I have comma separated list of regular expressions:
.{8},[0-9],[^0-9A-Za-z ],[A-Z],[a-z]
I have done a split on the comma. Now I'm trying to match this regex against a generated password. The problem is that Pattern.compile does not like square brackets that is not escaped.
Can some please give me a simple function that takes a string like so: [0-9] and returns the escaped string \[0-9\].
For some reason, the above answer didn't work for me. For those like me who come after, here is what I found.
I was expecting a single backslash to escape the bracket, however, you must use two if you have the pattern stored in a string. The first backslash escapes the second one into the string, so that what regex sees is \]. Since regex just sees one backslash, it uses it to escape the square bracket.
\\]
In regex, that will match a single closing square bracket.
If you're trying to match a newline, for example though, you'd only use a single backslash. You're using the string escape pattern to insert a newline character into the string. Regex doesn't see \n - it sees the newline character, and matches that. You need two backslashes because it's not a string escape sequence, it's a regex escape sequence.
You can use Pattern.quote(String).
From the docs:
public static String quote​(String s)
Returns a literal pattern String for the specified String.
This method produces a String that can be used to create a Pattern that would match the string s as if it were a literal pattern.
Metacharacters or escape sequences in the input sequence will be given no special meaning.
You can use the \Q and \E special characters...anything between \Q and \E is automatically escaped.
\Q[0-9]\E
Pattern.compile() likes square brackets just fine. If you take the string
".{8},[0-9],[^0-9A-Za-z ],[A-Z],[a-z]"
and split it on commas, you end up with five perfectly valid regexes: the first one matches eight non-line-separator characters, the second matches an ASCII digit, and so on. Unless you really want to match strings like ".{8}" and "[0-9]", I don't see why you would need to escape anything.
Related
Every time i try to split the string "hello*world" using s.split("*"); I get a PatternSyntaxException.
I have tried using s.split("\*"); but that gives me another error. Im sure this is something simple.
How do i stop this?
* is a meta-character in regular expressions used as a wildcard quantifier to match zero of more characters
Try using 2 backslash characters
s.split("\\*");
The split method takes a regular expression as the argument, not a normal string. The * has special meaning in regular expressions. If you want to split on a literal *, you have to escape it with a backslash. But the backslash is also an escape character in Java string literals, so you have to escape the backslash too by using two backslashes:
s.split("\\*")
I'm trying to compare following strings with regex:
#[xyz="1","2"'"4"] ------- valid
#[xyz] ------------- valid
#[xyz="a5","4r"'"8dsa"] -- valid
#[xyz="asd"] -- invalid
#[xyz"asd"] --- invalid
#[xyz="8s"'"4"] - invalid
The valid pattern should be:
#[xyz then = sign then some chars then , then some chars then ' then some chars and finally ]. This means if there is characters after xyz then they must be in format ="XXX","XXX"'"XXX".
Or only #[xyz]. No character after xyz.
I have tried following regex, but it did not worked:
String regex = "#[xyz=\"[a-zA-z][0-9]\",\"[a-zA-z][0-9]\"'\"[a-zA-z][0-9]\"]";
Here the quotations (in part after xyz) are optional and number of characters between quotes are also not fixed and there could also be some characters before and after this pattern like asdadad #[xyz] adadad.
You can use the regex:
#\[xyz(?:="[a-zA-z0-9]+","[a-zA-z0-9]+"'"[a-zA-z0-9]+")?\]
See it
Expressed as Java string it'll be:
String regex = "#\\[xyz=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\"\\]";
What was wrong with your regex?
[...] defines a character class. When you want to match literal [ and ] you need to escape it by preceding with a \.
[a-zA-z][0-9] match a single letter followed by a single digit. But you want one or more alphanumeric characters. So you need [a-zA-Z0-9]+
Use this:
String regex = "#\\[xyz(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")?\\]";
When you write [a-zA-z][0-9] it expects a letter character and a digit after it. And you also have to escape first and last square braces because square braces have special meaning in regexes.
Explanation:
[a-zA-z0-9]+ means alphanumeric character (but not an underline) one or more times.
(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")? means that expression in parentheses can be one time or not at all.
Since square brackets have a special meaning in regex, you used it by yourself, they define character classes, you need to escape them if you want to match them literally.
String regex = "#\\[xyz=\"[a-zA-z][0-9]\",\"[a-zA-z][0-9]\"'\"[a-zA-z][0-9]\"\\]";
The next problem is with '"[a-zA-z][0-9]' you define "first a letter, second a digit", you need to join those classes and add a quantifier:
String regex = "#\\[xyz=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\"\\]";
See it here on Regexr
there could also be some characters before and after this pattern like
asdadad #[xyz] adadad.
Regex should be:
String regex = "(.)*#\\[xyz(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")?\\](.)*";
The First and last (.)* will allow any string before the pattern as you have mentioned in your edit. As said by #ademiban this (=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")? will come one time or not at all. Other mistakes are also very well explained by Others +1 to all other.
I have this line of code to remove some punctuation:
str.replaceAll("[\\-\\!\\?\\.\\,\\;\\:\\\"\\']", "");
I don't know if all the chars in this regex need to be escaped, but I escaped only for safety.
Is there some way to build a regex like this in a more clear way?
Inside [...] you don't need to escape the characters. [.] for instance wouldn't make sense anyway!
The exceptions to the rule are
] since it would close the whole [...] expression prematurely.
^ if it is the first character, since [^abc] matches everything except abc.
- unless it's the first/last character, since [a-z] matches all characters between a to z.
Thus, you could write
str.replaceAll("[-!?.,;:\"']", "")
To quote a string into a regular expression, you could also use Pattern.quote which escapes the characters in the string as necessary.
Demo:
String str = "abc-!?.,;:\"'def";
System.out.println(str.replaceAll("[-!?.,;:\"']", "")); // prints abcdef
You might need to escape the double-quotes because you have the string in double-quotes; but as aioobe says, don't escape the rest. Put the - at the end of the group, however.
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.
I have a question about strings in Java. Let's say, I have a string like so:
String str = "The . startup trace ?state is info?";
As the string contains the special character like "?" I need the string to be replaced with "\?" as per my requirement. How do I replace special characters with "\"? I tried the following way.
str.replace("?","\?");
But it gives a compilation error. Then I tried the following:
str.replace("?","\\?");
When I do this it replaces the special characters with "\\". But when I print the string, it prints with single slash. I thought it is taking single slash only but when I debugged I found that the variable is taking "\\".
Can anyone suggest how to replace the special characters with single slash ("\")?
On escape sequences
A declaration like:
String s = "\\";
defines a string containing a single backslash. That is, s.length() == 1.
This is because \ is a Java escape character for String and char literals. Here are some other examples:
"\n" is a String of length 1 containing the newline character
"\t" is a String of length 1 containing the tab character
"\"" is a String of length 1 containing the double quote character
"\/" contains an invalid escape sequence, and therefore is not a valid String literal
it causes compilation error
Naturally you can combine escape sequences with normal unescaped characters in a String literal:
System.out.println("\"Hey\\\nHow\tare you?");
The above prints (tab spacing may vary):
"Hey\
How are you?
References
JLS 3.10.6 Escape Sequences for Character and String Literals
See also
Is the char literal '\"' the same as '"' ?(backslash-doublequote vs only-doublequote)
Back to the problem
Your problem definition is very vague, but the following snippet works as it should:
System.out.println("How are you? Really??? Awesome!".replace("?", "\\?"));
The above snippet replaces ? with \?, and thus prints:
How are you\? Really\?\?\? Awesome!
If instead you want to replace a char with another char, then there's also an overload for that:
System.out.println("How are you? Really??? Awesome!".replace('?', '\\'));
The above snippet replaces ? with \, and thus prints:
How are you\ Really\\\ Awesome!
String API links
replace(CharSequence target, CharSequence replacement)
Replaces each substring of this string that matches the literal target sequence with the specified literal replacement sequence.
replace(char oldChar, char newChar)
Returns a new string resulting from replacing all occurrences of oldChar in this string with newChar.
On how regex complicates things
If you're using replaceAll or any other regex-based methods, then things becomes somewhat more complicated. It can be greatly simplified if you understand some basic rules.
Regex patterns in Java is given as String values
Metacharacters (such as ? and .) have special meanings, and may need to be escaped by preceding with a backslash to be matched literally
The backslash is also a special character in replacement String values
The above factors can lead to the need for numerous backslashes in patterns and replacement strings in a Java source code.
It doesn't look like you need regex for this problem, but here's a simple example to show what it can do:
System.out.println(
"Who you gonna call? GHOSTBUSTERS!!!"
.replaceAll("[?!]+", "<$0>")
);
The above prints:
Who you gonna call<?> GHOSTBUSTERS<!!!>
The pattern [?!]+ matches one-or-more (+) of any characters in the character class [...] definition (which contains a ? and ! in this case). The replacement string <$0> essentially puts the entire match $0 within angled brackets.
Related questions
Having trouble with Splitting text. - discusses common mistakes like split(".") and split("|")
Regular expressions references
regular-expressions.info
Character class and Repetition with Star and Plus
java.util.regex.Pattern and Matcher
In case you want to replace ? with \?, there are 2 possibilities: replace and replaceAll (for regular expressions):
str.replace("?", "\\?")
str.replaceAll("\\?","\\\\?");
The result is "The . startup trace \?state is info\?"
If you want to replace ? with \, just remove the ? character from the second argument.
But when I print the string, it prints
with single slash.
Good. That's exactly what you want, isn't it?
There are two simple rules:
A backslash inside a String literal has to be specified as two to satisfy the compiler, i.e. "\". Otherwise it is taken as a special-character escape.
A backslash in a regular expresion has to be specified as two to satisfy regex, otherwise it is taken as a regex escape. Because of (1) this means you have to write 2x2=4 of them:"\\\\" (and because of the forum software I actually had to write 8!).
String str="\\";
str=str.replace(str,"\\\\");
System.out.println("New String="+str);
Out put:- New String=\
In java "\\" treat as "\". So, the above code replace a "\" single slash into "\\".