Question

我在mongodb中嵌入了文档如何只删除一个地址，其中pincode：140901并更新其中pincode为：152364

db.add_fun.insert({
"_id" : ObjectId("5a82e6dc1139b572569fa785"),
"name" : "Vikas",
"salary" : 72.0,
"address" : [ 
    {
        "address_id" : ObjectId("5a82f0e51139b572569fa78c"),
        "address" : "Mullanpur ",
        "pincode" : "140901",
        "country" : "India"
    }, 
    {
        "address_id" : ObjectId("5a82f0e51139b572569fa78d"),
        "address" : "mohali ",
        "pincode" : "152364",
        "country" : "India"
    }
]
})

我试试这个但没有工作

db.add_fun.update({},
        {
            $pull: {
                        address: {
                            $elemMatch: {
                                pincode: "140901"
                            }
                        }
            }
        },
        {
            multi:true
        }
    )

想要删除此

{
"address_id" : ObjectId("5a82f0e51139b572569fa78c"),
"address" : "Mullanpur ",
"pincode" : "140901",
"country" : "India"
},

并希望获得此结果

{
    "_id" : ObjectId("5a82e6dc1139b572569fa785"),
    "name" : "Vikas",
    "salary" : 72.0,
    "address" : [ 

        {
            "address_id" : ObjectId("5a82f0e51139b572569fa78d"),
            "address" : "mohali ",
            "pincode" : "152364",
            "country" : "India"
        }
    ]
}

Answer 1

不需要$elemMAtch，只是数组中子文档的条件。在你的情况下：

db.add_fun.update(
{},
{ $pull: {
    address: {
        pincode: "140901"
    }    
} },
{ multi:true }
);

文档：https://docs.mongodb.com/manual/reference/operator/update/pull/

Answer 2

只需在地址栏中添加一个查询，以确保您只更新数组：

db.add_fun.update(
    { address: { $type: 4 }},
    { $pull: {
        address: {
            pincode: "140901"
        }    
     } },
    { multi:true }
);

Answer 3

请允许我建议您通过ObjectId删除对象以保证唯一性。

db.add_fun.update(
    { },
    { $pull: {
        address: {
            "address_id" : ObjectId("5a82f0e51139b572569fa78c")
        }    
     } },
    { multi:true }
);

从mongo db $ pull中删除嵌入式文档无法正常工作

3 个答案: