Question

所以我正在尝试用MapDB做一些事情而且我遇到了障碍。我会尽力描述它：

我有四个数据，我们会说它是这样的：

1) String action; //the name of the action itself
2) String categoryOfAction; //the category of the action
3) Integer personWhoPerformedAction; //the person that did the action
4) Long timeOfOccurrence; //the time the action was done

可以在此数据库中，由不同的人，在不同的时间以及在不同的类别中多次执行相同的操作。我想要有三个单独的地图，每个地图将数据组织成这样的东西：

String[] actionOccurances = map1.get(action); //returns every occurrence of that action (possibly in an array), including who did that occurrence, time the occurrence occurred, and the category of that occurrence

Long latestOccurance = map2.get(action); //returns the latest occurrence of that action

String[] actionsPerformedByPerson = map3.get(personWhoPerformedAction); //returns every action that this person has done, including the category of that action, the time they performed that action, and the name of the action itself

所以，我想尽可能高效地做到这一点。我知道我可以做这样的事情：

DB thedb = DBMaker.newTempFileDB().make();

NavigableSet<Object[]> map1 = thedb.createTreeSet("actionOccurences").comparator(Fun.COMPARABLE_ARRAY_COMPARATOR).make();

HTreeMap<String, Long> map2 = thedb.getHashMap("lastOccurrence");

NavigableSet<Object[]> map3 = thedb.createTreeSet("actionsPerformedByPerson").comparator(Fun.COMPARABLE_ARRAY_COMPARATOR).make();

但我觉得那是错的。必须有一种更有效的方法，我不必多次存储相同的数据，是吗？

我在Bind类和它的函数（secondaryValues，mapInverse等等）上玩了很多，但我似乎无法找到一种方法来映射这组数据，我希望它是如何

有任何帮助吗？感谢。

Answer 1

啊！经过一段时间的努力，我找到了解决方案。我基本上为每条记录分配一个唯一的ID，然后使用MapDB的secondaryKey绑定。它是这样的：

static class Record implements Serializable
{
    final String action;
    final String categoryOfAction;
    final String personWhoPerformedAction;
    final Long timeOfOccurrence;

    public record(String actn, String cat, String person, Long time)
    {
        action = actn;
        categoryOfAction = cat;
        personWhoPerformedAction = person;
        timeOfOccurence = time;
    }

}

static void main(String[] args)
{
    DB thedb = DBMaker.newTempFileDB().make();

    //primaryMap maps each record to a unique ID
    BTreeMap<Integer,Record> primaryMap = thedb.createTreeMap("pri")
                                        .keySerializer(BTreeKeySerializer.INTEGER)
                                        .makeOrGet();;

    //this map holds the unique ID of every record in primaryMap with a common action
    NavigableSet<Object[]> map_commonAction = thedb.createTreeSet("com_a")
                                        .comparator(Fun.COMPARABLE_ARRAY_COMPARATOR)
                                        .makeOrGet();

    //this map holds the unique ID of every record in primaryMap with a common person
    NavigableSet<Object[]> map_commonPerson = thedb.createTreeSet("com_p")
                                        .comparator(Fun.COMPARABLE_ARRAY_COMPARATOR)
                                        .makeOrGet();

    //binding map_commonAction to primaryMap so it is updated with primary
    Bind.secondaryKey(primaryMap, map_commonAction, new Fun.Function2<String, Integer, Record>() {
        @Override
        public String run(Integer recordID, Record r) {
            return r.action;
        }
    });

    //binding map_commonPerson to primaryMap so it is updated with primary
    Bind.secondaryKey(primaryMap, map_commonPerson, new Fun.Function2<String, Integer, Record>() {
        @Override
        public String run(Integer recordID, Record r) {
            return r.personWhoPerformedAction;
        }
    });


    primaryMap.put(1, new Record("a", "abc", "person1", 123434L));
    primaryMap.put(2, new Record("a", "abc", "person2", 322443L));
    primaryMap.put(3, new Record("b", "def", "person2", 124243L));
    primaryMap.put(4, new Record("b", "abc", "person1", 983243L));
    primaryMap.put(5, new Record("c", "def", "person2", 999993L));


    //this is how we attain all records with some action
    for (Object[] k : Fun.filter(map_commonAction, "someAction"))
    {
        Record obtainedRecord = primary.get(k[1]);

    }

    //this is how we attain all records with some person
    for (Object[] k : Fun.filter(map_commonPerson, "somePerson"))
    {
        Record obtainedRecord = primary.get(k[1]);

    }

}

但是，如果您认为可以改进此解决方案，仍然会发出声响。谢谢！

在mapdb中存储它的最佳方法是什么？

1 个答案: