我在MongoDB中有这个数据:
{
"_id" : ObjectId("5c7e459f875ea5548de25722"),
"Autos" : [
{
"_id" : ObjectId("5cad9759e1c3895999adaceb"),
"deleted" : 1,
},
{
"_id" : ObjectId("5cad9a8be1c3895999adacef"),
"deleted" : 0,
},
{
"_id" : ObjectId("5cad9aa4e1c3895999adacf0"),
"deleted" : 0,
}
]
}
{
"_id" : ObjectId("5c7e45e9875ea5548de25724"),
"Shoemaking" : [
{
"_id" : ObjectId("5cad9770e1c3895999adacec"),
"deleted" : 1,
},
{
"_id" : ObjectId("5cad9a5de1c3895999adaced"),
"deleted" : 0,
},
]
我想基本上
select * from table where deleted = 0
显示删除记录等于0的位置。
以下是我迄今为止所做的尝试:
db.rental.find({"Autos.deleted":{$ne: 1}}).pretty()
db.rental.find({"Autos": {$elemMatch: {deleted: 1 } } } ).pretty()
db.rental.find({"Autos.deleted": 0},{"Autos": {$elemMatch:
{deleted:0}}});
但以上这些都不适合我。我做错什么了?
期望输出:
{
"_id" : ObjectId("5c7e459f875ea5548de25722"),
"Autos" : [
{
"_id" : ObjectId("5cad9a8be1c3895999adacef"),
"deleted" : 0,
},
{
"_id" : ObjectId("5cad9aa4e1c3895999adacf0"),
"deleted" : 0,
}
]
}
{
"_id" : ObjectId("5c7e45e9875ea5548de25724"),
"Shoemaking" : [
{
"_id" : ObjectId("5cad9a5de1c3895999adaced"),
"deleted" : 0,
},
]
}
我希望输出与上面类似,但是到目前为止我尝试的所有查询要么只选择一个数组记录,要么什么都不选择。
最佳答案
db.rental.aggregate([
{
$project: {
Autos: {
$filter: {
input: "$Autos",
as: "auto",
cond: { $eq:["$$auto.deleted",0] }
}
}
}
}
])
关于node.js - 如何在MongoDB中的Array字段中投影特定元素?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/55608467/