Mongodb聚合:如何仅返回数组的匹配元素

Question

在我的mongoDB书籍集中,我的文档结构如下:

/* 0 */
{
  "_id" : ObjectId("50485b89b30f1ea69110ff4c"),

  "publisher" : {
    "$ref" : "boohya",
    "$id" : "foo"
  },
  "displayName" : "Paris Nightlife",
  "catalogDescription" : "Some desc goes here",
  "languageCode" : "en",
  "rating" : 0,
  "status" : "LIVE",
  "thumbnailId" : ObjectId("50485b89b30f1ea69110ff4b"),
  "indexTokens" : ["Nightlife", "Paris"]
}

我执行以下正则表达式查询以查找具有以"Par"开头的一个indexToken的所有文档:

{ "indexTokens" : { "$regex" : "^Par" , "$options" : "i"}}

如果我只选择要返回的indexTokens字段,如下所示:

{ "indexTokens" : 1}

生成的DBObject是

{ "_id" : { "$oid" : "50485b89b30f1ea69110ff4c"} , "indexTokens" : [ "Nightlife" , "Paris"]}

我想得到的只是与正则表达式匹配的标记/标记(此时我不关心检索文档,我也不需要匹配文档的所有标记)

这是MongoDB v2.2下新的聚合框架的案例吗？？

如果是,我如何修改我的查询,以便实际结果如下所示:

{"indexTokens":["巴黎","天堂河","帕尔马"等......]

奖金问题(你有没有代码):我如何使用Java驱动程序？

现在我的java看起来像:

DBObject query = new BasicDBObject("indexTokens", java.util.regex.Pattern.compile("^"+filter+"", Pattern.CASE_INSENSITIVE));
    BasicDBObject fields = new BasicDBObject("indexTokens",1);
    DBCursor curs = getCollection()
                    .find(query, fields)
                    .sort( new BasicDBObject( "indexTokens" , 1 ))
                    .limit(maxSuggestionCount);

谢谢 :)

编辑:

根据您的回答,我修改了我的JAVA代码如下:

BasicDBObject cmdBody = new BasicDBObject("aggregate", "Book"); 
    ArrayList<BasicDBObject> pipeline = new ArrayList<BasicDBObject>(); 

    BasicDBObject match = new BasicDBObject("$match", new BasicDBObject("indexTokens", java.util.regex.Pattern.compile("^"+titleFilter+"", Pattern.CASE_INSENSITIVE)));
    BasicDBObject unwind = new BasicDBObject("$unwind", "$indexTokens");
    BasicDBObject match2 = new BasicDBObject("$match", new BasicDBObject("indexTokens", java.util.regex.Pattern.compile("^"+titleFilter+"", Pattern.CASE_INSENSITIVE)));
    BasicDBObject groupFilters = new BasicDBObject("_id",null);
    groupFilters.append("indexTokens", new BasicDBObject( "$push", "$indexTokens"));
    BasicDBObject group = new BasicDBObject("$group", groupFilters);

    pipeline.add(match);
    pipeline.add(unwind);
    pipeline.add(match2);
    pipeline.add(group);

    cmdBody.put("pipeline", pipeline); 



    CommandResult res = getCollection().getDB().command(cmdBody);
    System.out.println(res);

哪个输出

{ "result" : [ { "_id" :  null  , "indexTokens" : [ "Paris"]}] , "ok" : 1.0}

这才是天才!

非常感谢 !

Answer 1

您可以使用2.2聚合框架执行此操作.像这样的东西;

db.books.runCommand("aggregate", {
    pipeline: [
        {   // find docs that contain Par*
            $match: { "indexTokens" : { "$regex" : "^Par" , "$options" : "i"}},
        },
        {   // create a doc with a single array elemm for each indexToken entry
            $unwind: "$indexTokens" 
        },
        {   // now produce a list of index tokens
            $group: {
                _id: "$indexTokens",
            },
        },
    ],
})

或者如果你真的想要没有doc的数组,那么这可能更接近你所追求的;

db.books.runCommand("aggregate", {
    pipeline: [
        {   // find docs that contain Par*
            $match: { "indexTokens" : { "$regex" : "^Par" , "$options" : "i"}},
        },
        {   // create a doc with a single array elemm for each indexToken entry
            $unwind: "$indexTokens" 
        },
        {   // now throw out any unwind's that DON'T contain Par*
            $match: { "indexTokens": { "$regex": "^Par", "$options": "i" } },
        },
        {   // now produce the list of index tokens
            $group: {
                _id: null,
                indexTokens: { $push: "$indexTokens" },
            },
        },
    ],
})

Answer 2

根据 cirrus 的响应，我建议$unwind首先执行此操作以避免多余的$match. 就像是：

db.books.aggregate(
    {$unwind:"$indexTokens"},
    {$match:{indexTokens:/^Par/}},
    {$group:{_id:null,indexTokens:{$push:"$indexTokens"}}
})

在 Java 中如何做到这一点？您可以使用DBCollection.aggregate(...)MongoDB v2.9.0驱动程序的方法。每个管道操作员，例如。$unwind或$match, 对应于一个DBObject对象。