Mongodb find() 查询：仅返回唯一值（无重复）
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Mongodb find() query : return only unique values (no duplicates)
提问by Franckl
I have a collection of documents :
我有一组文件：
{
"networkID": "myNetwork1",
"pointID": "point001",
"param": "param1"
}
{
"networkID": "myNetwork2",
"pointID": "point002",
"param": "param2"
}
{
"networkID": "myNetwork1",
"pointID": "point003",
"param": "param3"
}
...
pointIDs are unique but networkIDs are not.
pointID 是唯一的，但 networkID 不是。
Is it possible to query Mongodb in such a way that the result will be : [myNetwork1,myNetwork2]
是否可以以这样的方式查询 Mongodb，结果将是：[myNetwork1,myNetwork2]
right now I only managed to return [myNetwork1,myNetwork2,myNetwork1]
现在我只能返回 [myNetwork1,myNetwork2,myNetwork1]
I need a list of unique networkIDs to populate an autocomplete select2 component. As I may have up to 50K documents I would prefer mongoDb to filter the results at the query level.
我需要一个唯一的 networkID 列表来填充自动完成 select2 组件。由于我可能有多达 50K 个文档，我更喜欢 mongoDb 在查询级别过滤结果。
回答by Laura Uzcategui
I think you can use db.collection.distinct(fields,query)
我想你可以用 db.collection.distinct(fields,query)
You will be able to get the distinct values in your case for NetworkID.
您将能够在您的案例中为 NetworkID 获得不同的值。
It should be something like this :
它应该是这样的：
Db.collection.distinct('NetworkID')
