MongoDB 在_id 数组中选择哪里？

可以在 mongodb 中选择 Collection 的文档，如 SQL:

SELECT * FROM collection WHERE _id IN (1,2,3,4);

or if i have a _id array i must select one by one and then recompose the array/object of results?

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最佳答案

简单:)

db.collection.find( { _id : { $in : [1,2,3,4] } } );

取自: https://www.mongodb.com/docs/manual/reference/operator/query/in/#mongodb-query-op.-in

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list is a array of ids

此代码列表中是用户集合中的 id 数组

var list = ["5883d387971bb840b7399130","5883d389971bb840b7399131","5883d38a971bb840b7399132"]


.find({ _id: {$in : list}})
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因为 mongodb 使用 bson，而 bson 是重要的属性类型。因为 _id是 ObjectId你必须这样使用:

db.collection.find( { _id : { $in : [ObjectId('1'),ObjectId('2')] } } );

在 mongodb compass中使用如下:

{ "_id" : { $in : [ObjectId('1'),ObjectId('2')] } }

注意: 字符串中的 objectId 具有 24长度。

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if you want to find by user and also by another field like conditionally, you can easily do it like beneath with spread and ternary operator using aggregate and match

 const p_id = patient_id;
let fetchingReports = await Reports.aggregate([
...(p_id
? [
{
$match: {
createdBy: mongoose.Types.ObjectId(id),
patient_id: p_id,
},
},
]
: [
{
$match: {
createdBy: mongoose.Types.ObjectId(id),
},
},
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你可以试试这个

var ids = ["5883d387971bb840b7399130","5883d389971bb840b7399131","5883d38a971bb840b7399132"];


var oids = [];
ids.forEach(function(item){
oids.push(new ObjectId(item));
});


.find({ _id: {$in : oids}})

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