Develop Reference

How to escape + character in java?

How to escape + character while using split function call in java?
split declaration
String[] split(String regularExpression)
thats what i did
services.split("+"); //says dongling metacharacter
services.split("\+"); //illegal escape character in string literal
But it allows to do something like this
String regExpr="+";

Since the + is a regex meta-character (denoting an occurrence of 1 or more times), you will have to escape it with \ (which also has to be escaped because it's a meta-character that's being used when describing the tab character, the new line character(s) \r\n and others), so you have to do:
services.split("\\+");

Java and Regex both have special escape sequences, and both of them begin with \.
Your issue lies in writing a string literal in Java. Java's escape sequences are resolved at compile time, long before the string is passed into your Regex engine for parsing.
The sequence "\+" would throw an error as this is not a valid Java string.
If you want to pass \+ into your Regex engine you have to explicitly let Java know you want to pass in a backslash character using "\\+".
All valid Java escape sequences are as follows:
\t Insert a tab in the text at this point.
\b Insert a backspace in the text at this point.
\n Insert a newline in the text at this point.
\r Insert a carriage return in the text at this point.
\f Insert a formfeed in the text at this point.
\' Insert a single quote character in the text at this point.
\" Insert a double quote character in the text at this point.
\\ Insert a backslash character in the text at this point.

should be like this :
services.split("\\+");

Underlined backslash IntelliJ

I am using a backslash as an escape character for a serialization format I am working on. I have it as a constant but IntelliJ is underlining it and highlighting it red. On hover it gives no error messages or any information as to why it does not like it.
What is the reason for this and how do I fix it?

IntelliJ is smarter than I am and realised that I was using this character in a regular expression where 2 backslashes would be needed, however, IntelliJ also assumed that my puny mind could find the problem without giving me any information about it.

If it's being used as a regular expression, then the "\" must be escaped.
If you're escaping a "\" as "\" like traditional regular expressions require, then you also need to add two more \\ for a total of \\\\.
This is because of the way Java interprets "\":
In literal Java strings the backslash is an escape character. The
literal string "\" is a single backslash. In regular expressions, the
backslash is also an escape character. The regular expression \
matches a single backslash. This regular expression as a Java string,
becomes "\\". That's right: 4 backslashes to match a single one.
The regex \w matches a word character. As a Java string, this is
written as "\w".
The same backslash-mess occurs when providing replacement strings for
methods like String.replaceAll() as literal Java strings in your Java
code. In the replacement text, a dollar sign must be encoded as \$ and
a backslash as \ when you want to replace the regex match with an
actual dollar sign or backslash. However, backslashes must also be
escaped in literal Java strings. So a single dollar sign in the
replacement text becomes "\$" when written as a literal Java string.
The single backslash becomes "\\". Right again: 4 backslashes to
insert a single one.

System.setProperty: Invalid Escape Sequence

Im trying to make a reference to a bin.
System.setProperty("mbrola.base", "C:\Users\Name\Desktop\FreeTTS\MBrola Project");
But Im getting this error:
Invalid escape sequence (valid ones are \b \t \n \f \r \" \' \\ )

You want actual backslashes, which are usually part of escape sequences. You must escape the backslashes themselves, with another backslash.
System.setProperty("mbrola.base", "C:\\Users\\Name\\Desktop\\FreeTTS\\MBrola Project");

Yes, because this isn't a valid string literal:
"C:\Users\Name\Desktop\FreeTTS\MBrola Project"
You need to escape the backslashes:
"C:\\Users\\Name\\Desktop\\FreeTTS\\MBrola Project"
The string itself will only have the single backslashes though - you're just escaping it in source code.

java regex expression escape characters

Hi i'm trying to split a string separated by vertical bars. for example:
String str = "a=1|b=2";
In java, we should do like this:
str.split("\\|");
If I use a single slash:
str.split("\|");
compiler gives errors:
Invalid escape sequence (valid ones are \b \t \n \f \r \" \' \ )
Can anyone explain me why this happens? thanks!

Backslash \ is a special character. In the Java world it is used to escape a character.
The pipe | is a special character in the Regex world, which means "OR".
To use the pipe as a separator you need to escape it(so it can be recognized during the regex parsing), so you need to get this in your regex: \|.
But as backshlash is a special character in Java and that you are using a String object, you have to escape the backslash so it can be interpreted as the final expected final result: \|
To do so, you simply escape backslash with another backslash: \\|
The first backslash escapes the second backslash (java requirement) which escapes the pipe (regex requirement).

In Java strings, a backslash needs to be escaped with another backslash. So, while the "canonical" form of the regex is indeed \|, as a Java string, this must be written "\\|".

regular-expression in java

I am using this regular expression in a java file to validate the password.
"^\w*(?=\w*\d)(?=\w*[a-z])(?=\w*[A-Z])\w*$"
It's showing the error :
Invalid escape sequence (valid ones are \b \t \n \f \r \" \' \\ )
Can anybody tell me what mistake I am doing in this?(I don't know anything about regular expressions. I copied it from google.)

In Java string literals you need to escape the backslashes.
"^\\w*(?=\\w*\\d)(?=\\w*[a-z])(?=\\w*[A-Z])\\w*$"
You can also simplify your regular expression by removing the first \\w* as it is not needed.