ' ' // catch everything as word until you see a space
     phon = '[' .+ ']' // then follows phonetic, which is everything in brackets
     (MultipleMeaning | UniqueMeaning)

MultipleMeaning : Int '.' definition= .+ ':' // a MultipleMeaning has a number 
                                             // before the definition

UniqueMeaning : definition= .+ ':'

import org.antlr.runtime.ANTLRStringStream;
import org.antlr.runtime.CharStream;
import org.antlr.runtime.Token;

public class TestLexer {


    public static void main(String[] args) {


        String str = "Word [phon]1.definition:";
        CharStream input = new ANTLRStringStream(str);
        DudenLexer lexer = new DudenLexer(input);
        Token token;
        while ((token = lexer.nextToken())!=Token.EOF_TOKEN) {
          System.out.println("Token: "+token);
        }
    }
}

我遇到的问题：

dictionary-entry : 
     word = .+ ' ' // catch everything as word until you see a space
     phon = '[' .+ ']' // then follows phonetic, which is everything in brackets
     (MultipleMeaning | UniqueMeaning)

MultipleMeaning : Int '.' definition= .+ ':' // a MultipleMeaning has a number 
                                             // before the definition

UniqueMeaning : definition= .+ ':'

import org.antlr.runtime.ANTLRStringStream;
import org.antlr.runtime.CharStream;
import org.antlr.runtime.Token;

public class TestLexer {


    public static void main(String[] args) {


        String str = "Word [phon]1.definition:";
        CharStream input = new ANTLRStringStream(str);
        DudenLexer lexer = new DudenLexer(input);
        Token token;
        while ((token = lexer.nextToken())!=Token.EOF_TOKEN) {
          System.out.println("Token: "+token);
        }
    }
}

• 我得到错误消息：行1:0规则定义失败谓词：{cs==2}
• 我不知道这样做是否正确

我被这件事耽搁了大约三天，非常感谢你的帮助和提示

谢谢,，

---

Tom

我认为你需要在语法中明确地处理换行符才能做到这一点，否则你将很难处理

2之类的事情。直到冒号的第二个定义：这是2。行

。