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 = "\\^";
Related
I have this regex :
^(([A-Z]:)|((\\|/){1,2}\w+)\$?)((\\|/)(\w[\w ]*.*))+\.([txt|exe]+)$
but every time I assign it to any string, Eclipse returns me invalid escape sequences, I have inserted a backward slash but it gives me the same error.
How to assign the above expression to string in java?
Replace all "\\" with "\\\\". Java has no language support for regular expressions. So you'll need "\\" to get a backslash from the Compiler into the String. If the regular expression shall contain an escaped backslash, you need "\\\\".
final String re = "^(([A-Z]:)|((\\\\|/){1,2}\\w+)\\$?)((\\\\|/)(\\w[\\w ]*.*))+\\.([txt|exe]+)$"
Try the following:
String regex = "^(([A-Z]:)|((\\\\|/){1,2}\\w+)\\$?)((\\\\|/)(\\w[\\w ]*.*))+\\.([txt|exe]+)$";
The backslash character itself needs to be escaped as well, so you would end up with four \ characters.
I don't know why this code doesn't work.
This is my code.
String value[] = pce.getPropertyName().toString().split(".");
the value of pce.getPropertyName is com.newbie.model.Names
when I debug it the size of value is 0.
Anyone encounter this problem?
. has a special meaning in regex-world (specifically, it matches any character), and recall that split() does indeed take a regular expression as an argument. You want
String value[] = pce.getPropertyName().toString().split("\\.");
i.e. escape the ..
You have to escape the dot character, since dot is a meta-character:
String value[] = pce.getPropertyName().toString().split("\\.");
If you want the dot or other characters with a special meaning in regexes to be a normal character, you have to escape it with a backslash. Since regexes in Java are normal Java strings, you need to escape the backslash itself, so you need two backslashes like \\.
Java docs for the same can be found here.
So, this is what you should do.
String value[] = pce.getPropertyName().toString().split("\\.");
I am having some problem with searching for a special character "(".
I got a java.util.regex.PatternSyntaxException exception has occurred.
It might have something to do with "(" being treated as special character.
I am not very good with pattern expression. Can someone help me properly search for the escape character?
// I need to split the string at the "("
String myString = "Room Temperature (C)";
String splitList[] = myString.split ("("); // i got an exception
// I tried this but got compile error
String splitList[] = myString.split ("\(");
Try one of these:
string.split("\\(");
string.split(Pattern.quote("("));
Since a string split takes a regular expression as an argument, you need to escape characters properly. See Jon Skeet's answer on this here:
The reason you got an exception the first time is because split() takes a regular expression as argument, and ( has a special meaning there, as you suggest. To avoid this, you need to escape it using a \, like you tried.
What you missed, is that you also need to escape your backslashes with an extra \ in Java, meaning you need a total of two:
String splitList[] = myString.split ("\\(");
You need to escape the character via backslashes: string.split("\\(");
( is one of regex special characters. To escape it you can use e.g.
split(Pattern.quote("(")),
split("\\Q(\\E"),
split("\\("),
split("[(]").
I have an android application where I have to find out if my user entered the special character '\' on a string. But i'm not obtaining success by using the string.replaceAll() method, because Java recognizes \ as the end of the string, instead of the " closing tag. Does anyone have suggestions of how can I fix this?
Here is an example of how I tried to do this:
private void ReplaceSpecial(String text) {
if (text.trim().length() > 0) {
text = text.replaceAll("\", "%5C");
}
It does not work because Java doesn't allow me. Any suggestions?
Try this: You have to use escape character '\'
text = text.replaceAll("\\\\", "%5C");
Try
text = text.replaceAll("\\\\", "%5C");
replaceAll uses regex syntax where \ is special character, so you need to escape it. To do it you need to pass \\ to regex engine but to create string representing regex \\ you need to write it as "\\\\" (\ is also special character in String and requires another escaping for each \)
To avoid this regex mess you can just use replace which is working on literals
text = text.replace("\\", "%5C");
The first parameter to replaceAll is interpreted as a regular expression, so you actually need four backslashes:
text = text.replaceAll("\\\\", "%5C");
four backslashes in a string literal means two backslashes in the actual String, which in turn represents a regular expression that matches a single backslash character.
Alternatively, use replace instead of replaceAll, as recommended by Pshemo, which treats its first argument as a literal string instead of a regex.
text = text.replaceAll("\", "%5C");
Should be:
text = text.replaceAll("\\\\", "%5C");
Why?
Since the backward slash is an escape character. If you want to represent a real backslash, you should use double \ (\\)
Now the first argument of replaceAll is a regular expression. So you need to escape this too! (Which will end up with 4 backslashes).
Alternatively you can use replace which doesn't expect a regex, so you can do:
text = text.replace("\\", "%5C");
First, since "\" is the escape character in Java, you need to use two backslashes to get one backslash. Second, since the replaceAll() method takes a regular expression as a parameter, you will need to escape THAT backslash as well. Thus you need to escape it by using
text = text.replaceAll("\\\\", "%5C");
I could be late but not the least.
Add \\\\ following regex to enable \.
Sample regex:
private val specialCharacters = "-#%\\[\\}+'!/#$^?:;,\\(\"\\)~`.*=&\\{>\\]<_\\\\"
private val PATTERN_SPECIAL_CHARACTER = "^(?=.*[$specialCharacters]).{1,20}$"
Hope it helps.
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.