我需要知道,我可以进行复杂的查询,它会返回数组板的结果,但只有参数Object在里面:
我在mongodb中很弱,所以我问是否有可能这样做,因为我有订单,我不知道是在查询中还是在angularjs中的循环中
这是来自db结果的查询db.getCollection('boards')。find({})(2results)
/* 1 */
{
"_id" : ObjectId("5a0b5c6da0502e2174fd849f"),
"name" : "example 1",
"users" : [
ObjectId("59cd114cea98d9326ca1c421")
],
"lists" : [
{
"list" : "example",
"_id" : ObjectId("5a0b5c7ba0502e2174fd84ae"),
"cards" : [
{
"name" : "1",
"_id" : ObjectId("5a0b5c80a0502e2174fd84b4")
},
{
"name" : "2",
"_id" : ObjectId("5a0b5c80a0502e2174fd84b3")
},
{
"name" : "3",
"_id" : ObjectId("5a0b5c80a0502e2174fd84b2")
}
]
},
{
"list" : "example 1",
"_id" : ObjectId("5a0b5c7ba0502e2174fd84ad"),
"cards" : [
{
"name" : "1",
"_id" : ObjectId("5a0b5c83a0502e2174fd84b5")
}
]
},
{
"list" : "example",
"_id" : ObjectId("5a0b5c7ba0502e2174fd84ac"),
"cards" : [
{
"name" : "2",
"_id" : ObjectId("5a0b5c85a0502e2174fd84b6")
}
]
}
],
"__v" : 0
}
/* 2 */
{
"_id" : ObjectId("5a0b5c71a0502e2174fd84a4"),
"name" : "example 2",
"users" : [
ObjectId("59cd114cea98d9326ca1c421")
],
"lists" : [
{
"list" : "example next 1",
"_id" : ObjectId("5a0b5c93a0502e2174fd84bc"),
"cards" : [
{
"name" : "1",
"_id" : ObjectId("5a0b5c95a0502e2174fd84c2")
},
{
"name" : "2",
"_id" : ObjectId("5a0b5c95a0502e2174fd84c1")
},
{
"name" : "3",
"_id" : ObjectId("5a0b5c95a0502e2174fd84c0")
}
]
},
{
"list" : "example next 2",
"_id" : ObjectId("5a0b5c93a0502e2174fd84bb"),
"cards" : [
{
"name" : "1",
"_id" : ObjectId("5a0b5c98a0502e2174fd84c5")
},
{
"name" : "2",
"_id" : ObjectId("5a0b5c98a0502e2174fd84c4")
}
]
},
{
"list" : "example next 3",
"_id" : ObjectId("5a0b5c93a0502e2174fd84ba"),
"cards" : [
{
"name" : "1",
"_id" : ObjectId("5a0b5c9aa0502e2174fd84c6")
}
]
}
],
"__v" : 0
}
答案 0 :(得分:0)
是,
您可以使用聚合框架构建自己的响应。
https://docs.mongodb.com/manual/aggregation/
我使用的框架中最有用的阶段是:
$project: Structure what I want.
$match: Very similiar to select in sql.
$group: to make groups
$sort: to sort :D
$unwind: to unwind arrays.
我认为还有更多,但取决于你想要的结构,这里有一个很好的教程:
https://www.tutorialspoint.com/mongodb/mongodb_aggregation.htm