Q: How to apply the escape characters in java string My problem is
I read a text from a file
In the text file there is some text like that
Base64.valueOf("\030)?8*\"q1;16=\0331:?5/8~:8(6y", 42 - -33))

I able to get the text using regular expression
\030)?8*\"q1;16=\0331:?5/8~:8(6y

but the String is equal below statement
regText = "\\030)?8*\\"q1;16=\\0331:?5/8~:8(6y"

But the exact string I want is
expectedText = "\030)?8*\"q1;16=\0331:?5/8~:8(6y"

So how can I turn the regText to expectedText
In here it do almost what I expected
Except it is not turn the "\030" to "\030" instead it turn to "030"

A: Your problem becomes easier if the program writing to the file writes the data properly to begin with. If that is not possible use the StringEscapeUtils from apache commons lang.
System.out.println(StringEscapeUtils.unescapeJava("New line: \\n Tab: \\t"));

That call manually processes \n and converts it to a newline. It only handles literals. You are on your own for octals and other escapes. You can always fork the code and work on the octals too.

A: By using String#replaceAll you can do the following :
expectedText = regText.replaceAll("\\\\", "\\");

Which will replace the double backslashes in the regText string, to the simple backslashes that you expect.
