Antlr3 matching tokens without whitespace

问题

Given the input "term >1", the number(1) and comparison operator(>) should generate seperate nodes in an AST. How can this be achieved?

In my tests matching only occured if "c" and "1" where seperated with a space like so "term < 1".

Current grammar:

startExpression  : orEx;

expressionLevel4    
: LPARENTHESIS! orEx RPARENTHESIS! | atomicExpression;
expressionLevel3    
: (fieldExpression) | expressionLevel4 ;
expressionLevel2    
: (nearExpression) | expressionLevel3 ;
expressionLevel1    
: (countExpression) | expressionLevel2 ;
notEx   : (NOT^)? expressionLevel1;
andEx   : (notEx        -> notEx)
(AND? a=notEx -> ^(ANDNODE $andEx $a))*;
orEx    : andEx (OR^  andEx)*;

 countExpression  : COUNT LPARENTHESIS WORD RPARENTHESIS RELATION NUMBERS -> ^(COUNT WORD RELATION NUMBERS);

nearExpression  : NEAR LPARENTHESIS (WORD|PHRASE) MULTIPLESEPERATOR (WORD|PHRASE) MULTIPLESEPERATOR NUMBERS RPARENTHESIS -> ^(NEAR WORD* PHRASE* ^(NEARDISTANCE NUMBERS));

fieldExpression : WORD PROPERTYSEPERATOR WORD -> ^(FIELDSEARCH ^(TARGETFIELD WORD) WORD );

atomicExpression 
: WORD
| PHRASE
;

fragment NUMBER : ('0'..'9');
fragment CHARACTER : ('a'..'z'|'A'..'Z'|'0'..'9'|'*'|'?');
fragment QUOTE     : ('"');
fragment LESSTHEN : '<';
fragment MORETHEN: '>';
fragment EQUAL: '=';
fragment SPACE     : ('\u0009'|'\u0020'|'\u000C'|'\u00A0');
fragment UNICODENOSPACES:  ('\u0021'..'\u0027'|'\u0030'..'\u0039'|'\u003B'..'\u007E'|'\u00A1'..'\uFFFF');
//fragment UNICODENOSPACES  :  ('\u0021'..'\u0039'|'\u003B'..'\u007E'|'\u00A1'..'\uFFFF');

LPARENTHESIS : '(';
RPARENTHESIS : ')';

AND    : ('A'|'a')('N'|'n')('D'|'d');
OR     : ('O'|'o')('R'|'r');
ANDNOT : ('A'|'a')('N'|'n')('D'|'d')('N'|'n')('O'|'o')('T'|'t');
NOT    : ('N'|'n')('O'|'o')('T'|'t');
COUNT:('C'|'c')('O'|'o')('U'|'u')('N'|'n')('T'|'t');
NEAR:('N'|'n')('E'|'e')('A'|'a')('R'|'r');
PROPERTYSEPERATOR : ':';
MULTIPLESEPERATOR : ',';

WS     : (SPACE) { $channel=HIDDEN; };
RELATION : LESSTHEN? MORETHEN? EQUAL?;
NUMBERS : (NUMBER)+;
PHRASE : (QUOTE)(CHARACTER)+((SPACE)+(CHARACTER)+)+(QUOTE);
WORD   : (UNICODENOSPACES)+;

回答1:

That is because your WORD rule matches too much: it also matches ">" so when ">1" are written together, these 2 chars are tokenized as a single WORD-token.

Whenever I'm unsure what my lexer is doing, I simple let the parser match zero or more tokens of any type, and print the type and text of all tokens:

parse
 : (t=. {System.out.printf("\%-15s '\%s'\n", tokenNames[$t.type], $t.text);})* EOF
 ;

When you let the rule above match your input "term > 1", the following gets printed:

WORD            'term'
RELATION        '>'
WORD            '1'

and of the input "term" >1

WORD            'term'
WORD            '>1'

There's no way around this: when the lexer can match 2 (or more) characters (the WORD rule), it will choose that path over a rule defined before it which will only match a single char (the RELATION rule).

Also note that your RELATION rule:

RELATION : LESSTHEN? MORETHEN? EQUAL?;

potentially matches the empty string. Make sure every lexer rule matches at least 1 character, otherwise your lexer might get into an infinite loop.

Better do something like this:

RELATION
 : (LESSTHEN | MORETHEN)? EQUAL // '<=', '>=', or '='
 | (LESSTHEN | MORETHEN)        // '<' or '>'
 ;

来源：https://stackoverflow.com/questions/13683803/antlr3-matching-tokens-without-whitespace

标签

antlr

whitespace

tokenize

antlr3