java - 使用 $elemMatch 时如何投影第一个子文档以外的内容

Question

我有一个包含以下形式文档的集合:

{
    "_id" : { "$oid" : "67bg............"},
    "ID"  : "xxxxxxxx",
    "senses" : [
        {
            "word"   : "hello",
            "lang"   : "EN",
            "source" : "EN_DICTIONARY"
        },
        {
            "word"   : "coche",
            "lang"   : "ES",
            "source" : "ES_DICTIONARY"
        },
        {
            "word"   : "bye",
            "lang"   : "EN",
            "source" : "EN_DICTIONARY"
        }
    ]
}

我想查找与 lang=X 和 source=Y 至少一种含义匹配的所有文档，并返回仅包含这些 sense< 的匹配文档 匹配 lang=X 和 source=Y。

我试过这个:

DBObject sensesQuery = new BasicDBObject();
sensesQuery.put("lang", "EN");
sensesQuery.put("source", "EN_DICTIONARY");
DBObject matchQuery = new BasicDBObject("$elemMatch",sensesQuery);

DBObject fields = new BasicDBOject();
fields.put("senses",matchQuery);

DBObject projection = new BasicDBObject();
projection.put("ID",1)
projection.put("senses",matchQuery);
DBCursor cursor = collection.find(fields,projection)

while(cursor.hasNext()) {
    ...
}

我的查询适用于匹配文档，但不适用于投影。以上面的文档为例，如果我运行查询，我会得到以下结果:

{
    "_id" : { "$oid" : "67bg............"},
    "ID"  : "xxxxxxxx",
    "senses" : [
        {
            "word"   : "hello",
            "lang"   : "EN",
            "source" : "EN_DICTIONARY"
        }
    ]
}

但我想要这个:

{
    "_id" : { "$oid" : "67bg............"},
    "ID"  : "xxxxxxxx",
    "senses" : [
        {
            "word"   : "hello",
            "lang"   : "EN",
            "source" : "EN_DICTIONARY"
        },
        {
            "word"   : "bye",
            "lang"   : "EN",
            "source" : "EN_DICTIONARY"
        }
    ]
}

我读到了有关聚合的内容，但我不明白如何在 MongoDB Java 驱动程序中使用它。

谢谢

Answer 1

您正在投影和过滤器上使用 $elemMatch 运算符。

来自the docs

The $elemMatch operator limits the contents of an field from the query results to contain only the first element matching the $elemMatch condition.

因此，您所看到的行为是 elemMatch-in-a-projection 的预期行为。

如果您想投影符合过滤条件的文档中 senses 数组中的所有子文档，那么您可以使用:

projection.put("senses", 1);

但是，如果您只想投影那些与您的过滤条件匹配的子文档，那么 $elemMatch 将不适合您，因为它只返回与 $elemMatch< 匹配的第一个元素 条件。您的替代方案是使用聚合框架，例如:

db.collection.aggregate([
  // matches documents with a senses sub document having the given lang and source values
  {$match: {'senses.lang': 'EN', 'senses.source': 'EN_DICTIONARY'}},

  // projects on the senses sub document and filters the output to only return sub 
  // documents having the given lang and source values
  {$project: {
      senses: {
        $filter: {
            input: "$senses",
            as: "sense",
            cond: { $eq: [ "$$sense.lang", 'EN' ], $eq: [ "$$sense.source", 'EN_DICTIONARY' ] }
          }
        }
      }
  }
])

这是使用 MongoDB Java 驱动程序的聚合调用:

Document filter = new Document("senses.lang", "EN").append("senses.source", "EN_DICTIONARY");

DBObject filterExpression = new BasicDBObject();
filterExpression.put("input", "$senses");
filterExpression.put("as", "sense");
filterExpression.put("cond", new BasicDBObject("$and", Arrays.<Object>asList(
        new BasicDBObject("$eq", Arrays.<Object>asList("$$sense.lang", "EN")),
        new BasicDBObject("$eq", Arrays.<Object>asList("$$sense.source", "EN_DICTIONARY")))
));

BasicDBObject projectionFilter = new BasicDBObject("$filter", filterExpression);

AggregateIterable<Document> documents = collection.aggregate(Arrays.asList(
        new Document("$match", filter),
        new Document("$project", new Document("senses", projectionFilter))));

for (Document document : documents) {
    logger.info("{}", document.toJson());
}

结果输出是:

2017-10-01 17:15:39 [main] INFO  c.s.mongo.MongoClientTest - { "_id" : { "$oid" : "59d10cdfc26584cd8b7a0d3b" }, "senses" : [{ "word" : "hello", "lang" : "EN", "source" : "EN_DICTIONARY" }, { "word" : "bye", "lang" : "EN", "source" : "EN_DICTIONARY" }] }

更新 1:以下评论:

After a long period of testing, trying to understand why the query was slow, I noticed that the "$match" parameter does not work, the query should select only records that have at least one sense with source = Y AND lang = X and project them , but the query also returns me documents with senses = []

此过滤器:new Document("senses.lang", "EN").append("senses.source", "EN_DICTIONARY") 不会匹配没有 Senses 的文档 属性也不会匹配具有空 senses 属性的文档。为了验证这一点，我将以下文档添加到我自己的集合中:

{
    "_id" : ObjectId("59d72a24c26584cd8b7b70a5"),
    "ID" : "yyyyyyyy"
}

{
    "_id" : ObjectId("59d72a3ac26584cd8b7b70ae"),
    "ID" : "zzzzzzzzz",
    "senses" : []
}

重新运行上面的代码，我仍然得到想要的结果。

我怀疑您关于上述代码不起作用的说法要么是误报，要么您正在查询的文档与我一直在使用的示例不同。

为了帮助您自己诊断此问题，您可以...

• 与其他运营商合作，例如无论有没有 $exists 运算符，$match 阶段的行为都是相同的:

  new Document("senses", new BasicDBObject("$exists", true))
          .append("senses.lang", new BasicDBObject("$eq", "EN"))
          .append("senses.source", new BasicDBObject("$eq", "EN_DICTIONARY"))
  

• 删除 $project 阶段以准确查看 $match 阶段生成的内容。