Illegal escape character in string literal java ошибка

I’m working on a solution to a previous question, as best as I can, using regular expressions. My pattern is

"d{4}w{3}(0[1-9]|[12][0-9]|3[01])([01][0-9]|2[0-3])([0-5][0-9]){2}"

According to NetBeans, I have two illegal escape characters. I’m guessing it has to do with the d and w, but those are both valid in Java. Perhaps my syntax for a Java regular expression is off…

The entire line of code that is involved is:

userTimestampField = new FormattedTextField(
  new RegexFormatter(
    "d{4}w{3}(0[1-9]|[12][0-9]|3[01])([01][0-9]|2[0-3])([0-5][0-9]){2}"
));

What about the following: \d{4}\w{3}(0[1-9]|[12][0-9]|3[01])([01][0-9]|2[0-3])([0-5][0-9]){2}

Have you tried this?

\d{4}\w{3}(0[1-9]|[12][0-9]|3[01])([01][0-9]|2[0-3])([0-5][0-9]){2}

all you need to do is to put

 *
 ex: string ex = 'this is the character: *\s';

before your invalid character and not 8 !!!!!

answered Feb 12, 2015 at 1:12

Dear Sir,
While complile this code the error message display that «Illegal escape character».Please guide me to rectify this code.

The data.txt file contains:->Sumanta,Sagar,Harish

Thanks and Regards
Sumanta Panda
[ November 21, 2008: Message edited by: Martijn Verburg ]

Hi,

I think you should write

If you want to write a in a String literal you have to write \, anyway Java compiler interprets it as an escape character. Since no such escape characters as T or d the java compiler reports an error.
Regards,
Miki

lowercase baba

Posts: 13086

posted 14 years ago

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There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

This simple regex program

import java.util.regex.*;
class Regex {
    public static void main(String [] args) {
        System.out.println(args[0]); // #1
        Pattern p = Pattern.compile(args[0]); // #2
        Matcher m = p.matcher(args[1]);
        boolean b = false;
        while(b = m.find()) {
            System.out.println(m.start()+" "+m.group());
        }
    }
}

invoked by java regex "d" "sfdd1" compiles and runs fine.

But if #1 is replaced by Pattern p = Pattern.compile("d");, it gives compiler error saying illegal escape character. In #1 I also tried printing the pattern specified in the command line arguments. It prints d, which means it is just getting replaced by d in #2.

So then why won’t it throw any exception? At the end it’s string argument that Pattern.compile() is taking, doesn’t it detect illegal escape character then? Can someone please explain why is this behaviour?

asked Sep 24, 2012 at 21:33

A backslash character in a string literal needs to be escaped (preceded by a backslash). When passed in from the command line the string is not a string literal. The compiler complains because "d" is not a valid escape sequence (see Escape Sequences for Character and String Literals ).

The backslash in Java results in an escape in strings. For example, the string "t" results in a tab character in java. This is also why "n" produces a newline.

In regular expressions, d is an escape with respect to the regular expression, not Java. This means in order to get d in a string literal, you have to type "\d" in the string. Basically, you have to escape the to get the literal value d, and then when Pattern compiles the regex, it further escapes the d to be parsed as a digit.

This can be confusing, but long story short, you should never have a single in a string literal for a regular expression since even the string literal "\n" gets parsed properly.

answered Sep 24, 2012 at 21:47

answered Sep 24, 2012 at 21:36

This simple regex program

import java.util.regex.*;
class Regex {
    public static void main(String [] args) {
        System.out.println(args[0]); // #1
        Pattern p = Pattern.compile(args[0]); // #2
        Matcher m = p.matcher(args[1]);
        boolean b = false;
        while(b = m.find()) {
            System.out.println(m.start()+" "+m.group());
        }
    }
}

invoked by java regex "d" "sfdd1" compiles and runs fine.

But if #1 is replaced by Pattern p = Pattern.compile("d");, it gives compiler error saying illegal escape character. In #1 I also tried printing the pattern specified in the command line arguments. It prints d, which means it is just getting replaced by d in #2.

So then why won’t it throw any exception? At the end it’s string argument that Pattern.compile() is taking, doesn’t it detect illegal escape character then? Can someone please explain why is this behaviour?

asked Sep 24, 2012 at 21:33

A backslash character in a string literal needs to be escaped (preceded by a backslash). When passed in from the command line the string is not a string literal. The compiler complains because "d" is not a valid escape sequence (see Escape Sequences for Character and String Literals ).

The backslash in Java results in an escape in strings. For example, the string "t" results in a tab character in java. This is also why "n" produces a newline.

In regular expressions, d is an escape with respect to the regular expression, not Java. This means in order to get d in a string literal, you have to type "\d" in the string. Basically, you have to escape the to get the literal value d, and then when Pattern compiles the regex, it further escapes the d to be parsed as a digit.

This can be confusing, but long story short, you should never have a single in a string literal for a regular expression since even the string literal "\n" gets parsed properly.

answered Sep 24, 2012 at 21:47

answered Sep 24, 2012 at 21:36

Почему я не могу использовать u000D и u000A как CR и LF в Java? При компиляции кода выдает ошибку:

String x = "u000A hello";//Error - Illegal escape character in string literal.

Экраны Unicode предварительно обрабатываются перед запуском компилятора. Следовательно, если поставить u000A в строковом литерале вроде этого:

String someString = "foou000Abar";

Он будет скомпилирован точно так, как если бы вы написали:

String someString = "foo
bar";

Придерживаться r (возврат каретки; 0x0D) и расширение n (перевод строки; 0x0A)

Бонус: Вы всегда можете получить от этого удовольствие, особенно с учетом ограничений большинства выделителей синтаксиса. В следующий раз, когда у вас будет секунда, попробуйте запустить этот код:

public class FalseIsTrue {
    public static void main(String[] args) {
        if ( false == true ) { //these characters are magic: u000au007du007b
            System.out.println("false is true!");
        }
    }
}

ответ дан 05 окт ’10, 18:10

Потому что он попадает в диапазон Управляющие символы Unicode

Который U+0000–U+001F и U+007F.

Управляющие символы Unicode используются для управления интерпретацией или отображением текста, но сами эти символы не имеют визуального или пространственного представления.

Их можно избежать, используя как описано в ответе @Mark выше

ИЗ RFC:

2.5. Струны

Представление строк аналогично соглашениям, используемым в семействе языков программирования C. Строка начинается и заканчивается кавычками. Все символы Unicode могут быть помещены в кавычки, за исключением символов, которые необходимо экранировать: кавычки, обратная косая черта и управляющие персонажи
(От U + 0000 до U + 001F).

Любой персонаж май сбежать.

ответ дан 20 апр.