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Java - escaping double quotes in string from file

I have html string from file. I need to escape all double quotes. So I do this way:
String content=readFile(file.getAbsolutePath(), StandardCharsets.UTF_8);
content=content.replaceAll("\"","\\\"");
System.out.println(content);
However, the double quotes are not escaped and the string is the same as it was before replaceAll method. When I do
String content=readFile(file.getAbsolutePath(), StandardCharsets.UTF_8);
content=content.replaceAll("\"","^^^");
System.out.println(content);
All double quotes are replaced with ^^^.
Why content.replaceAll("\"","\\\""); doesn't work?

You need to use 4 backslashes to denote one literal backslash in the replacement pattern:
content=content.replaceAll("\"","\\\\\"");
Here, \\\\ means a literal \ and \" means a literal ".
More details at Java String#replaceAll documentation:
Note that backslashes (\) and dollar signs ($) in the replacement string may cause the results to be different than if it were being treated as a literal replacement string; see Matcher.replaceAll
And later in Matcher.replaceAll documentation:
Dollar signs may be treated as references to captured subsequences as described above, and backslashes are used to escape literal characters in the replacement string.
Another fun replacement is replacing quotes with dollar sign: the replacement is "\\$". The 2 \s turn into 1 literal \ for the regex engine and it escapes the special character $ used to define backreferences. So, now it is a literal inside the replacement pattern.

You need to do :
String content = "some content with \" quotes.";
content = content.replaceAll("\"", "\\\\\"");
Why will this work?
\" represents the " symbol, while you need \".
If you add a \ as a prefix (\\") then you'll have to escape the prefix too, i.e. you'll have a \\\". This will now represent \", where \ is not the escaping character, but the symbol \.
However in the Java String the " character will be escaped with a \ and you will have to replace it as well. Therefore prefixing again with \\ will do fine:
x = x.replaceAll("\"", "\\\\\"");

It took me way too long in Java to discover Pattern.quote and Matcher.quoteReplacement. These will you achieve what you are trying to do here - which is a simple "find" and "replace" - without any regex and escape logic. The Pattern.quote here would not be necessary but it shows how you can ensure that the "find" part is not interpreted as a regex string:
#Test
public void testEscapeQuotes()
{
String content="some content with \"quotes\".";
content=content.replaceAll(Pattern.quote("\""), Matcher.quoteReplacement("\\\""));
Assert.assertEquals("some content with \\\"quotes\\\".", content);
}
Remember that you can also use the simple .replace method which will also "replaceAll" but will not interpret your parameters as regular expressions:
#Test
public void testEscapeQuotes()
{
String content="some content with \"quotes\".";
content=content.replace("\"", "\\\"");
Assert.assertEquals("some content with \\\"quotes\\\".", content);
}

Much easier with Apache Commons Text-
System.out.println(StringEscapeUtils.escapeJava("\""));
Output:
\"

Honestly, I am surprised by the behaviour, but it seems like you need to double-escape the backslash:
System.out.println("\"Hello world\"".replaceAll("\"", "\\\\\""));
which outputs:
\"Hello world\"
Demo

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('"');

How does string.replaceAll() work?

I am making a program that replaces a certain part of the string.
String x = "hello";
x=x.replaceAll("e","\\\\s");
System.out.println(x);
output: h\sllo
but for
System.out.println("\\s");
output: \s
why do we need extra escape characters in the first case.

You need \\ for a single \ character in regex
But Java string also interprets backslash therefore you need to escape each \ for String hence you need 2+2=4 backslashes to match a single \ (2 for String and 2 for regex engine)
Also note that 2nd argument to String#replaceAll method is also interpreted by regex engine due to potential presence of back-references and that is the reason same regex rules apply for replacement string also.
Your regex is using replacement string of a literal \ followed by a literal s

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"