ip字段 phone 字段 排序

Question

public static List <Record> sortip(List <Record> recordList){
compareip comparator=new compareip();
Collections.sort(recordList, comparator);
return recordList;
}调用接口为
class compareipimplements Comparator{
public int compare(Object arg0, Object arg1) {
Record user0=(Record)arg0;
Record user1=(Record)arg1;
  return user0.getip().compareTo(user1.getip());
}   
}1程序执行没有错误 瞎奶有个 字段是ip地址  （如12。121。233。2）
等。 发现compareTo没有能力对 ip字段进行排序。  
现在如何修改代码才鞥让ip字段的 list 排序？？
2另外有个字段是 phone 
表里这个字段值不规范 如
[phone]
[phone]
809002
[phone]
。
还有很多空值
compareto也没偶能力为这个字段进行排序
（我的想法先判度是否有＋ 如果有去掉
号码大小来排序）

Answer 1

import java.util.Comparator;
public class Record  implements Comparator<Record>{
private String ip;
public Record(String ip){
this.ip=ip;
}
public int compare(Record ra, Record rb) {
int raIp=Integer.parseInt(ra.ip.split("\\.")[0]);
int rbIp=Integer.parseInt(rb.ip.split("\\.")[0]);
return raIp>rbIp?1:(ra.ip.equals(rb.ip)?1:-1);
}
public String getIp(){
return this.ip;
}
}
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import com.state.cn.Record;
public class Test {
public static void main(String[] args){
List<Record> list=new ArrayList<Record>();
list.add(new Record("192.168.2.254"));
list.add(new Record("172.168.2.254"));
list.add(new Record("123.168.2.254"));
list.add(new Record("91.168.2.254"));
list.add(new Record("27.168.2.254"));
list.add(new Record("36.168.2.254"));
list.add(new Record("54.168.2.254"));
Collections.sort(list,new Comparator<Record>(){
public int compare(Record ra, Record rb) {
int raIp=Integer.parseInt(ra.getIp().split("\\.")[0]);
int rbIp=Integer.parseInt(rb.getIp().split("\\.")[0]);
return raIp>rbIp?1:(ra.getIp().equals(rb.getIp())?1:-1);
}
});
for(Record r:list)
System.out.println(r.getIp());
}
}

Answer 2

当然Record类中没有必要实现接口了，我写在test类中了public class Record{
private String ip;
public Record(String ip){
this.ip=ip;
}
public String getIp(){
return this.ip;
}
}

Answer 3

getip()获取的数据是什么格式的？
我建议把ip存成整数排序会更好看一些

Answer 4

感谢　closewbq
第２个问题　2另外有个字段是 phone 
表里这个字段值不规范 如 
[phone] 
[phone] 
809002 
[phone] 
。 
还有很多空值 
compareto也没偶能力为这个字段进行排序 
（我的想法先判度是否有＋ 如果有去掉 
号码大小来排序） 
－－－－－－－－－－－－有人建议用正则表达式做　（从字符串中数字开头的地方取数字
如　pattern.compile  (+(\\d))?? 不知道如何写这个　并应用
）

Answer 5

 Collections.sort(list,new Comparator<Record>(){
            public int compare(Record ra, Record rb) {
                int raIp=Integer.parseInt(ra.getIp().split("\\.")[0]);
                int rbIp=Integer.parseInt(rb.getIp().split("\\.")[0]);
                return raIp>rbIp?1:(ra.getIp().equals(rb.getIp())?1:-1);
            }
        });这只是比较每个IP的第一个字节比如我输入input　
list.add(new Record("1","1","1","1","1","1","1","1","1","192.168.2.254"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","192.168.2.25"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","123.168.2.254"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","91.168.2.254"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","192.168.112.25"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","36.168.2.254"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","54.168.2.254"));my结果
36.168.2.254
54.168.2.254
91.168.2.254
123.168.2.254
192.168.2.254
192.168.2.25
192.168.112.25实际上第２个　第３　第４都需要判断的　
比如
192.168.2.25
应该在192.168.2.254
的前面
192.168.112.25
应该在１９２的最前面
36.168.2.254
54.168.2.254
91.168.2.254
192.168.112.25
192.168.2.25
123.168.2.254
192.168.2.254
请问代码如何再改写哦

Answer 6

private static long ipToLong(String strIP)    {
         int j=0;
         int i=0;
         long [] ip=new long[4];
         int position1=strIP.indexOf(".");
         int position2=strIP.indexOf(".",position1+1);
         int position3=strIP.indexOf(".",position2+1);  
         ip[0]=Long.parseLong(strIP.substring(0,position1));
         ip[1]=Long.parseLong(strIP.substring(position1+1,position2));
         ip[2]=Long.parseLong(strIP.substring(position2+1,position3));
         ip[3]=Long.parseLong(strIP.substring(position3+1));
         return (ip[0]<<24)+(ip[1]<<16)+(ip[2]<<8)+ip[3]; //ip1*256*256*256+ip2*256*256+ip3*256+ip4
    } public static void main(String[] args) throws Exception {
List<Record> list=new ArrayList<Record>();
        list.add(new Record("1","1","1","1","1","1","1","1","1","192.168.2.254"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","192.168.2.25"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","123.168.2.254"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","91.168.2.254"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","192.168.112.25"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","36.168.2.254"));
        list.add(new Record("1","1","1","1","1","1","1","1","1","54.168.2.254"));
        Collections.sort(list,new Comparator<Record>(){
            public int compare(Record ra, Record rb) {
               // if   ((startDate==null)   ||   (startDate.equals("")))  
            
             //int raIp=Integer.parseInt(ra.getIpStr().split("\\.")[0]);
                //int rbIp=Integer.parseInt(rb.getIpStr().split("\\.")[0]);
             long a;
             long b;            
             a=ipToLong(ra.getIpStr());
             b=ipToLong(rb.getIpStr());
                return a>b?1:-1;
             //if(a>b)

Answer 7

int rbIp=Integer.parseInt(rb.getIpStr().split("\\.")[0]); 
这个仅仅比较了第１个int只是还不是很理想的那种，需要４个int都进行比较的那种如何写？