我的split（）及其正则表达式出了什么问题？

Question

Part of my application I encountered this problem. The String line variable contains 12.2 Andrew and I'm trying to split them separately but it doesn't work and comes with a NumberFormatException error. Could you guys help me on that please?

String line = "12.2 Andrew";
String[] data = line.split("(?<=\\d)(?=[a-zA-Z])");

System.out.println(Double.valueOf.(data[0]));

Answer 1

Did you look at your data variable? It didn't split anything at all, since the condition never matches. You are looking for a place in the input immediately after a number and before a letter, and since there is a space in between this doesn't exist.

Try adding a space in the middle, that should fix it:

String[] data = line.split("(?<=\\d) (?=[a-zA-Z])");

Answer 2

If you print content of data[0] you will notice that it still contains 12.2 Andrew so you actually didn't split anything. That is because your regex says:
split on place which has digit before and letter after it

which for data like

123foo345bar 123 baz

effectively can only split in places marked with |

123|foo345|bar 123 baz
                  ^it will not split `123 baz` like
                   `123| baz` because after digit is space (not letter)
                   `123 |baz` before letter is space (not digit)
                   so regex can't match it

What you need is to " split on space which has digit before and letter after it " so use

String[] data = line.split("(?<=\\d)\\s+(?=[a-zA-Z])");
//                                  ^^^^ - this represent one ore more whitespaces

Answer 3

Your split is not working, and not splitting the String . Therefore Double.parseDouble is parsing the whole input.

Try the following:

String line = "12.2 Andrew";
String[] data = line.split("(?<=\\d)(?=[a-zA-Z])");
System.out.println(Arrays.toString(data));
// System.out.println(Double.valueOf(data[0]));
// fixed
data = line.split("(?<=\\d).(?=[a-zA-Z])");
System.out.println(Arrays.toString(data));
System.out.println(Double.valueOf(data[0]));

Output

[12.2 Andrew]
[12.2, Andrew]
12.2