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Escape any non-alphanumeric characters in a string using Regex

I would like to escape non-alphanumeric characters occurring in a string as follows:
Say, the original string is: "test_", I would like to transform as "test\_".
In order to do this, one approach I can take by scanning the original string, and constructing a new string and while a non-alphanumeric character is found, append a '\' in front of this character.
But I am wondering if there is any cleaner approach to do the same using regular expression.

You can use the replaceable parameter as shown below:
public class Main {
public static void main(String[] args) {
String s = "test_";
s = s.replaceAll("[^\\p{Alnum}]", "\\\\$0");
System.out.println(s);
}
}
Output:
test\_
Notes:
$0 represents the string matched by the complete regex pattern, [^\\p{Alnum}].
\p{Alnum} specifies alphanumeric character and ^ inside [] is used to negate the pattern. Learn more about patterns from the documentation.
Notice the extra pair of \\ which is to escape \ that has been used to escape \.

How to check weather string contains backslash?

I want to check if a string contains "\" or not. I used the string method contains(), but its not working
String path = "D:\steve\";
if(path.contains("\"){
path = path.replaceAll("\\\\","");
}

Use escape character \
public class test{
public static void main(String[] args) {
String str = "a\\b";
System.out.println(str.contains("\\"));
}
}
A character preceded by a backslash (\) is an escape sequence and has
special meaning to the compiler.
https://docs.oracle.com/javase/tutorial/java/data/characters.html

Representing ^A (Unicode \u0001) correctly in Java when passed as argument

public class TestU {
public static void main(String[] args) {
String str = "\u0001";
System.out.println("str-->"+str);
System.out.println("arg[0]-->"+args[0]);
}
}
Output :
str-->^A
arg[0]-->\u0001
I am passing arg[0] as \u0001
I executed this code in linux, the command line variable is not taken as unicode special character.

The argument you pass from command line is not actually unicode character but it's a String of unicode character which is escaped with \. Ultimately, your String will become \\u0001 and that's why it is printing \u0001. Same way, if you enter \ as a command line argument it will become \\ to escape your backward slash.
While the String you have declared in main is actually unicode character.
String escapedstring = "\\u0001";//in args[0]
String unicodeChar = "\u0001";// in str
So, now you want \\u0001 to be converted into \u0001 and there are lot of ways to achieve that. i.e. you can use StringEscapeUtils#unescapeJava method of utility or you can also try following way.
String str = "\\u0001";
char unicodeChar = (char) Integer.parseInt(str.substring(2));
System.out.println(unicodeChar);
NOTE : You can find other ways to convert unicode String to unicode characters in following question.(Already provided in comment by Marcinek)
How to convert a string with Unicode encoding to a string of letters

Regular expression for password at least 8 characters with alphanumeric/special characters

My test method is here and all the other inputs are fine but the problem is can't test on the string containing \ backslash character,
public static void main(String[] args) {
String reg2 = "^[a-zA-Z0-9#\\#$%&*()_+\\]\\[';:?.,!^-]{8,}$";
System.out.println("".matches(reg2));
System.out.println("Pass".matches(reg2));
System.out.println("P#ssw0r\\".matches(reg2)); //not work
System.out.println("P#ssw0r$".matches(reg2));
System.out.println("P#ssw0r(".matches(reg2));
System.out.println("P#ssw0r)".matches(reg2));
System.out.println("P#ssw0r_".matches(reg2));
System.out.println("P#ssw0r+".matches(reg2));
System.out.println("P#ssw0r'".matches(reg2));
System.out.println("P#ssw0r;".matches(reg2));
System.out.println("P#ssw0r:".matches(reg2));
System.out.println("P#ssw0r?".matches(reg2));
System.out.println("P#ssw0r.".matches(reg2));
System.out.println("P#ssw0r.][".matches(reg2));
}
Please suggest my regular expression to work on the string with \ character, and guide me if my approach is wrong.

Escapse \ character twice (one for java and one for regex)
String reg2 = "^[a-zA-Z0-9#\\\\#$%&*()_+\\]\\[';:?.,!^-]{8,}$";

How to replace \ followed by letters in java String?

I need to delete all tokens that are started with \ and followed by any characters.
I created such a pattern:
input.replaceAll("\\[a-zA-Z0-9]*", "");
But it doesn't work because it doesn't delete \rad from string 5 4\rad.
EDIT:
public static void main(String[] args)
{
String input="Wolf 3 3\4par";
String replaceAll = input.replaceAll("\\\\[a-zA-Z0-9]*", "");
System.out.println("replaceAll=" + replaceAll);
}
Thank you!

The \ is special both in string literals and in regular expressions. To put an actual \ in a regular expression, you have to escape it twice. You also have to assign the result somewhere, which it wasn't clear from your question you were doing. So:
input = input.replaceAll("\\\\[a-zA-Z0-9]*", "");
Complete example: Live Copy
import java.util.*;
public class Temp {
public static void main(String[] args) {
String input = "4 5 \\rad";
input = input.replaceAll("\\\\[a-zA-Z0-9]*", "");
System.out.println(input);
}
}
Output:
4 5

To create \ literal in regex you need to pass \\ to regex engine. But to create \ literal in String you also have to escape it so you need to write it as "\\".
\ literal in regex engine
\\ regex pattern
"\\\\" String representing regex pattern
Now you are using one \ in your regex pattern regex engine sees it as \[ which escapes [ making it simple literal.
Try this way
input.replaceAll("\\\\[a-zA-Z0-9]*", "");

From
Sorry, but my String is exactly 5 4\rad. Indeed how to delete \rad? – Volodymyr Levytskyi
Try
String k= "5 4\rad";
System.out.println(k.replaceAll("\r\\w*", ""));
Output
5 4