Develop Reference

How Java replaceAll operation works with backslashes?

Why do I need four backslashes (\) to add one backslash into a String?
String replacedValue = neName.replaceAll(",", "\\\\,");
Here in above code you can check I have to replace all commas (,) from \, but I have to add three more backslash (\) ?
Can anybody explain this concept?

Escape once for Java, and a second time for regexp.
\ -> \\ -> \\\\
Or since you're not actually using regular expressions, take khelwood's advice and use replace(String,String) so you need to only escape once.

The documentation of String.replaceAll(regex, replacement) states:
Note that backslashes (\) and dollar signs ($) in the replacement string may cause the results to be different than if it were being treated as a literal replacement string; see Matcher.replaceAll.
The documentation of Matcher.replaceAll(replacement) then states:
backslashes are used to escape literal characters in the replacement string
So to put this more clearly, when you replace with \,, it is as if you were escaping the comma. But what you want is really the \ character, so you should escape it with \\,. Since that in Java, \ also needs to be escaped, the replacement String becomes \\\\,.
If you are having a hard time remembering all this, you can use the method Matcher.quoteReplacement(s), whose goal is to correctly escape the replacement part. Your code would become:
String replacedValue = neName.replaceAll(",", Matcher.quoteReplacement("\\,"));

\ is used for escape sequence
For example
go to next line then use \n or \r
for tab \t
likewise to print \ which is special in string literal you have to escape it with another \ which gives us \\
Now replaceAll should be used with a regex, since you're not using a regex, use replace as suggested in the comments.
String s = neName.replace(",", "\\,");

You have to first escape the backslash because it's a literal (giving \\), and then escape it again because of the regular expression (giving \\\\).
Therefore this -
String replacedValue = neName.replaceAll(",", "\\\\,"); // you need ////
You can use replace instead of replaceAll-
String replacedValue = neName.replace(",", "\\,");

proper way to pattern match with escaped char

Pattern.matches("123$45","123$45") returns false, I presume because of the special $ char.
My suspicion was that escaping the $ would make it pass
e.g. Pattern.matches("123\$45","123\$45")
But this also fails.
What is the proper way to make sure they match?

This is the "canonical" regex which is \$, but here this is a Java string. And in a Java string, a \ is written "\\". Therefore:
"123\\$45"
As to your target string, it just needs to be "123$45".

If the pattern you are looking for is fixed pattern, then manually escape the '$' character so that it isn't treated as a regex metacharacter; i.e.
boolean itMatches = Pattern.matches("123\\$45", "123$45");
The '$' is escaped at the level of the String object using a single backslash. However, since we are expressing this using a String literal, and backslash is the escape character for string literals, we need to (string) escape the (regex) escape character. Hence, we need two backslashes ... here.
If you don't escape the escape, the Java compiler says in effect "I don't recognize "\$" as a valid String literal escape sequence. ERROR!".
On the other hand, if the pattern input or generated, then you can use Pattern.quote() to quote it; i.e.
String literal = "123$45"; // ... or any literal string you want to match.
boolean itMatches = Pattern.matches(Pattern.quote(literal), "123$45");

Why does .split("\\") generate an exception?

I have a String representing a directory, where \ is used to separate folders. I want to split based on "\\":
String address = "C:\\saeed\\test";
String[] splited = address.split("\\");
However, this is giving me a java.util.regex.PatternSyntaxException.

As others have suggested, you could use:
String[] separated = address.split("\\\\");
or you could use:
String[] separated = address.split(Pattern.quote("\\"));
Also, for reference:
String address = "C:\saeed\test";
will not compile, since \s is not a valid escape sequence. Here \t is interpreted as the tab character, what you actually want is:
String address = "C:\\saeed\\test";
So, now we see that in order to get a \ in a String, we need "\\". The regular expression \\ matches a single backslash since \ is a special character in regex, and hence must be escaped. Once we put this in quotes, aka turn it into a String, we need to escape each of the backslashes, yielding "\\\\".

String#split() method takes a regex. In regex, you need to escape the backslashes. And then for string literals in Java, you need to escape the backslash. In all, you need to use 4 backslashes:
String[] splited = address.split("\\\\");

\ has meaning as a part of the regex, so it too must be quoted. Try \\\\.
The Java will have at \\\\, and produce \\ which is what the regex processor needs to obtain \.

You need to use \\\\ instead of \\.
The backslash(\) is an escape character in Java Strings.If you want to use backslash as a literal you have to type \\\\ ,as \ is also a escape character in regular expressions.
For more details click here

Use separators:
String address = "C:\saeed\test";
String[] splited = address.split(System.getProperty("file.separator"));

String's replaceAll() method and escape characters

The line
System.out.println("\\");
prints a single back-slash (\). And
System.out.println("\\\\");
prints double back-slashes (\\). Understood!
But why in the following code:
class ReplaceTest
{
public static void main(String[] args)
{
String s = "hello.world";
s = s.replaceAll("\\.", "\\\\");
System.out.println(s);
}
}
is the output:
hello\world
instead of
hello\\world
After all, the replaceAll() method is replacing a dot (\\.) with (\\\\).
Can someone please explain this?

When replacing characters using regular expressions, you're allowed to use backreferences, such as \1 to replace a using a grouping within the match.
This, however, means that the backslash is a special character, so if you actually want to use a backslash it needs to be escaped.
Which means it needs to actually be escaped twice when using it in a Java string. (First for the string parser, then for the regex parser.)

The javadoc of replaceAll says:
Note that backslashes ( \ ) and dollar signs ($) in the replacement
string may cause the results to be different than if it were being
treated as a literal replacement string; see Matcher.replaceAll. Use
Matcher.quoteReplacement(java.lang.String) to suppress the special
meaning of these characters, if desired.

This is a formatted addendum to my comment
s = s.replaceAll("\\.", Matcher.quoteReplacement("\\"));
IS MORE READABLE AND MEANINGFUL THAN
s = s.replaceAll("\\.", "\\\\\\");

If you don't need regex for replacing and just need to replace exact strings, escape regex control characters before replace
String trickyString = "$Ha!I'm tricky|.|";
String safeToUseInReplaceAllString = Pattern.quote(trickyString);

The backslash is an escape character in Java Strings. e.g. backslash has a predefined meaning in Java. You have to use "\ \" to define a single backslash. If you want to define " \ w" then you must be using "\ \ w" in your regex. If you want to use backslash you as a literal you have to type \ \ \ \ as \ is also a escape character in regular expressions.

I believe in this particular case it would be easier to use replace instead of replace all.
Reverend Gonzo Has the correct answer when he talks about escaping the character.
Using replaceAll:
s = s.replaceAll("\\.", "\\\\\\\\");
Using replace:
s = s.replaceAll(".", "\\");
replace just takes a string to match to, not a regular expression.

I don't like this implementation of regex. We should be able to escape characters with a single '\' , not '\'. But anyway if you want to get THIS.Out_Of_That you can do:
String prefix = role.replaceFirst("(\\.).*", "");
So you get prefix = THIS;

String format using java

I have to make below statement as string.i am trying,but it's giving invalid character sequence.I know it is basic,But not able to do this.any help on this appreciated.
String str="_1";
'\str%' ESCAPE '\'
Output should be: '\_1%' ESCAPE '\'.
Thanks,
Chaitu

String result = "'\\" + str + "%' ESCAPE '\\'";

Inside a string, a backslash character will "escape" the character after it - which causes that character to be treated differently.
Since \ has this special meaning, if you actually want the \ character itself in the string, you need to put \\. The first backslash escapes the second, causing it to be treated as a literal \ inside the string.
Knowing this, you should be able to construct the resulting string you need. Hope this helps.

String str="_1";
String source = "'\\str%' ESCAPE '\\'";
String result = source.replaceAll("str", str);
Another way to implement string interpolation. The replaceAll function finds all occurrences of str in the source string and replaces them by the passed argument.
To encode the backslash \ in a Java string, you have to duplicate it, because a single backslash works as an escape character.
Beware that the first argument if replaceAll is actually a regular expression, so some characters have a special meaning, but for simple words it will work as expected.

String str="_1";
String output = String.format("'\\%s%%' ESCAPE '\\'",str);
System.out.println(output);//prints '\_1%' ESCAPE '\'