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我有一个忘恩负义的任务来修复用于解析 edifact 文件的旧 antlr2 解析器中的错误。不幸的是，我对 antlr2 或解析器一点也不熟悉，我无法让它工作。

edifact 文件如下所示：

ABC+Name+Surname+zip+city+street+country+1961219++0037141008'
XYZ+Company+++XYZ+zip+street'
LMN+20081010+1100'

有几个不同的段，以关键字开头。例如 XYZ 或 ABC。关键字后面跟着不同的属性值，都用' + '分隔。属性值可能为空。每个段都以'结尾。

问题是，只要数据属性包含关键字，解析器就会抛出错误：

意外令牌：XYZ

XYZ +公司+++ XYZ +zip+street'

这是语法文件的摘录：

// $ANTLR 2.7.6


xyz: "XYZ"       ELT_SEP! 
     (xyz1_1a:ANUM|xyz1_1b:NUM)          {lq(90,xyz1_1a,xyz1_1b,"XYZ1-1"+LQ90)}?  ELT_SEP!
     (xyz1_2a:ANUM|xyz1_2b:NUM)?         {lq_(90,xyz1_2a,xyz1_2b,"XYZ1-2"+LQ90)}? ELT_SEP!
     (xyz1_3a:ANUM|xyz1_3b:NUM)?         {lq_(90,xyz1_3a,xyz1_3b,"XYZ1-3"+LQ90)}? ELT_SEP! 
     (xyz2a:ANUM|xyz2b:NUM)?             {lq_(3,xyz2a,xyz2b,"XYZ2"+LQ3)}?         ELT_SEP! 
     (xyz3a:ANUM|xyz3b:NUM)?             {lq_(6,xyz3a,xyz3b,"XYZ3"+LQ6)}?         ELT_SEP! 
     (xyz4a:ANUM|xyz4b:NUM)              {lq(30,xyz4a,xyz4b,"XYZ4"+LQ30)}?
     (ELT_SEP! (xyz5a:ANUM|xyz5b:NUM)?)?  {lq_(46,xyz5a,xyz5b,"XYZ5"+LQ46)}?       SEG_TERM!
     {
        if (skipNachricht()) return;
        Xyz xyz = new Xyz();
        xyz.xyz1_1 = getText(nn(xyz1_1a, xyz1_1b));
        xyz.xyz1_2 = getText(nn(xyz1_2a, xyz1_2b));
        xyz.xyz1_3 = getText(nn(xyz1_3a, xyz1_3b));
        xyz.xyz2 = getText(nn(xyz2a, xyz2b));
        xyz.xyz3 = getText(nn(xyz3a, xyz3b));
        xyz.xyz4 = getText(nn(xyz4a, xyz4b));
        xyz.xyz5 = getText(nn(xyz5a, xyz5b));
        handleXyz(xyz);
     }
   ;  



/*
 * Lexer
 */
class EdifactLexer extends Lexer;

options { 
          k=2; 
          filter=true; 
          charVocabulary = '\3'..'\377'; // Latin
}

DEZ_SEP: ',' 
    {
          //System.out.println("Found dez_sep: " + getText()); 
        }
    ;

ELT_SEP: '+' 
    {
          //System.out.println("Found elt_sep: " + getText()); 
        }
    ;

SEG_TERM: '\''
    {
          // System.out.println("Found seg_term: " + getText()); 
        }
    ;

NUM:   (('0'..'9')+ (',' ('0'..'9')+)? ('+' | '\'')) 
          => ('0'..'9')+ (',' ('0'..'9')+)? 
            {
                //System.out.println("num_: " + getText());
            }
       | 
       ((ESCAPED | ~('?' | '+' | '\'' | ',' | '\r' | '\n'))+ ) 
          => ( ESCAPED | ~('?' | '+' | '\'' | ',' | '\r' | '\n'))+
                {
                        $setType(ANUM); 
            //System.out.println("anum: " + getText());
        } 
       |
       (WRONGLY_ESCAPED) => WRONGLY_ESCAPED 
                {$setType(WRONGLY_ESCAPED); }
       ;

protected
WRONGLY_ESCAPED: '?' ~('?' | ':' | '+' | '\'' | ',') 
    {
          //System.out.println("Found wrong_escaped: " + getText()); 
        }
        ;

protected
ESCAPED: '?' 
      ( ','  {$setText(","); }
      | '?'  {$setText("?"); }
          | '\'' {$setText("'"); }
          | ':'  {$setText(":"); }
          | '+'  {$setText("+"); }
      ) 
    {
          //System.out.println("Found escaped: " + getText()); 
        }
    ;

NEWLINE   :  ( "\r\n" // DOS
               | '\r'   // MAC
               | '\n'   // Unix
             )
             { newline(); 
               $setType(Token.SKIP);
             }
          ;

非常感谢任何帮助:)。

4

1 回答 1

0

这可能不是最好的解决方案，但我终于找到了解决问题的方法。所以，如果有人遇到类似的问题，这是我的解决方案：

我编写了一个方法，如果当前令牌类型与我的任何关键字匹配，则将令牌类型更改为ANUM ：

void ckt() throws TokenStreamException, SemanticException {
    if (mKeywordList.contains(LT(1).getType())) {
        LT(1).setType(ANUM);
    }
}

在尝试访问ANUM -Token 之前，在我的解析器规则中调用了该方法：

xyz: "XYZ"       ELT_SEP! 
     {ckt();}(xyz1_1a:ANUM|xyz1_1b:NUM)          {lq(90,xyz1_1a,xyz1_1b,"XYZ1-1"+LQ90)}?  ELT_SEP!
     {ckt();}(xyz1_2a:ANUM|xyz1_2b:NUM)?         {lq_(90,xyz1_2a,xyz1_2b,"XYZ1-2"+LQ90)}? ELT_SEP!
     {ckt();}(xyz1_3a:ANUM|xyz1_3b:NUM)?         {lq_(90,xyz1_3a,xyz1_3b,"XYZ1-3"+LQ90)}? ELT_SEP! 
     {ckt();}(xyz2a:ANUM|xyz2b:NUM)?             {lq_(3,xyz2a,xyz2b,"XYZ2"+LQ3)}?         ELT_SEP! 
     {ckt();}(xyz3a:ANUM|xyz3b:NUM)?             {lq_(6,xyz3a,xyz3b,"XYZ3"+LQ6)}?         ELT_SEP! 
     {ckt();}(xyz4a:ANUM|xyz4b:NUM)              {lq(30,xyz4a,xyz4b,"XYZ4"+LQ30)}?
     (ELT_SEP! {ckt();}(xyz5a:ANUM|xyz5b:NUM)?)?  {lq_(46,xyz5a,xyz5b,"XYZ5"+LQ46)}?       SEG_TERM!
     {
        if (skipNachricht()) return;
        Xyz xyz = new Xyz();
        xyz.xyz1_1 = getText(nn(xyz1_1a, xyz1_1b));
        xyz.xyz1_2 = getText(nn(xyz1_2a, xyz1_2b));
        xyz.xyz1_3 = getText(nn(xyz1_3a, xyz1_3b));
        xyz.xyz2 = getText(nn(xyz2a, xyz2b));
        xyz.xyz3 = getText(nn(xyz3a, xyz3b));
        xyz.xyz4 = getText(nn(xyz4a, xyz4b));
        xyz.xyz5 = getText(nn(xyz5a, xyz5b));
        handleXyz(xyz);
     }
   ;
于 2019-08-20T14:24:40.997 回答