command = command.replaceAll("\\{player}", e.getPlayer().getName());
What is the purpose of the "\\" in "\\{player}" within that line?
Why doesn't the programmer just do "{player}" ?
That's an escape sequence. Since the { character has a special meaning in a regex, it needs to be escaped to signify you actually meant the literal {.
Backslash are usually to "escape" characters. Characters that usually can be used for regex. In this case, the {} can be interpreted as regex (which you can utilize for string manipulations), hence why you'd have to escape them for your {} to be interpreted as a literal brackets instead of reg ex.
"\ with some other characters have special meanings.
For example, \n moves to next line \t writes "tab" character and so on.
Because of this, if you really want to write \ in your code like, "c:\someFolder\someFile" you should use "C:\\someFolder\\someFile"
\ called escape character
If you write
String text ="firstname\lastname";
compiler gives you an error "illegal escape character"
It means that you should not use"\" alone in your code.
Happy coding!
Related
I was wondering about regex in Java and stumbled upon the use of backslashes. For instance, if I wanted to look for occurences of the words "this regex" in a text, I would do something like this:
Pattern.compile("this regex");
Nonetheless, I could also do something like this:
Pattern.compile("this\\sregex");
My question is: what is the difference between the two of them? And why do I have to type the backslash twice, I mean, why isn't \s an escape sequence in Java? Thanks in advance!
\s means any whitespace character, including tab, line feed and carriage return.
Java string literals already use \ to escape special characters. To put the character \ in a string literal, you need to write "\\". However regex patterns also use \ as their escape character, and the way to put that into a string literal is to use two, because it goes through two separate escaping processes. If you read your regex pattern from a plain text file for example, you won't need double escaping.
The reason you need two backslashes is that when you enter a regex string in Java code you are actually dealing with two parsers:
The first is the Java compiler, which is converting your string literal to a Java String.
The second is the regex parser, which is interpreting your regex, after it has been converted to a Java string and then passed to the regex parse when you call Pattern.compile.
So when you input "this\\sregex", it will be converted to the Java string "this\sregex" by the Java compiler. Then when you call Pattern.compile with the string, the backslash will be interpreted by the regex compiler as a special character.
The difference is that \s denotes a whitespace character, which can be more than just a blank space. It can be a tab, newline, line feed, to name a few.
With regular expressions in Java, why I should write "\n" to define a new line character and "\\s" to define whitespace character?
Why does the quantity of backslashes differs?
Java does its own string parsing, converting it from your code to an internal string in memory and before it sends the string to the regex parser.
Java converts the 2 characters \n to a linefeed (ASCII code 0x0A) and the first 2 (!) characters in \\s to a single backslash: \s. Now this string is sent to the regex parser, and since regular expressions recognize their own special escaped characters, it treats the \s as "any whitespace".
At this point, the code \n is already stored as a single character "linefeed", and the regular expression does not process it again.
Since regular expressions also recognize the set \n as "a linefeed", you can also use \\n in your Java string -- Java converts the escaped \\ to a single \, and the regular expression module then finds \n, which (again) gets translated into a linefeed.
A Java string has a certain set of allowed escape sequences, of which "\n" is one, but "\s" is not. A string doesn't understand the regexp shorthand for whitespace. You're probably passing a Java string to the RegExp constructor, so in order to pass "\s" as a string, you have to escape the "\" by doubling it.
\ is special character in many languages (in Java it is special in String or char) or tools like regex.
In String or char it is used to create other special characters which you normally couldn't write. By using \x where x is representation of that special character you are able to create
\t tab
\b backspace
\n newline
\r carriage return
\f formfeed
or to escape other special characters
\' single quote (' is special in char because it represents where char starts and ends, so to actually write ' character you need to escape it and write it as
here we start creating character
| here we end creating character
↓ ↓
'\''
↑↑
here we created literal of '
\" double quote - similarly to \' in char, in String " represents where it starts and ends, so to put " literal into string (to actually be able to write it) you need to escape it
here we start creating String
| here we end creating String
↓ ↓
"\""
↑↑
here we created literal of "
\\ backslash - since \ is special character used to create others special character there has to be a way to un-special it so we could actually print \ as simple literal.
Problem: how to write string representing day\night? If you write it such string in a way "day\night" it will be interpreted asday[newline]ight`.
So in many languages to represent \ literal another \ is added before it to escape it. So String which represent day\night needs to be written as "day\\night" (now \ in \n is escaped so it no longer represents \n - newline - but concatenation of \ and n characters)
In case of regex to represent character class which will accept any whitespace you need to actually pass \s.
But string which will represent \s needs to be written as "\\s" because as mentioned earlier in String \ is special and needs escaping.
If you would write \s as "\s" you would get
I am new to regular expressions (and to java), so this is probably a simple question.
I am trying to match the character { at the end of a line. My attempts are simply this:
row.matches("{$")
row.matches("\{$")
But both just give
Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal repetition
What am I doing wrong?
row.matches("^.*\\{$");
You simply need to escape the {, since it's a metacharacter. Because Java reserves a single backslash for special contexts (\n, \r, etc.), two backslashes are required to generate one backslash for the Pattern. Therefore,
\\{
will properly evaluate to
\{
Not only this, but the matches method checks to see iff the entire string matches, instead of just a subset. Hence, the ^.* part
You must escape the { character as it is a special char for regex
row.matches("\\{$")
Did escaping the angle bracket work?
as in \\{$
Tried it against
hello world{
whatever{
hello{dontmatch
}
}
It matched world{ and whatever{ but not hello{dontmatch
you need to escape the { with an \ but to prevent that the \{ is read as special character (like \n for line-feed) you need to escape also the \ with an additional \ resulting to:
row.matches("\\{$");
In Java, I am trying to split on the ^ character, but it is failing to recognize it. Escaping \^ throws code error.
Is this a special character or do I need to do something else to get it to recognize it?
String splitChr = "^";
String[] fmgStrng = aryToSplit.split(splitChr);
The ^ is a special character in Java regex - it means "match the beginning" of an input.
You will need to escape it with "\\^". The double slash is needed to escape the \, otherwise Java's compiler will think you're attempting to use a special \^ sequence in a string, similar to \n for newlines.
\^ is not a special escape sequence though, so you will get compiler errors.
In short, use "\\^".
The ^ matches the start of string. You need to escape it, but in this case you need to escape it so that the regular expression parser understands which means escaping the escape, so:
String splitChr = "\\^";
...
should get you what you want.
String.split() accepts a regex. The ^ sign is a special symbol denoting the beginning of the input sequence. You need to escape it to make it work. You were right trying to escape it with \ but it's a special character to escape things in Java strings so you need to escape the escape character with another \. It will give you:
\\^
use "\\^". Use this example as a guide:
String aryToSplit = "word1^word2";
String splitChr = "\\^";
String[] fmgStrng = aryToSplit.split(splitChr);
System.out.println(fmgStrng[0]+","+fmgStrng[1]);
It should print "word1,word2", effectively splitting the string using "\\^". The first slash is used to escape the second slash. If there were no double slash, Java would think ^ was an escape character, like the newline "\n"
None of the above answers makes no sense. Here is the right explanation.
As we all know, ^ doesn't need to be escaped in Java String.
As ^ is special charectar in RegulalExpression , it expects you to pass in \^
How do we make string \^ in java? Like this String splitstr = "\\^";
I have Java string:
String b = "/feedback/com.school.edu.domain.feedback.Review$0/feedbackId");
I also have generated pattern against which I want to match this string:
String pattern = "/feedback/com.school.edu.domain.feedback.Review$0(.)*";
When I say b.matches(pattern) it returns false. Now I know dollar sign is part of Java RegEx, but I don't know how should my pattern look like. I am assuming that $ in pattern needs to be replaced by some escape characters, but don't know how many. This $ sign is important to me as it helps me distinguish elements in list (numbers after dollar), and I can't go without it.
Use
String escapedString = java.util.regex.Pattern.quote(myString)
to automatically escape all special regex characters in a given string.
You need to escape $ in the regex with a back-slash (\), but as a back-slash is an escape character in strings you need to escape the back-slash itself.
You will need to escape any special regex char the same way, for example with ".".
String pattern = "/feedback/com\\.navteq\\.lcms\\.common\\.domain\\.poi\\.feedback\\.Review\\$0(.)*";
In Java regex both . and $ are special. You need to escape it with 2 backslashes, i.e..
"/feedback/com\\.navtag\\.etc\\.Review\\$0(.*)"
(1 backslash is for the Java string, and 1 is for the regex engine.)
Escape the dollar with \
String pattern =
"/feedback/com.navteq.lcms.common.domain.poi.feedback.Review\\$0(.)*";
I advise you to escape . as well, . represent any character.
String pattern =
"/feedback/com\\.navteq\\.lcms\\.common\\.domain\\.poi\\.feedback\\.Review\\$0(.)*";
The ans by #Colin Hebert and edited by #theon is correct. The explanation is as follows. #azec-pdx
It is a regex as a string literal (within double quotes).
period (.) and dollar-sign ($) are special regex characters (metacharacters).
To make the regex engine interpret them as normal regex characters period(.) and dollar-sign ($), you need to prefix a single backslash to each. The single backslash ( itself a special regex character) quotes the character following it and thus escaping it.
Since the given regex is a string literal, another backslash is required to be prefixed to each to avoid confusion with the usual visible-ASCII escapes(character, string and Unicode escapes in string literals) and thus avoid compiler error.
Even if you use within a string literal any special regex construct that has been defined as an escape sequence, it needs to be prefixed with another backslash to avoid compiler error.For example, the special regex construct (an escape sequence) \b (word boundary) of regex would clash with \b(backspace) of the usual visible-ASCII escape(character escape). Thus another backslash is prefixed to avoid the clash and then \\b would be read by regex as word boundary.
To be always safe, all single backslash escapes (quotes) within string literals are prefixed with another backslash. For example, the string literal "\(hello\)" is illegal and leads to a compile-time error; in order to match the string (hello) the string literal "\\(hello\\)" must be used.
The last period (.)* is supposed to be interpreted as special regex character and thus it needs no quoting by a backslash, let alone prefixing a second one.