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nifi regex failed to escape backslash "\" [duplicate]

How to write a regular expression to match this \" (a backslash then a quote)? Assume I have a string like this:
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I need to replace all the \" with a ", so the result would look like:
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This one does not work: str.replaceAll("\\\"", "\"") because it only matches the quote. Not sure how to get around with the backslash. I could have removed the backslash first, but there are other backslashes in my string.

If you don't need any of regex mechanisms like predefined character classes \d, quantifiers etc. instead of replaceAll which expects regex use replace which expects literals
str = str.replace("\\\"","\"");
Both methods will replace all occurrences of targets, but replace will treat targets literally.
BUT if you really must use regex you are looking for
str = str.replaceAll("\\\\\"", "\"")
\ is special character in regex (used for instance to create \d - character class representing digits). To make regex treat \ as normal character you need to place another \ before it to turn off its special meaning (you need to escape it). So regex which we are trying to create is \\.
But to create string literal representing text \\ so you could pass it to regex engine you need to write it as four \ ("\\\\"), because \ is also special character in String literals (part of code written using "...") since it can be used for instance as \t to represent tabulator.
That is why you also need to escape \ there.
In short you need to escape \ twice:
in regex \\
and then in String literal "\\\\"

You don't need a regular expression.
str.replace("\\\"", "\"")
should work just fine.
The replace method takes two substrings and replaces all non-overlapping occurrences of the first with the second. Per the javadoc:
public String replace(CharSequence target,
CharSequence replacement)
Replaces each substring of this string that matches the literal target sequence with the specified literal replacement sequence. The replacement proceeds from the beginning of the string to the end, for example, replacing "aa" with "b" in the string "aaa" will result in "ba" rather than "ab".

try this: str.replaceAll("\\\\\"", "\\\"")
because Java will replace \ twice:
(1) \\\\\" --> \\" (for string)
(2) \\" --> \" (for regex)

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(",", "\\,");

How to split a string with double quotes " as the delimiter?

I tried splitting like this-
tableData.split("\\"")
but it does not work.

It seems that you tried to escape it same way as you would escape | which is "\\|". But difference between | and " is that
| is metacharacter in regex engine (it represents OR operator)
" is metacharacter in Java language in string literal (it represents start/end of the string)
To escape any String metacharacter (like ") you need to place before it other String metacharacter responsible for escaping which is \1. So to create String which would contain " like this is "quote" you would need to write it as
String s = "this is \"quote\"";
// ^^ ^^ these represent " literal, not end of string
Same idea is applied if we would like to create \ literal (we would need to escape it by placing another \ before it). For instance if we would want to create string representing c:\foo\bar we would need to write it as
String s = "c:\\foo\\bar";
// ^^ ^^ these will represent \ literal
So as you see \ is used to escape metacharacters (make them simple literals).
This character is used in Java language for Strings, but it also is used in regex engine to escape its metacharacters:
\, ^, $, ., |, ?, *, +, (, ), [, {.
If you would like to create regex which will match [ character you will need to use regex \[ but String representing this regex in Java needs to be written as
String leftBracketRegex = "\\[";
// ^^ - Remember what was said earlier?
// To create \ literal in String we need to escape it
So to split on [ we would need to invoke split("\\[") because regex representing [ is \[ which needs to be written as "\\[" in Java.
Since " is not special character in regex but it is special in String we need to escape it only in string literal by writing it as
split("\"");
1) \ is also used to create other characters line separators \n, tab \t. It can also be used to create Unicode characters like \uXXXX where XXXX is index of character in Unicode table in hexadecimal form.

You have escaped the \ by putting in \ twice, try
tableData.split("\"")
Why does this happen?
A backslash escapes the following character. Since the next character is another backslash, the second backslash will be escaped, thus the doublequote won't.
Your resulting escaped string is \", where it should really be just ".
Edit:
Also keep in mind, that String.split() interprets its pattern parameter as a regular expression, which has several special characters, which have to be escaped in the resulting string.
So if you want split by a .(which is a special regex character), you need to specify it as String.split("\\."). The first backslash escapes the escaping function of the second backlash and would result in "\.".
In case of regex characters you could also just use Pattern.quote(); to escape your desired delimiter, but this is far out of the scope the question orignally had.

Try with single backslash \
tableData.split("\"")

Try like this by escaping " with single backslash \ :
tableData.split("\"")

You are not escaping properly. The snippet code will not even compile because of it. The correct way to do it is
tableData.split("\"");

A single backslash will do the trick.
Like this:
tableData.split("\"");

You can actually split without the backward slash. You only have to use single quote
tableData.split('"');

Java Regex Escape Characters

I'm learning Regex, and running into trouble in the implementation.
I found the RegexTestHarness on the Java Tutorials, and running it, the following string correctly identifies my pattern:
[\d|\s][\d]\.
(My pattern is any double digit, or any single digit preceded by a space, followed by a period.)
That string is obtained by this line in the code:
Pattern pattern =
Pattern.compile(console.readLine("%nEnter your regex: "));
When I try to write a simple class in Eclipse, it tells me the escape sequences are invalid, and won't compile unless I change the string to:
[\\d|\\s][\\d]\\.
In my class I'm using`Pattern pattern = Pattern.compile();
When I put this string back into the TestHarness it doesn't find the correct matches.
Can someone tell me which one is correct? Is the difference in some formatting from console.readLine()?

\ is special character in String literals "...". It is used to escape other special characters, or to create characters like \n \r \t.
To create \ character in string literal which can be used in regex engine you need to escape it by adding another \ before it (just like you do in regex when you need to escape its metacharacters like dot \.). So String representing \ will look like "\\".
This problem doesn't exist when you are reading data from user, because you are already reading literals, so even if user will write in console \n it will be interpreted as two characters \ and n.
Also there is no point in adding | inside class character [...] unless your intention is to make that class also match | character, remember that [abc] is the same as (a|b|c) so there is no need for | in "[\\d|\\s]".

If you want to represent a backslash in a Java string literal you need to escape it with another backslash, so the string literal "\\s" is two characters, \ and s. This means that to represent the regular expression [\d\s][\d]\. in a Java string literal you would use "[\\d\\s][\\d]\\.".
Note that I also made a slight modification to your regular expression, [\d|\s] will match a digit, whitespace, or the literal | character. You just want [\d\s]. A character class already means "match one of these", since you don't need the | for alternation within a character class it loses its special meaning.

My pattern is any double digit or single digit preceded by a space, followed by a period.)
Correct regex will be:
Pattern pattern = Pattern.compile("(\\s\\d|\\d{2})\\.");
Also if you're getting regex string from user input then your should call:
Pattern.quote(useInputRegex);
To escape all the regex special characters.
Also you double escaping because 1 escape is handled by String class and 2nd one is passed on to regex engine.

What is happening is that escape sequences are being evaluated twice. Once for java, and then once for your regex.
the result is that you need to escape the escape character, when you use a regex escape sequence.
for instance, if you needed a digit, you'd use
"\\d"

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;