Mongodb find()查询:仅返回唯一值(无重复) [英] Mongodb find() query : return only unique values (no duplicates)

问题描述

我有一些文件:

{
    "networkID": "myNetwork1",
    "pointID": "point001",
    "param": "param1"
}
{
    "networkID": "myNetwork2",
    "pointID": "point002",
    "param": "param2"
}
{
    "networkID": "myNetwork1",
    "pointID": "point003",
    "param": "param3"
}
...

pointID是唯一的，但networkID不是.

pointIDs are unique but networkIDs are not.

是否有可能以这样的方式查询Mongodb: [myNetwork1，myNetwork2]

Is it possible to query Mongodb in such a way that the result will be : [myNetwork1,myNetwork2]

现在我只设法返回了[myNetwork1，myNetwork2，myNetwork1]

right now I only managed to return [myNetwork1,myNetwork2,myNetwork1]

我需要一个唯一的networkID列表来填充自动完成的select2组件. 由于我可能有多达5万个文档，因此我更希望mongoDb在查询级别过滤结果.

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.

推荐答案

我认为您可以使用db.collection.distinct(fields,query)

在您的情况下，您将能够获取NetworkID的不同值.

You will be able to get the distinct values in your case for NetworkID.

应该是这样的:

Db.collection.distinct('NetworkID')

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