I'm working with some really old Java. 1.3 to be exact.
I'm attempting to sanitize some String input by removing non alphabet characters (punctuation and numbers, etc)
Normally I'd do something like:
String.replaceAll("[^A-Za-z]", "");
However, .replaceAll() was introduced in Java 1.4! So it won't compile! http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replaceAll(java.lang.String, java.lang.String)
How did we accomplish this prior to Java 1.4?
Well probably you can write a simple loop like this:
char[] origArr = str.toCharArray();
char[] destArr = new char[origArr.length];
int j = 0;
for (int i=0; i < origArr.length; i++) {
char c = origArr[i];
if ((c >= 65 && c <= 90) || (c >= 97 && c <= 122))
destArr[j++] = c;
}
String dest = new String(destArr, 0, j);
Sorry don't have JDK1.3 to test it out.
String dest = new String(destArr, 0, j) to make sure dest isn't doing anything with those extra Array elements.
String dest = new String(destArr, 0, j); should be used in the end.See if this works:
public static String replaceAll(
String haystack, // String to search in
String needle, // Substring to find
String replacement) { // Substring to replace with
int i = haystack.lastIndexOf( needle );
if ( i != -1 ) {
StringBuffer buffer = new StringBuffer( haystack );
buffer.replace( i, i+needle.length(), replacement );
while( (i=haystack.lastIndexOf(needle, i-1)) != -1 ) {
buffer.replace( i, i+needle.length(), replacement );
}
haystack = buffer.toString();
}
return haystack;
}
EDIT: This won't support regular expressions. As you're looking to erase more than just a single character, I would suggest you either tweak this code to allow an array of needles or (the ugly method) loop through the disallowed characters and repeatedly call the function.
lastIndexOf(String) does not handle regular expressions, I believe. The Pattern class was not even in java until 1.4You can use jakarta by the Apache Software Foundation.
Jakarta Regexp is a 100% Pure Java Regular Expression package
It's not maintained but the last version is not so old (2011).
The documentation: http://jakarta.apache.org/regexp/
For the replaceAll you can use subst with the REPLACE_ALL flag.
PS: The link is dead, here a mirror to download the lib.
replaceAll under the hood.
Assuming you need only alphabets and anything else to be replaced with blank. Character also got isDigit method. please refer to http://docs.oracle.com/javase/1.3/docs/api/java/lang/Character.html if it helps.
public static void main (String[] args)
{
String yourstring = "2323ABF!@CD24";
char[] check = yourstring.toCharArray();
StringBuffer str = new StringBuffer();
for(int i=0; i < check.length; i++){
if(!Character.isLetter(check[i])){
str.append("");
}
else{
str.append(check[i]);
}
}
System.out.println(str.toString());
}
replaceAll()method, were it me, I would iterate over atoCharArray()version of theStringand build a new string, skipping any characters that fall outside of the alphabet range