How to get rid of " around my string in ANTLR? - antlr

Question

When I get the token with these rules
STRINGA : '"' (options {greedy=false;}: ESC | .)* '"';
STRINGB : '\'' (options {greedy=false;}: ESC | .)* '\'';
it ends up grabbing 'text' instead of just text. I can easily remove the ' and ' myself but was wondering how I can get ANTLR to remove it?

Answer 1

You will need some custom code for that. Also, you shouldn't be using a . (dot) inside the rule: you should explicitly define you want to match everything except a backslash (assuming that is what your ESQ starts with), a quote and line break chars probably.
Something like this would do it:
grammar T;
parse
: STRING EOF {System.out.println($STRING.text);}
;
STRING
: '"' (ESQ | ~('"' | '\\' | '\r' | '\n'))* '"'
{
String matched = getText();
StringBuilder builder = new StringBuilder();
for(int i = 1; i < matched.length() - 1; i++) {
char ch = matched.charAt(i);
if(ch == '\\') {
i++;
ch = matched.charAt(i);
switch(ch) {
case 'n': builder.append('\n'); break;
case 't': builder.append('\t'); break;
default: builder.append(ch); break;
}
}
else {
builder.append(ch);
}
}
setText(builder.toString());
}
;
fragment ESQ
: '\\' ('n' | 't' | '"' | '\\')
;
If you now parse the input "tabs:'\t\t\t'\nquote:\"\nbackslash:\\", the following will be printed to the console:
tabs:' '
quote:"
backslash:\
To keep the grammar clean, you could of course move the code in a custom method:
grammar T;
#lexer::members {
private String fix(String str) {
...
}
}
parse
: STRING EOF {System.out.println($STRING.text);}
;
STRING
: '"' (ESQ | ~('"' | '\\' | '\r' | '\n'))* '"' {setText(fix(getText()));}
;
fragment ESQ
: '\\' ('n' | 't' | '"' | '\\')
;

Answer 2

One approach is to define the string contents as a separate category, for example
STRINGA : '"' STRINGCONTENTS '"';
STRINGB : '\'' STRINGCONTENTS '\'';
then capture the STRINGCONTENTS value.