日期:2014-05-20  浏览次数:20786 次

一个数组的条件判断
Java code



public class Test6 {
    
    public static void main(String[] args) {
        String[] arrStr1={"00000004", "暂无姓名"};
        String[] arrStr4={"00000004", "暂无姓名"};
        String[] arrStr2={"2011-10-10", "08:08:59"};
        String[] arrStr3={"2011-10-10", "18:08:59"};
        
        
        Map<String[], String[]> map=new HashMap<String[], String[]>();
        map.put(arrStr1, arrStr2);
        map.put(arrStr4, arrStr3);
        
        
        Iterator it=map.keySet().iterator();
        while (it.hasNext()) {
            String[] key=(String[]) it.next();
            String[] value=(String[]) map.get(key);
            System.out.println(Arrays.toString(key)+" "+Arrays.toString(value));
            
        }
        
    }
    
}




如何在key一样的情况下。并且value的第零个元素 也就是日期相等的情况下 计算时间差











------解决方案--------------------
for example
Java code

import java.util.*;
public class Test6 {
    
    public static void main(String[] args) {
        String[] arrStr1={"00000004", "暂无姓名"};
        String[] arrStr4={"00000004", "暂无姓名"};
        String[] arrStr2={"2011-10-10", "08:08:59"};
        String[] arrStr3={"2011-10-10", "18:08:59"};
        
        
        Map<String[], String[]> map=new HashMap<String[], String[]>();
        map.put(arrStr1, arrStr2);
        map.put(arrStr4, arrStr3);

        SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

        List<String[][]> list = new ArrayList<String[][]>();
        for (Map.Entry<String[], String[]> entry : map.entrySet()) {
            list.add(new String[][]{entry.getKey(), entry.getValue()});
        }

        for (int i=0; i<list.size(); i++) {
            String[] key1 = list.get(i)[0];
            String[] value1 = list.get(i)[1];

            for (int j=i+1; j<list.size(); j++) {
                String[] key2 = list.get(j)[0];
                String[] value2 = list.get(j)[1];

                if (Arrays.equals(key1, key2) && value1[0].equals(value2[0])) {
                    Date d1 = sdf.parse(value1[0] + " " + value1[1]);
                    Date d2 = sdf.parse(value2[0] + " " + value2[1]);
                    long dif = Math.abs(d1.getTime() - d2.getTime());
                    System.out.printf("key1=%s, value1=%s\n", Arrays.toString(key1), Arrays.toString(value1));
                    System.out.printf("key2=%s, value2=%s\n", Arrays.toString(key2), Arrays.toString(value2));
                    System.out.printf("hour:%d, minute:%d, second:%d\n", 
                                      (dif/1000)/3600, ((dif/1000)%3600)/60, (dif/1000)%60);
                }
            }
        }                
    }
    
}