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javascript - 如何使用 mongodb 查询返回数组中不可用的给定输入值

转载 作者:可可西里 更新时间:2023-11-01 10:45:16 27 4
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我有一个输入数组,例如:让 UPID = ["0","1","10"]。我必须检查 members.regularStudent 给定的输入值是否可用?假设不可用意味着我必须推送一个数组并返回结果

My documents:

{
"_id" : "5bb20d7556db6915846da67f",
"members" : {
"regularStudent" : [
"3",
"4"
]
}
},

{
"_id" : "5bb20d7556db6915846da55f",
"members" : {
"regularStudent" : [
"1",
"2"
]
}
}

My Expected Output

    [
"0",
"10"
]

My Code:

    let UPID = ["0","1","10"]
db.Groups.find(
/*{
"members.regularStudent": { $nin: UPIDs }
}*/
)
.forEach(function(objects){
print(objects)
})

I had updated mycode, kindly see top on my question section, print(objects) means i am having the my objects, based on this variable can you update your answer,

** 打印(对象)**

{
"_id" : "5bb20d7556db6915846da67f",
"members" : {
"regularStudent" : [
"3",
"4"
]
}
},

{
"_id" : "5bb20d7556db6915846da55f",
"members" : {
"regularStudent" : [
"1",
"2"
]
}
}

最佳答案

您可以使用map方法组合filter

let UPID = ["0","1","10"]; 
let docs = [{ "_id" : "5bb20d7556db6915846da67f", "members" : { "regularStudent" : [ "3", "4" ] } },
{ "_id" : "5bb20d7556db6915846da55f", "members" : { "regularStudent" : [ "1", "2" ] } }]

let ids = [].concat(...docs.map(elem => elem.members.regularStudent));
console.log(UPID.filter(id => !ids.includes(id)));

关于javascript - 如何使用 mongodb 查询返回数组中不可用的给定输入值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54107046/

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