Question

我有两个数组：

SELECT * RECORDS
WHERE date >= date('2017-01-23') AND date <= date('2017-01-24')
ORDER BY DATE,TIME

数组E中的元素如下：float [] E; float [] Location;

数组位置中的元素为{1900, 16400, 77666, 8000, 13200, 15600}

这些数据是从我们关联的数据库中提取的，意思是：

{Birmingham, Blackburn, London, Luton, Manchester, Newcastle}

我希望能够确保位置与正确的数据相关联，如上所示，如果有意义的话。有没有办法做到这一点？

提前致谢！

Answer 1

嗯，他们已经通过索引关联了。 E中索引0处的条目与Location中索引0处的条目相关。

但是我会通过不来解决这个问题。相反，我有一个存储一个对象的数组，其中（例如）浮动1900（我不会使用float;至少使用double）和字符串{ {1}}。

"Birmingham"

然后：

class SomeAppropriateName {
    private double distance;
    private String location;

    SomeAppropriateName(double distance, String _location) {
        this.distance = _distance;
        this.location = _location;
    }

    // ...add getters and setters as appropriate...
}

Answer 2

您可以使用android.util.Pair<F,S>。

List<Pair<String,Double>> pairs = Arrays.asList(
    new Pair("Birmingham", 1900),
    new Pair("Blackburn", 16400),
    new Pair("London", 77666),
    new Pair("Luton", 8000),
    new Pair("Manchester", 13200),
    new Pair("Newcastle", 15600)
);

for (Pair<String, Double> pair : pairs) {
    Log.d("log", pair.first + " - " + pair.second);
}

Answer 3

只需使用HashMap！

HashMap是一个具有键值的数组。你可以迭代它等等。例如这段代码：

 HashMap<Integer, String> hmap = new HashMap<Integer, String>();

  /*Adding elements to HashMap*/
  hmap.put(12, "Chaitanya");
  hmap.put(2, "Rahul");
  hmap.put(7, "Singh");
  hmap.put(49, "Ajeet");
  hmap.put(3, "Anuj");

  /* Display content using Iterator*/
  Set set = hmap.entrySet();
  Iterator iterator = set.iterator();
  while(iterator.hasNext()) {
     Map.Entry mentry = (Map.Entry)iterator.next();
     System.out.print("key is: "+ mentry.getKey() + " & Value is: ");
     System.out.println(mentry.getValue());
  }

  /* Get values based on key*/
  String var= hmap.get(2);
  System.out.println("Value at index 2 is: "+var);

  /* Remove values based on key*/
  hmap.remove(3);
  System.out.println("Map key and values after removal:");
  Set set2 = hmap.entrySet();
  Iterator iterator2 = set2.iterator();
  while(iterator2.hasNext()) {
      Map.Entry mentry2 = (Map.Entry)iterator2.next();
      System.out.print("Key is: "+mentry2.getKey() + " & Value is: ");
      System.out.println(mentry2.getValue());
   }

祝你好运

Answer 4

i don't know how you're querying your database, but i assume you're making one query to get both the location and the distance. 

//1. create a model to represent the Location and its corresponding distance
public class LocationDistance {
    private float distance;
    private String location;

    public LocationDistance(float distance, String location) {
        this.distance = distance;
        this.location = location;
    }

    public String getLocation(){
        return this.location;
    } 

    public float getDistance(){
        return this.distance;
    }

    public void setLocation(String loc){
         this.location=loc;
    } 

    public void setDistance(float distance){
         this.distance=distance;
    }
}

   //2. Loop over the two arrays and create a list of LocationDistance
    List<LocationDistance> locationDistance=new ArrayList<LocationDistance>();
    float[] e = { 1900f, 16400f, 77666f, 8000f, 13200f, 15600f };

    String[] location = { "Birmingham", "Blackburn", "London", "Luton", "Manchester", "Newcastle" };

    for(int i=0; i<e.length;i++){
      locationDistance.add(new LocationDistance(e[i],location[i]));
    }

Answer 5

是浮点数，与位置相同吗？

你应该为那个

定义一个Pojo类

class Pojo {
    private final float e;
    private final String city;

    public Pojo(float e, String city) {
        this.e = e;
        this.city = city;
    }
    public static void main(String[] args) {
        final float[] e = { 1900f, 16400f, 77666f, 8000f, 13200f, 15600f };
        final String[] location = { "Birmingham", "Blackburn", "London", "Luton", "Manchester", "Newcastle" };
        final Pojo[] p = new Pojo[e.length];

        for (int i = 0; i < e.length; i++) {
            p[i] = new Pojo(e[i], location[i]);
        }
    }

}

如何将一个Array中的元素与另一个数组中的元素相关联（Android Studio）

5 个答案: