how to split the string into string and integer in java

问题

I have the String a="abcd1234" and i want to split this into String b="abcd" and Int c=1234 This Split code should apply for all king of input like ab123456 and acff432 and so on. How to split this kind of Strings. Is it possible?

回答1:

You could try to split on a regular expression like (?<=\D)(?=\d). Try this one:

String str = "abcd1234";
String[] part = str.split("(?<=\\D)(?=\\d)");
System.out.println(part[0]);
System.out.println(part[1]);

will output

abcd
1234

You might parse the digit String to Integer with Integer.parseInt(part[1]).

回答2:

You can do the next:

Split by a regex like split("(?=\\d)(?<!\\d)")
You have an array of strings with that and you only have to parse it.

回答3:

Use a regular expression:

Pattern p = Pattern.compile("([a-z]+)([0-9]+)");
Matcher m = p.matcher(string);
if (!m.find())
{ 
  // handle bad string
}
String s = m.group(1);
int i = Integer.parseInt(m.group(2));

I haven't compiled this, but you should get the idea.

回答4:

A brute-force solution.

String a = "abcd1234";
int i;
for(i = 0; i < a.length(); i++){
    char c = a.charAt(i);
    if( '0' <= c && c <= '9' )
        break;
}
String alphaPart = a.substring(0, i);
String numberPart = a.substring(i);

回答5:

public static void main(String... s) throws Exception {
        Pattern VALID_PATTERN = Pattern.compile("([A-Za-z])+|[0-9]*");
    List<String> chunks = new ArrayList<String>();
    Matcher matcher = VALID_PATTERN.matcher("ab1458");
    while (matcher.find()) {
        chunks.add( matcher.group() );
    }
}