我有这个任务。 输入是TreeMap中的TreeMap,无法从第二张地图获取值
IP=192.23.30.40 message='Hello&derps.' user=destroyer
IP=192.23.30.41 message='Hello&yall.' user=destroyer
IP=192.23.30.40 message='Hello&hi.' user=destroyer
IP=192.23.30.42 message='Hello&Dudes.' user=destroyer
end
输出是:
child0:
192.23.33.40 => 1.
child1:
192.23.30.40 => 1.
destroyer:
192.23.30.42 => 2.
mother:
FE80:0000:0000:0000:0202:B3FF:FE1E:8329 => 1.
unknown:
192.23.155.40 => 1.
yetAnotherUsername:
192.23.50.40 => 2.
基本上我有打印每IP和多少邮件每一个用户,他们在格式已将
username:
IP => count, IP => count… (last must be with dot in end)
的问题是我不知道如何添加(第27行代码为trow异常,也不知道为什么)+1给IP。 以及如何捕获最后一个键的最后一个值。
package notReady;
import java.util.Scanner;
import java.util.TreeMap;
/**
* Created by Philip on 10-Jun-16.
*/
public class Problem09_02_UserLogs {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
TreeMap<String, TreeMap<String, Integer>> map = new TreeMap<>();
while (true) {
String in = scanner.nextLine();
if (in.equals("end")) {
break;
}
String[] input = in.split(" ");
String ip = input[0].replaceAll("IP=", "");
String user = input[2].replaceAll("user=", "");
if (!map.containsKey(user)) {
map.put(user, new TreeMap<>());
map.get(user).put(ip, 1);
}else {
Integer tempCount = (map.get(user).get(ip)) + 1;
map.get(user).put(ip, tempCount);
}
}
for (String person : map.keySet()) {
System.out.printf("%s: \n", person);
for (String ips : map.get(person).keySet()) {
System.out.printf("%s => %d.", ips, map.get(person).get(ips));
}
System.out.println();
}
}
}
这是下一个测试,这里一切正常。 输入:
IP=FE80:0000:0000:0000:0202:B3FF:FE1E:8329 message='Hey&son' user=mother
IP=192.23.33.40 message='Hi&mom!' user=child0
IP=192.23.30.40 message='Hi&from&me&too' user=child1
IP=192.23.30.42 message='spam' user=destroyer
IP=192.23.30.42 message='spam' user=destroyer
IP=192.23.50.40 message='' user=yetAnotherUsername
IP=192.23.50.40 message='comment' user=yetAnotherUsername
IP=192.23.155.40 message='Hello.' user=unknown
end
输出:
destroyer:
192.23.30.40 => 2, 192.23.30.41 => 1, 192.23.30.42 => 1.
感谢您的帮助:)对不准确的错误。 – Phil