i have to create a regular expression to put into the method Pattern.compile(regex); My regex has to allow integer (without 0 at the beginning of the number),a sequence of numbers and char (A-Za-z),but the problem is the third point: a string that must begin and end with the char ' " ', must avoid the chars backslash (unless there isn't one other backslash) and the char '"' (unless there is before a backslash)
but i don't understand how to do the second point ( i have met a lot of errors), this is my java code:
public static void main(String[] args) {
if (args.length == 0)
throw new IllegalArgumentException();
Matcher matcher = Pattern.compile("([a-zA-Z]+[0-9a-zA-Z_]*)|"
+ "(0(?![0-9])|([1-9]+)([0-9]*))|" //"?!" è una asserzione : " se la condizione tra parentesi è vera non considerare lo zero
+ "([\"]{1}(([\\\\][^\"\\][\\\"])*)[\"]{1})"
+ "|(\\s+)").matcher(args[0]);// \s = [ \t\n\x0B\f\r]
System.out.println("Input: " + args[0]); //println va a capo dopo la stampa
while (matcher.lookingAt()) {
System.out.print("Lexeme '" + matcher.group() + "'"); //non va a capo dopo la stampa
System.out.println(" group " + ExampleLexer.getGroup(matcher));
matcher.region(matcher.end(), matcher.regionEnd());
}
//attenzione: matcher.hitEnd() restituisce true se il matcher arriva in fondo
//all'input anche se l'ultimo match non ha avuto successo, quindi funziona solo
//per espressioni regolari "semplici"
if (matcher.regionStart() == matcher.regionEnd())
System.out.println("All lexems succesfully matched");
else {
System.err.print("Unmatched lexem ");
matcher.usePattern(Pattern.compile(".*"));
matcher.lookingAt();
System.err.println(matcher.group());
}
}
"\"(\\\\.|[^\"\\\\])*\"" // (1)
"\"(\\\\[\"\\\\]|[^\"\\\\])*\"" // (2)
(
... )*
\\
followed by (1) any char .
/ (2) any non-quote/non-backslash |
not a quote, not a backslash [^
... ]
On the escapes:
\\"
is a String escape representing one char, the double quote. \\\\
is the String escape representing a backslash, one char. \\\\\\\\
is the regex escape \\\\
to represent the backslash itself instead of being a regex escape sequence itself. s = s.replace("\\\\", "\\\\\\\\");
doubles every backslash. Guess how that is written in regex with replaceAll
.
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