mongodb 具有多个条件的MongoDB查询
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MongoDB query with multiple conditions
提问by Amrut Gaikwad
I have data with multiple documents :
我有多个文件的数据：
{
"_id" : ObjectId("57b68dbbc19c0bd86d62e486"),
"empId" : "1"
"type" : "WebUser",
"city" : "Pune"
}
{
"_id" : ObjectId("57b68dbbc19c0bd86d62e487"),
"empId" : "2"
"type" : "Admin",
"city" : "Mumbai"
}
{
"_id" : ObjectId("57b68dbbc19c0bd86d62e488"),
"empId" : "3"
"type" : "Admin",
"city" : "Pune"
}
{
"_id" : ObjectId("57b68dbbc19c0bd86d62e489"),
"empId" : "4"
"type" : "User",
"city" : "Mumbai"
}
I want to get data according to my multiple conditions :
我想根据我的多个条件获取数据：
condition 1:- {"type" : "WebUser", "city" : "Pune"}
condition 2:- {"type" : "WebUser", "city" : "Pune"} & {"type" : "User", "city" : "Mumbai"}
I want below result when run condition 1 :
当运行条件 1 时，我想要以下结果：
{
"_id" : ObjectId("57b68dbbc19c0bd86d62e486"),
"empId" : "1"
"type" : "WebUser",
"city" : "Pune"
}
When I run second condition :
当我运行第二个条件时：
{
"_id" : ObjectId("57b68dbbc19c0bd86d62e486"),
"empId" : "1"
"type" : "WebUser",
"city" : "Pune"
}
{
"_id" : ObjectId("57b68dbbc19c0bd86d62e489"),
"empId" : "4"
"type" : "User",
"city" : "Mumbai"
}
I want above result by one query,
我想通过一个查询获得以上结果，
Currently I am using below aggregate query,
目前我正在使用以下聚合查询，
db.emp.aggregate([
{ $match: { '$and': [
{"type" : "WebUser", "city" : "Pune"},
{"type" : "User", "city" : "Mumbai"}
] } },
{ $group: { _id: 1, ids: { $push: "$empId" } } }
])
Above query work for first condition & fails for other. Please help me.
以上查询适用于第一个条件，而其他查询失败。请帮我。
回答by chridam
For the second condition, you can use the $inoperator in your query as:
对于第二个条件，您可以$in在查询中使用运算符作为：
db.emp.find({
"type" : { "$in": ["WebUser", "User"] },
"city" : { "$in": ["Pune", "Mumbai"] }
})
If you want to use in aggregation:
如果要在聚合中使用：
db.emp.aggregate([
{
"$match": {
"type" : { "$in": ["WebUser", "User"] },
"city" : { "$in": ["Pune", "Mumbai"] }
}
},
{ "$group": { "_id": null, "ids": { "$push": "$empId" } } }
])
or simply use the distinct()method to return an array of distinct empIds that match the above query as:
或者简单地使用该distinct()方法返回与上述查询匹配的不同 empId 的数组，如下所示：
var employeeIds = db.emp.distinct("empId", {
"type" : { "$in": ["WebUser", "User"] },
"city" : { "$in": ["Pune", "Mumbai"] }
});
