我在mongo的世界里是个全新的人。到了我不能正确表达我的问题的程度。但是现在开始了..。所以这是我最好的尝试。
db.t2.insertOne( { animal: "cat", action: "meow"});
db.t2.insertOne( { animal: "cat", action: "stalk"});
db.t2.insertOne( { animal: "cat", action: "claw"});
db.t2.insertOne( { animal: "dog", action: "woof"});
db.t2.insertOne( { animal: "dog", action: "beg"});
db.t2.insertOne( { animal: "fish", action: "swim"});数据相当平坦,对吧?
db.t2.find();
{ _id: ObjectId("61142a398f4cfec27b46e89d"),
animal: 'cat',
action: 'meow' }
{ _id: ObjectId("61142a438f4cfec27b46e89e"),
animal: 'cat',
action: 'stalk' }
{ _id: ObjectId("61142a4a8f4cfec27b46e89f"),
animal: 'cat',
action: 'claw' }
{ _id: ObjectId("61142a7e8f4cfec27b46e8a0"),
animal: 'dog',
action: 'woof' }
{ _id: ObjectId("61142a838f4cfec27b46e8a1"),
animal: 'dog',
action: 'beg' }
{ _id: ObjectId("61142a918f4cfec27b46e8a2"),
animal: 'fish',
action: 'swim' }我正在寻找一种方法来查询更压缩的数据。就像..。
[{
"animal":"cat",
"action":["meow", "stalk", "claw"]
},
"animal":"dog",
"action":["woof", "beg"]
},
"animal":"fish",
"action":["swim"]
}]发布于 2021-08-11 20:29:58
您可以对它们进行分组
db.collection.aggregate([
{
"$group": {
"_id": "$animal",
"actions": {
"$push": "$action"
}
}
},
{
"$project": {
"_id": 0,
"animal": "$_id",
"actions": 1
}
}
])如果您希望将数据保存在新表单中,您可以执行上述操作,然后使用$out stage创建一个具有新模式的新集合。
https://stackoverflow.com/questions/68748394
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