java mongodb 3.0:在文档的内部数组中查找值
我正在使用java和mongodb v.3.0.7。 我有一个玩家列表,里面有一系列的游戏和分数。这是一个插入文档的测试:java mongodb 3.0:在文档的内部数组中查找值,java,mongodb,mongodb-java,Java,Mongodb,Mongodb Java,我正在使用java和mongodb v.3.0.7。 我有一个玩家列表,里面有一系列的游戏和分数。这是一个插入文档的测试: public void insertPlayer(String id_device) throws ParseException{ DateFormat format = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss'Z'", Locale.ENGLISH); db.getCollection("player").i
public void insertPlayer(String id_device) throws ParseException{
DateFormat format = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss'Z'", Locale.ENGLISH);
db.getCollection("player").insertOne(
new Document()
.append("nikname", "Guest")
.append("password", "Guest")
.append("my_id", "")
.append("id_device", id_device)
.append("language", "Italian")
.append("games", asList(
new Document()
.append("gamename", "PPA")
.append("banned", false)
.append("date", format.parse("2014-10-01T00:00:00Z"))
.append("score", 11),
new Document()
.append("gamename", "Test2game")
.append("banned", false)
.append("date", format.parse("2014-01-16T00:00:00Z"))
.append("score", 17)))
);
}
new Document()
.append("gamename", "PPA")
.append("banned", false)
.append("date", format.parse("2014-10-01T00:00:00Z"))
.append("score", 11),
为了确定某个玩家是否被禁止参加某个特定的游戏,我将这样做:
public boolean isBanned(String id_device){
FindIterable<Document> iterable = db.getCollection("player").find(eq("id_device", "machine1"));
System.out.println(iterable.first());
List<Document> dl = (List<Document>)iterable.first().get("games");
for(int i=0;i<dl.size();i++){
Document d = dl.get(i);
System.out.println(d.getString("gamename"));
if(d.getString("gamename").equals("PPA")){
boolean ban = d.getBoolean("banned");
return ban;
}
}
提供id_设备和游戏名?
谢谢要实现这一点,您需要聚合数据。 根据您的用例,聚合可能会改变(更多查询,更多管道步骤) 以下是您的数据:
{
"_id" : ObjectId("569a30a30586bcb40f7d2531"),
"my_id" : "",
"id_device" : "machine1",
"language" : "Italian",
"games" : [
{
"gamename" : "PPA",
"banned" : true,
"date" : ISODate("2014-10-01T00:00:00.000Z"),
"score" : 11
},
{
"gamename" : "Test2game",
"banned" : false,
"date" : ISODate("2014-01-16T00:00:00.000Z"),
"score" : 17
}
]
}
我假设:
You have a list of unique players. Each of them play unique games.
(Example: player1 never plays PPA twice)
So, you need to search a game where the player is banned and return
all information for that game.
汇总如下:
db.player.aggregate([
{$match:{ "id_device" : "machine1"}},
{$unwind: "$games"},
{$match:{ "games.gamename" : "PPA", "games.banned" : true}}
])
结果
有些不同,如果您的玩家可能多次玩同一个游戏(不同的日期),您可以更改聚合管道
{
"_id" : ObjectId("569a30a30586bcb40f7d2531"),
"my_id" : "",
"id_device" : "machine1",
"language" : "Italian",
"games" : [
{
"gamename" : "PPA",
"banned" : false,
"date" : ISODate("2014-10-01T00:00:00.000Z"),
"score" : 11
},
{
"gamename" : "Test2game",
"banned" : false,
"date" : ISODate("2014-01-16T00:00:00.000Z"),
"score" : 17
},
{
"gamename" : "PPA",
"banned" : true,
"date" : ISODate("2014-04-18T00:00:00.000Z"),
"score" : 23
},
{
"gamename" : "Foo",
"banned" : true,
"date" : ISODate("2015-03-03T00:00:00.000Z"),
"score" : 2
},
{
"gamename" : "Foo",
"banned" : false,
"date" : ISODate("2015-04-28T00:00:00.000Z"),
"score" : 2
}
]
}
因此,为了查询id_设备和游戏名“PPA”,我们通过以下方式定义聚合:
db.player.aggregate([
{$match:{ "id_device" : "machine1"}},
{$unwind: "$games"},
{$match:{ "games.gamename" : "PPA"}},
{$group: {_id:{"_id":"$_id", "my_id":"$my_id", "id_device":"$id_device","language":"$language"}, "games" : {$push:"$games"}}},
{$project: {"_id":"$_id._id", "my_id":"$_id.my_id", "id_device": "$_id.id_device", "language":"$_id.language", "games":"$games"}}
])
结果:
如您所见,您可以添加/修改管道步骤并获得所需的结果。要实现这一点,您需要聚合数据。 根据您的用例,聚合可能会改变(更多查询,更多管道步骤) 以下是您的数据:
{
"_id" : ObjectId("569a30a30586bcb40f7d2531"),
"my_id" : "",
"id_device" : "machine1",
"language" : "Italian",
"games" : [
{
"gamename" : "PPA",
"banned" : true,
"date" : ISODate("2014-10-01T00:00:00.000Z"),
"score" : 11
},
{
"gamename" : "Test2game",
"banned" : false,
"date" : ISODate("2014-01-16T00:00:00.000Z"),
"score" : 17
}
]
}
我假设:
You have a list of unique players. Each of them play unique games.
(Example: player1 never plays PPA twice)
So, you need to search a game where the player is banned and return
all information for that game.
汇总如下:
db.player.aggregate([
{$match:{ "id_device" : "machine1"}},
{$unwind: "$games"},
{$match:{ "games.gamename" : "PPA", "games.banned" : true}}
])
结果
有些不同,如果您的玩家可能多次玩同一个游戏(不同的日期),您可以更改聚合管道
{
"_id" : ObjectId("569a30a30586bcb40f7d2531"),
"my_id" : "",
"id_device" : "machine1",
"language" : "Italian",
"games" : [
{
"gamename" : "PPA",
"banned" : false,
"date" : ISODate("2014-10-01T00:00:00.000Z"),
"score" : 11
},
{
"gamename" : "Test2game",
"banned" : false,
"date" : ISODate("2014-01-16T00:00:00.000Z"),
"score" : 17
},
{
"gamename" : "PPA",
"banned" : true,
"date" : ISODate("2014-04-18T00:00:00.000Z"),
"score" : 23
},
{
"gamename" : "Foo",
"banned" : true,
"date" : ISODate("2015-03-03T00:00:00.000Z"),
"score" : 2
},
{
"gamename" : "Foo",
"banned" : false,
"date" : ISODate("2015-04-28T00:00:00.000Z"),
"score" : 2
}
]
}
因此,为了查询id_设备和游戏名“PPA”,我们通过以下方式定义聚合:
db.player.aggregate([
{$match:{ "id_device" : "machine1"}},
{$unwind: "$games"},
{$match:{ "games.gamename" : "PPA"}},
{$group: {_id:{"_id":"$_id", "my_id":"$my_id", "id_device":"$id_device","language":"$language"}, "games" : {$push:"$games"}}},
{$project: {"_id":"$_id._id", "my_id":"$_id.my_id", "id_device": "$_id.id_device", "language":"$_id.language", "games":"$games"}}
])
结果:
如您所见,您可以添加/修改管道步骤并获得所需的结果。是。您的查找查询应该是这样的:
db.player.find({“id\u设备”:“machine1”,“games.gamename”:“PPA”,“games.banked”:true})
。如果存在,这将返回被禁止的玩家列表(.size()
>0)。感谢您的回答,但我需要知道是否可能有嵌套文档或值,因为,例如,在将来,我需要使用id\u设备和gamename获取玩家的分数……是的,您可以。但是,您需要将find
更改为aggregate
。看看这篇文章,问题类似于如何通过查询获取/排除子文档。如果您愿意,我可以帮助您进行MongoDB shell查询,但是您需要更改为java。我尝试了AggregateIterable=db.getCollection(“player”).aggregate(asList(新文档($match),新文档(“id_设备”,id_设备)。append(“games.gamename”,“PPA”)),新文档($group),新文档(“id”,“id_设备”).append(“games.gamename”、“PPA”);但是给我错误,如果我删除append(“games.gamename”,“PPA”)只给我一条记录{{{id:machine1}},你能为你的用例提供更多细节吗?对于给定的玩家,游戏可以重复吗?你希望得到什么样的产出?是的。您的查找查询应该是这样的:db.player.find({“id\u设备”:“machine1”,“games.gamename”:“PPA”,“games.banked”:true})
。如果存在,这将返回被禁止的玩家列表(.size()
>0)。感谢您的回答,但我需要知道是否可能有嵌套文档或值,因为,例如,在将来,我需要使用id\u设备和gamename获取玩家的分数……是的,您可以。但是,您需要将find
更改为aggregate
。看看这篇文章,问题类似于如何通过查询获取/排除子文档。如果您愿意,我可以帮助您进行MongoDB shell查询,但是您需要更改为java。我尝试了AggregateIterable=db.getCollection(“player”).aggregate(asList(新文档($match),新文档(“id_设备”,id_设备)。append(“games.gamename”,“PPA”)),新文档($group),新文档(“id”,“id_设备”).append(“games.gamename”、“PPA”);但是给我错误,如果我删除append(“games.gamename”,“PPA”)只给我一条记录{{{id:machine1}},你能为你的用例提供更多细节吗?对于给定的玩家,游戏可以重复吗?您期望的具体输出是什么?