Thanks . Got ur point.
Just changed the regular expression to
LIKE_TEXT = <<'[%]*[a-zA-Z0-9][a-zA-Z0-9_/]*[a-zA-Z0-9][%]*'>>
Its working.
On 2/7/08, Leszek Doniec <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am not sure if I am right, but I think that when you use string with % at
> first position it will not be parsed as LIKE_TEXT, since LIKE_TEXT starts
> with a letter or digit. So send us the definition of ANY_TEXT or change
> LIKE_TEXT definition.
>
> Cheers,
> Leszek
>
>
> 2008/2/7, paskal sanil <[EMAIL PROTECTED]>:
> >
> > Hi,
> >
> > I have the following grammar (Included only the relevant grammar):
> >
> > Production
> > -------------
> > PExpression = PTerm [PRest];
> > PRest = "AND" PExpression | "OR" PExpression ;
> > PTerm = FORM_PARAM | ....
> >
> > Formula
> > ---------
> > FORM_PARAM = "Z" ":" "(" ZBooleanExpr ")"
> >
> > ZBooleanExpr = IDENTIFIER Comparator YIELD_IDENTIFIER
> > Comparator = LIKE | EQ | NE;
> > YIELD_IDENTIFIER = "'" LIKE_TEXT "'" | "'" ANY_TEXT "'" |
> > ....
> >
> > Regular Expression
> > -----------------------------
> >
> > LIKE_TEXT = <<[a-zA-Z0-9][a-zA-Z0-9%_/]*[a-zA-Z0-9]+>>
> >
> > Problem is that when i give a String like
> >
> > Str1 -> Z:(Id1 LIKE '%abc%' ) OR Z:(Id1 LIKE '%xyz%')
> >
> > It gives the error:
> >
> > Expected: " ' "
> > Found : "') OR Z:(Id1 LIKE '"
> >
> >
> > However the above works fine if use a <Space> in the following String:
> > Str2 -> Z:(Id1 LIKE '%abc%'<space> ) OR Z:(Id1 LIKE '%xyz%')
> > or don't use '%'
> > Str3 -> Z:(Id1 LIKE 'abc' ) OR Z:(Id1 LIKE 'xyz')
> >
> > How can i make the ")" to be recognized as a separate token?
> >
> > Please suggest some work around in the "grammar" to make Str1 work.
> >
> > Thanks in advance,
> > paskal sanil
> >
> >
> > _______________________________________________
> > Grammatica-users mailing list
> > Grammatica-users@nongnu.org
> > http://lists.nongnu.org/mailman/listinfo/grammatica-users
> >
>
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