Question

我在mongodb中有一个数据库，它有两个集合： cards 和 teams 。我的卡集合结构是：

{"Card":{"_id":{"$oid":"5a46626e150ed3f847f01bac"},"Number":1,"Page":1,"Team":[{"$oid":"5a465ca9150ed3f847f01b92"}]}}
{"Card":{"_id":{"$oid":"5a466d96150ed3f847f01bad"},"Number":18,"Page":6,"Team":[{"$oid":"5a465ca9150ed3f847f01b92"}]}}
{"Card":{"_id":{"$oid":"5a466de6150ed3f847f01bae"},"Number":35,"Page":8, "Team":[{"$oid":"5a465ca9150ed3f847f01b92"}]}}

和我的团队集合结构是：

{"Team":{"_id":{"$oid":"5a465ca9150ed3f847f01b92"},"Name":"NA","Coach":"NA"}}

我想打印所有卡片信息，也就是说，对于每张卡片，我想要打印它的ID，编号，页面和团队信息（id，name和coach）。我试过了：

var team = db.cards.find()
while(team.hasNext()){team_data = db.teams.find({"Team._id":{$in:team.Cards.Team}}).toArray()}

但它给了我：

2018-01-02T18:35:57.588+0000 E QUERY    [thread1] TypeError: team.Cards is undefined :
@(shell):1:62

然后我试了试：

var team = db.cards.find()
while(team.hasNext()){team_data = db.teams.find({"Team._id":{$in:team.Team}}).toArray()}

它给了我：

2018-01-02T18:36:21.620+0000 E QUERY    [thread1] Error: error: {
        "ok" : 0,
        "errmsg" : "$in needs an array",
        "code" : 2,
        "codeName" : "BadValue"
} :
_getErrorWithCode@src/mongo/shell/utils.js:25:13
DBCommandCursor@src/mongo/shell/query.js:717:1
DBQuery.prototype._exec@src/mongo/shell/query.js:117:28
DBQuery.prototype.hasNext@src/mongo/shell/query.js:288:5
DBQuery.prototype.toArray@src/mongo/shell/query.js:337:12
@(shell):1:35

最后我尝试了：

var team = db.cards.find()
team.forEach(team_data = db.teams.find({"Team._id":{$in:team.Team}}).toArray())

但它也不起作用：

2018-01-02T18:28:28.345+0000 E QUERY    [thread1] Error: error: {
        "ok" : 0,
        "errmsg" : "$in needs an array",
        "code" : 2,
        "codeName" : "BadValue"
} :
_getErrorWithCode@src/mongo/shell/utils.js:25:13
DBCommandCursor@src/mongo/shell/query.js:717:1

如何正确打印每张卡的所有信息？感谢

Answer 1

您可以通过以下方式之一进行操作。

迭代光标

var teamCursor = db.cards.find();

while (teamCursor.hasNext()) {
   var team = teamCursor.next();
   var team_data = db.teams.find({"Team._id":{$in:team.Card.Team}}).toArray();
   // format output here 
}

Cursor的foreach

db.cards.find().forEach( function(team) {        
  var team_data = db.teams.find({"Team._id":{$in:team.Card.Team}}).toArray();
  // format output here 
});

在3.6（首选）中使用聚合

db.cards.aggregate([
{"$lookup":{
    "from":"teams",
    "localField":"Card.Team",
    "foreignField":"Team._id",
    "as":"team_data"}
},
{"$replaceRoot":{
    "newRoot":{
      "$mergeObjects":[
        {"$arrayElemAt":["$team_data",0]},
        "$$ROOT"
    ]}
  }
},
{"$project":{"team_data":0}}
])

mongodb找到多个参考文件

1 个答案: