Question

What regular expression can get a number sequence from the input string, contains backslashes and not a numbers, for example –

"12\34a56ss7890"

I need to –

1234567890

Answer 1

The following example:

String a ="1\2sas";
String b ="1\\2sas";

System.out.println(a.replaceAll("[a-zA-Z\\\\]",""));
System.out.println(b.replaceAll("[a-zA-Z\\\\]",""));

gives output:

1X
12

where X is not a X but a little rectangle – a symbol which is shown when the text showing control does not know how to draw it, a so called non printable character.

It is because in String a the “\2” part obviously tries to be interpreted as a single escaped sign “\u0002”- similar to “\n” “\t” – you can see this in debugger (i tried it using NetBeans)

Since the first argument of a replaceAll method is passed to [Pattern.compile](http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#replaceAll(java.lang.String, java.lang.String)) it needs to be escaped twice as opposed to String literal (like b).

So if the String “12\34a56ss7890” looks like this on screen you have printed it out like this:

System.out.println("12\\34a56ss7890");

which is solved in the second example.

However if the literal is given as “12\34a56ss7890” then I think you can’t handle it with a single regexp, because if the backslash is followed by a number it gets interpreted as as \u0000 -\u0009 so the best I can think of is a very ugly solution:

str.replaceAll("\u0000","0").replaceAll("\u0001","1") ... .replaceAll("\u0009","9").replaceAll("[^\\d]")

the first then replacements (\u0000-\u0009) might be rewritten as a for loop to make it look elegant.

+1 for an EXCELLENT question 🙂

EDIT:
actually if a backslash is followed by more than one number they all get interpreted as a single sign – up to three numbers after a backslash, the fourth number will be treated as a single number.

Therefore, my solution is not generally correct, but could be extended to be. I would recommend Robin’s solution below as it is far more efficient.