设置存储的重复记录

Question

好的，这可能是一个简单的问题，但是当我读到 set 和我得到的回复时，我感到很困惑。

我正在从数据库中获取重复记录，并使用 set 为一个条件（代码）只存储一条记录以忽略剩余的重复记录。

Set<DiagnosisInfo> diagnosisInfoSet = patientDiagnosisHistoryRepository
                    .findBydiagnosisdesc(UUID.fromString(patientID.toUpperCase()));

我得到的回应是

 "diagnosisInfo": [
    {
      "diagnosisCode": "T49.1X6S",
      "diagnosisCodeDescription": "Underdosing of antipruritics, sequela"
    },
    {
      "diagnosisCode": "T49.1X6S",
      "diagnosisCodeDescription": "Underdosing of antipruritics, sequela"
    },
    {
      "diagnosisCode": "T49.1X6S",
      "diagnosisCodeDescription": "Underdosing of antipruritics, sequela"
    },
    {
      "diagnosisCode": "T49.1X6S",
      "diagnosisCodeDescription": "Underdosing of antipruritics, sequela"
    },
    {
      "diagnosisCode": "V09.20",
      "diagnosisCodeDescription": "Pedestrian injured in traffic accident involving unspecified motor vehicles*"
    },
    {
      "diagnosisCode": "V09.20",
      "diagnosisCodeDescription": "Pedestrian injured in traffic accident involving unspecified motor vehicles*"
    },
    {
      "diagnosisCode": "V09.20",
      "diagnosisCodeDescription": "Pedestrian injured in traffic accident involving unspecified motor vehicles*"
    },
    {
      "diagnosisCode": "E09.8",
      "diagnosisCodeDescription": "Drug or chemical induced diabetes mellitus with unspecified complications"
    },
    {
      "diagnosisCode": "E09.8",
      "diagnosisCodeDescription": "Drug or chemical induced diabetes mellitus with unspecified complications"
    },
    {
      "diagnosisCode": "E09.8",
      "diagnosisCodeDescription": "Drug or chemical induced diabetes mellitus with unspecified complications"
    },
    {
      "diagnosisCode": "E09.8",
      "diagnosisCodeDescription": "Drug or chemical induced diabetes mellitus with unspecified complications"
    },
    {
      "diagnosisCode": "E09.8",
      "diagnosisCodeDescription": "Drug or chemical induced diabetes mellitus with unspecified complications"
    }
  ]

正如我们所见，相同的诊断代码记录会显示多次，尽管它们是重复的。我想看到的响应如下

 "diagnosisInfo": [
    {
      "diagnosisCode": "T49.1X6S",
      "diagnosisCodeDescription": "Underdosing of antipruritics, sequela"
    },
    {
      "diagnosisCode": "V09.20",
      "diagnosisCodeDescription": "Pedestrian injured in traffic accident involving unspecified motor vehicles*"
    },
    {
      "diagnosisCode": "E09.8",
      "diagnosisCodeDescription": "Drug or chemical induced diabetes mellitus with unspecified complications"
    }
  ]

一个代码一个响应。我不知道我是否必须更改我的查询，或者关于 set 的一些事情。

Answer 1

重点是：

正如我们所见，相同的诊断代码记录虽然重复显示了多次。

不，它们不相同。您必须为您的类 DiagnosisInfo 实现正确的 equals() 和 hashCode() 方法！

为了判断两个对象是否“相同”； Set 实现将使用这些对象提供的 equals 方法。最有可能的是，您要么忘记在课堂上使用 @Override equals()；您提供了一个导致false 的实现；即使对于那些你认为“相同”的对象。

有关更多详细信息，请参阅here。

鉴于您的最新评论，您的逻辑似乎仍然倒退。你仍然认为你“知道”哪些对象是重复的，哪些不是。但是您认为不重要。唯一重要的是您的 equals() 方法中的实现。 操作的结果决定了对象是否相等/相似；或者它们是否不同。

所以：当你向一个类添加新字段时；并使用您的 IDE 重新生成这些方法，现在您对结果“不满意”；那么您的问题是您在生成过程中包含您实际上认为“不应该重要”的字段。代码按照您告诉它执行的操作。

因此，真正的答案在这里：

• 您应该退后一步，重新考虑您的对象模型。在生成 equals/hashCode 方法时，您不会创建/添加字段并包含（或不包含）它们，因为您可以这样做。不，您决定需要做什么；然后你就实现了。
• 然后：这样的代码对于单元测试和 TDD 来说是完美的。您无需等待读取一些真实数据；不：您有各种单元测试来验证您的 equals 实现提供了预期的结果。然后，当您添加新内容或更改代码时，您会重新运行这些测试，它们会告诉您您破坏了哪些内容。

Answer 2

是的，这就是 DiagnosisInfo 不覆盖 Equals 和 haschCode 时得到的结果

您的 Set 不知道如何检查此对象的时间

{
  "diagnosisCode": "V09.20",
  "diagnosisCodeDescription": "Pedestrian injured in traffic accident involving unspecified motor vehicles*"
},

还有这个

{
  "diagnosisCode": "V09.20",
  "diagnosisCodeDescription": "Pedestrian injured in traffic accident involving unspecified motor vehicles*"
},

都是一样的……

---

假设您调用的方法实现了HashSet<E>，那么源代码：

/**
 * Adds the specified element to this set if it is not already present.
 * More formally, adds the specified element <tt>e</tt> to this set if
 * this set contains no element <tt>e2</tt> such that
 * <tt>(e==null&nbsp;?&nbsp;e2==null&nbsp;:&nbsp;e.equals(e2))</tt>.
 * If this set already contains the element, the call leaves the set
 * unchanged and returns <tt>false</tt>.
 *
 * @param e element to be added to this set
 * @return <tt>true</tt> if this set did not already contain the specified
 * element
 */
public boolean add(E e) {
    return map.put(e, PRESENT)==null;
}

map.put 在哪里调用 putVal

而 putval 使用 equals 和 hashcode...哈哈！

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
               boolean evict) {
    Node<K,V>[] tab; Node<K,V> p; int n, i;
    if ((tab = table) == null || (n = tab.length) == 0)
        n = (tab = resize()).length;
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    else {
        Node<K,V> e; K k;
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
        else if (p instanceof TreeNode)
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else {
            for (int binCount = 0; ; ++binCount) {
                if ((e = p.next) == null) {
                    p.next = newNode(hash, key, value, null);
                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                        treeifyBin(tab, hash);
                    break;
                }
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                p = e;
            }
        }
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)
                e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
    }
    ++modCount;
    if (++size > threshold)
        resize();
    afterNodeInsertion(evict);
    return null;
}