I'm getting a date from a web (html): " abril 2013 Viernes 19"
I've tried all normal regex with no success.
Finally I discovered the string bytes (str.getBytes()), and this are the values:
[-96, 97, 98, 114, 105, 108, -96, -96, -96, 50, 48, 49, 51, -96, -96, 86, 105, 101, 114, 110, 101, 115, -96, 49, 57]
What are this -96?
how to replace 1 or more -96 or whatever empty space is by 1 space?
The byte -96 (A0 in hexadecimal, or 160 as an unsigned byte), is the non-breaking space in the ISO-8859-1 character encoding, which is probably the encoding you used to transform the string to bytes.
The first byte (-96) is negative because in Java bytes are signed. It corresponds to character 160 (256 - 96), which is a non-breaking space. You'll need to specify that character directly in your regular expression.
str = str.replaceAll(String.valueOf((char) -96), " ");
You should be able to use the Character.isSpaceChar function to do this. As mentioned in a response to a related question, you can use it in a java regex like this:
String sampleString = "\u00A0abril\u00A0\u00A02013\u00A0Viernes\u00A019";
String result = sampleString.replaceAll("\\p{javaSpaceChar}", " ");
I think that will do exactly what you want while avoiding any need to deal with raw bytes.
I fixed this way (please if anyone have a better answer I'll appreciate it):
byte[] b=str.getBytes();
for (int i = 0; i < b.length; i++) {
if (b[i]==-96)
b[i]=" ".getBytes()[0];
}
String strOut=new String(b).trim();
Pattern blank=Pattern.compile("\\s+|\b+|\t+|\n+|\f+|\r+");
strOut=blank.matcher(strOut).replaceAll(" ");
Thanks every body for help!