Hi Richard,
The answer is to use 2 lexer rules instead of one, and change the parser
rule. It is generally not a good idea to mark whitespace as hidden (as
you have done), and also use whitespace in your lexer rules. Here is
what I would try:
grammar testing;
start :
( DOLLAR? INT CENTS?
| NEW_LINE
)+
;
fragment DIGIT : '0'..'9';
DOLLAR : '$';
CENTS : '.' DIGIT DIGIT;
INT : DIGIT+ (',' DIGIT+)*;
NEW_LINE
: ('\r'?'\n');
WS : (SPACE |'\t'|'\u000C')
{$channel=HIDDEN;};
fragment SPACE
: ' ';
- Justin
On 2/15/2011 6:05 AM, Richard Druce wrote:
> Hi,
>
> I'm sure this question has been answered before but I'm not sure how
> to phrase it well enough to find it in the archives.
>
> I want to match to inputs an INT such as ' 170 \n'
> and an AMOUNT such as ' $ 170.00 ' (unfortunately there is
> occasionally a space before the amount)
>
> The following grammar works for ' $ 170.00 ' but not for ' 170 ' it
> sees the space and tries to match the AMOUNT and the output
> 'mismatched character ' ' expecting '.' ' is provided. Reading
> through the antlr reference book, it looks like there are a number of
> ways to possibly handle this but I'm not sure how best to approach it.
>
> grammar testing;
>
> start :
> (AMOUNT
> | INT
> | NEW_LINE)+;
>
> fragment DIGIT : '0'..'9';
>
> AMOUNT : '$'? SPACE? DIGIT+ (','DIGIT+)? '.' DIGIT DIGIT;
>
> INT : DIGIT+ (','DIGIT+)?;
>
> NEW_LINE
> : ('\r'?'\n');
>
> WS : (SPACE |'\t'|'\u000C')
> {$channel=HIDDEN;};
>
> fragment SPACE
> : ' ';
>
> Thanks,
>
> Richard
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