I have a path in Windows:
assert f.toString() == 'C:\\path\\to\\some\\dir'
I need to convert the backslashes \ to forward slashes /. Using Java syntax, I would write:
assert f.toString().replaceAll('\\\\', '/') == 'C:/path/to/some/dir'
But I am studying Groovy, so I thought I would write a literal regular expression:
assert f.toString().replaceAll(/\\/, '/') == 'C:/path/to/some/dir'
This throws a compilation error:
unexpected token: ) == at line: 4, column: 42
I started looking on the internet, and found several comments suggesting that this particular regex literal would not work, instead you would have to use a workaround like /\\+/. But this obviously changes the semantics of the regex.
I cannot really understand why /\\/ does not work. Maybe somebody does?
The \ at the end of the slashy string ruins it.
The main point is that you need to separate the \ from the / trailing slashy string delimiter.
It can be done in several ways:
println(f.replaceAll('\\\\', '/')) // Using a single-quoted string literal with 4 backslashes, Java style
println(f.replaceAll(/[\\]/, '/')) // Wrapping the backslash with character class
println(f.replaceAll(/\\{1}/, '/')) // Using a {1} limiting quantifier
println(f.replaceAll(/\\(?:)/, '/')) // Using an empty group after it
See the Groovy demo.
However, you may use dollar slashy strings to use the backslash at the end of the string:
f.replaceAll($/\\/$, '/')
See the demo and check this thread:
Slashy strings: backslash escapes end of line chars and slash, $ escapes interpolated variables/closures, can't have backslash as last character, empty string not allowed. Examples: def a_backslash_b = /a\b/; def a_slash_b = /a\/b/;
Dollar slashy strings: backslash escapes only EOL, $ escapes interpolated variables/closures and itself if required and slash if required, use $$ to have $ as last character or to have a $ before an identifier or curly brace or slash, use $/ to have a slash before a $, empty string not allowed. Examples: def a_backslash_b = $/a\b/$; def a_slash_b = $/a/b/$; def a_dollar_b = $/a$$b/$;
Related
I am trying to check if a string of length one is any of the following characters: "[", "\", "^", "_", single back-tick "`", or "]".
Right now I am trying to accomplish this with the following if statement:
if (character.matches("[[\\]^_`]")){
isValid = false;
}
When I run my program I get the following error for the if statement:
java.util.regex.PatternSyntaxException: null (in
java.util.regex.Pattern)
What is the correct syntax for a regex with escape characters?
Your list has four characters that need special attention:
^ is the inversion character. It must not be the first character in a character class, or it must be escaped.
\ is the escape character. It must be escaped for direct use.
[ starts a character class, so it must be escaped.
] ends a character class, so it must be escaped.
Here is the "raw" regex:
[\[\]_`\\^]
Since you represent your regex as a Java string literal, all backslashes must be additionally escaped for the Java compiler:
if (character.matches("[\\[\\]_`\\\\^]")){
isValid = false;
}
You need to escape the [, ] and \\ - the [ and ] so that the pattern compiler knows that they are not the special character class delimiters, and \\ because it's already being converted to one backslash because it's in a string literal, so to represent an escaped backslash in a pattern, you need to have no less than four consecutive backslashes.
So the resulting regex should be
"[\\[\\]\\\\^_`]"
(Test it on RegexPlanet - click on the "Java" button to test).
I know that we use escape characters like \n for next line and \t for tab.
But today while working on few string I came across \\$.
I had to print "nike$" so to print it I had to modify the string as "nike\\$".
I want to know what is the exact difference between \ and \\.
Inside a string literal, \ is an escape: The next character that follows tells us what it will do, as in your \n example for newline.
This means you can't put \ in a string on its own, since it's half of an escape sequence. Instead, to have a \ actually in a string, you use \\.
I had to print "nike$" so to print it I had to modify the string as "nike\\$"
"nike\\$" will result in a string that outputs (for instance, via System.out.println) as nike\$, not nike$.
Your use of \\$ suggests to me that you were feeding a regular expression pattern into something, e.g.:
p = Pattern.compile("nike\\$");
In that situation, we have two levels of escaping going on: The string literal, and the regular expression. To have a literal $ in a regular expression, it has to be escaped by \ because otherwise it's an end-of-input assertion. To get that \$ actually to the regular expression parser when using a string literal, we have to escape the backslash in the literal so we actually have a backslash in the string for the regular expression engine to see, thus \\$.
I am converting unicode characters stored a String into unicode text.
For example, here is a String -
String unicode = "\u0041\u006e\u0064\u0072\u006f\u0069\u0064";
Now from this string, I want to get separate unicode character -
u0041 u006e u0064 u0072 u006f u0069 u0064
So for that, I use the following code -
String[] parts = "\u0041\u006e\u0064\u0072\u006f\u0069\u0064".split("\");
But now since the " after \ is ignored in split("\"), I am getting a error.
How to not ignore a character after \?
The \ character is an escape character. You are getting a syntax error because \" is the escape sequence for placing a " character in a String literal. To place a \ inside a String literal, you need to use \\ (the first \ escapes the special meaning of the second \). So a syntactically correct statement would be:
String[] parts = "\u0041\u006e\u0064\u0072\u006f\u0069\u0064".split("\\");
But that is not going to give you what you want, because the first argument does not contain any \ characters. (Also, the split() method expects a regular expression and \ is not a valid regular expression.) Instead, it contains seven characters with code points U+0041, etc. Perhaps you want:
String[] parts = "\\u0041\\u006e\\u0064\\u0072\\u006f\\u0069\\u0064".split("\\\\");
or perhaps you want
char[] parts = "\u0041\u006e\u0064\u0072\u006f\u0069\u0064".toCharArray();
and you can then convert each element of parts to a Unicode code point string.
You need to escape the backslash. You also need to escape the backslash again because split() treats the string as a regular expression. Use .split("\\\\");
I'm trying to escape a RegExp metacharacter in Java. Below is what I want:
INPUT STRING: "This is $ test"
OUTPUT STRING: "This is \$ test"
This is what I'm currently doing but it's not working:
String inputStr= "This is $ test";
inputStr = inputStr.replaceAll("$","\\$");
But I'm getting wrong output:
"This is $ test$"
You'll need:
inputStr.replaceAll("\\$", "\\\\\\$");
The String to be replaced needs 2 backslashes because $ has a special meaning in the regexp. So $ must be escaped, to get: \$, and that backslash must itself be escaped within the java String: "\\$".
The replacement string needs 6 backslashes because both \ and $ have special meaning in the replacement strings:
\ can be used to escape characters in the replacement string.
$ can be used to make back-references in the replacement string.
So if your intended replacement string is "\$", you need to escape each of those two characters to get: \\\$, and then each backslash you need to use - 3 of them, 1 literal and 2 for escapes - must also be escaped within the java String: "\\\\\\$".
See: Matcher.replaceAll
As you said, $ is a reserved character for Regex. Then, you need to escape it. You can use a backslash character to do this:
inputStr.replaceAll("\\$", ...);
In the replacement, the $ and \ characters also have a special meaning:
Note that backslashes () and dollar signs ($) in the replacement string may cause the results to be different than if it were being treated as a literal replacement string; see Matcher.replaceAll
Then, the replacement will be the backslash character and the dollar sign, both of them being escaped by a '\' character (which needs to be doubled toi build the String):
inputStr.replaceAll("\\$", "\\\\\\$");
You have to put 6 backslashes so you escape the backslash and escape the metachar:
inputStr.replaceAll("\\$","\\\\\\$");
The first argument to replaceAll is infact a regexp, and the $ actually means "match the end of the string". You can just use replace instead, which doesn't use regexp, just a normal string replace, to achieve what you want in this case. If you want to use a regexp, just escape the $ in the first argument.
I want to replace any one of these chars:
% \ , [ ] # & # ! ^
... with empty string ("").
I used this code:
String line = "[ybi-173]";
Pattern cleanPattern = Pattern.compile("%|\\|,|[|]|#|&|#|!|^");
Matcher matcher = cleanPattern.matcher(line);
line = matcher.replaceAll("");
But it doesn't work.
What do I miss in this regular expression?
Some of the characters are special characters that are being interpreted differently. You can either escape them all with backslashes, or better yet put them in a character class (no need to escape the non-CC characters, eases readability):
Pattern cleanPattern = Pattern.compile("[%\\\\,\\[\\]#&#!^]");
There are several reasons why your solution doesn't work.
Several of the characters you wish to match have special meanings in regular expressions, including ^, [, and ]. These must be escaped with a \ character, but, to make matters worse, the \ itself must be escaped so that the Java compiler will pass the \ through to the regular expression constructor. So, to sum up step one, if you wish to match a ] character, the Java string must look like "\\]".
But, furthermore, this is a case for character classes [], rather than the alternation operator |. If you want to match "any of the characters a, b, c, that looks like [abc]. You character class would be [%\,[]#&#!^], but, because of the Java string escaping rules and the special meaning of certain characters, your regex will be [%\\\\,\\[\\]#&#!\\^].
You'd define your pattern as a character group enclosed in [ and ] and escape special chars, e.g.
String n = "%\\,[]#&#!^".replaceAll("[%\\\\,\\[\\]#&#!^]", "");