Question

我有一个带属性的测试

--> Integer id1
--> Integer id2
--> String stringValue.

因此，每个（id1和id2）组合的唯一ID都有stringValues列表。我希望能够将其存储在一个集合中。例如：

  test1 : id1 = 1, id2 = 2 , stringValue = one
  test11 : id1 = 1, id2 = 2 , stringValues =two
  test111 : id1 = 1, id2 = 3 , stringValues =three
  test2 : id1 = 5, id2 = 3, stringValues = four
  test22: id1 = 5, id2 = 2, stringValues = five
  test222: id = 5, id2 = 3, stringValues = six

所需的最终结果是将对象存储为

-> {id = 1, id = 2, String = {one, two} } 
-> {id = 1 , id =3 , String = {three}}
-> {id = 5, id = 3 , String = {four,six}}
-> {id = 5, id = 2 , String = {five}}

我希望能够将'test'对象列表存储在集合中。我试过Set＆lt;＆gt; ，重写equals和hashcode，但它没有按预期工作。你去吧

@Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((id1 == null) ? 0 : id1.hashCode());
        result = prime * result + ((id2 == null) ? 0 : id2.hashCode());
        return result;
    }


        @Override
    public boolean equals(Object obj) {
        if( obj instanceof Test){
            Test test = (Test) obj;
            return (test.id1 == this.id1 && test.id2 == this.id2);
        } else {
            return false;
        }
    }

我遍历每个对象并将其存储在SET中。它确实消除了（[id1和id2]的冗余组合，但它也消除了相应的字符串值。）哎呀我忘了提到我从数据库中检索到的结果集中的数据。所以每一行都包含id1，id2和一个字符串值（所有行都不同）。某些行将具有共同的id1和id2值

有人可以请指点我吗？

提前谢谢

Answer 1

这能解决您的问题吗？

import java.util.*;

public class Playground1 {
    public static void main(String[] args) {

        // fake resultset
        List<DBResultSetRowEmulation> resultSetEmulation = new ArrayList<>();
        resultSetEmulation.add(new DBResultSetRowEmulation(1, 2, "one"));
        resultSetEmulation.add(new DBResultSetRowEmulation(1, 2, "two"));
        resultSetEmulation.add(new DBResultSetRowEmulation(1, 3, "three"));
        resultSetEmulation.add(new DBResultSetRowEmulation(5, 3, "four"));
        resultSetEmulation.add(new DBResultSetRowEmulation(5, 2, "five"));
        resultSetEmulation.add(new DBResultSetRowEmulation(5, 3, "six"));

        Map<DoubleIndex, List<String>> resultData = new HashMap<>();

        // iterate through resultset
        for(DBResultSetRowEmulation row: resultSetEmulation) {
            DoubleIndex curRecordIdx = new DoubleIndex(row.a, row.b);
            if (resultData.containsKey(curRecordIdx)) {
                // append current string to some existing id1+id2 combination
                resultData.get(curRecordIdx).add(row.c);
            } else {
                // create a new list for new id1+id2 combination
                resultData.put(curRecordIdx, new ArrayList<>(Collections.singletonList(row.c)));
            }
        }

        System.out.println(resultData);
    }

    private static class DBResultSetRowEmulation {
        Integer a, b;
        String c;

        DBResultSetRowEmulation(Integer a, Integer b, String c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }
    }

    private static class DoubleIndex {
        private Integer a,b;

        public DoubleIndex(Integer a, Integer b) {
            this.a = a;
            this.b = b;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;

            DoubleIndex that = (DoubleIndex) o;

            return a.equals(that.a) && b.equals(that.b);
        }

        @Override
        public int hashCode() {
            int result = a.hashCode();
            result = 31 * result + b.hashCode();
            return result;
        }

        @Override
        public String toString() {
            return "id{" +
                    "a=" + a +
                    ", b=" + b +
                    '}';
        }
    }
}

输出：

{id {a = 1，b = 2} = [one，two]，

id {a = 1，b = 3} = [three]，

id {a = 5，b = 2} = [five]，

id {a = 5，b = 3} = [four，six]}

你必须弄清楚如何自己对结果进行排序。

用于将ID组合为唯一

1 个答案: