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如何从所有 mongodb 文档中仅获取特定值并将它们作为 json 返回?

[英]How to get only specific values from all mongodb documents and return them as json?

I have user documents like this for each individual user:对于每个用户,我都有这样的用户文档:

{
  "_id": {
    "$oid": "638df6dd4774e9573010b138"
  },
  "username": "abc",
  "email": "abc@xyz.com",
  "stats": {
    "ranking": "1",
    "match": "0.214"
  },
  "stats_extra": {
    "pre_ranking": "10",
    "pre_match": "0.290"
  }
}

and I am trying to fetch only "username" and "stats" for each individual user and return them as JSON api response.我试图为每个用户只获取“用户名”和“统计信息”,并将它们作为 JSON api 响应返回。

I can print usernames and stats for each individual user like this:我可以像这样为每个用户打印用户名和统计信息:

@app.get("/Stats",  tags=["userstats"])
def get_stats():
    for doc in app.Users.find():
        print(doc["username"],doc["stats"])
    return { } 

**but I am struggling to find the right way to send all user's usernames and stats as json response like this: ** **但我正在努力寻找将所有用户的用户名和统计信息作为 json 响应发送的正确方法,如下所示:**

{"data": [ 
{"username":"abc", "stats":{"ranking": "1","match": "0.214"}} ,
{"username":"xyz", "stats":{"ranking": "10","match": "0.2104"}} ,
{"username":"ijk", "stats":{"ranking": "12","match": "0.2014"}}]
} 

You can use the projection parameter to indicate which fields of the documents have to be returned.您可以使用投影参数来指示必须返回文档的哪些字段。 Check out this link .查看此链接

In your case something like this should work:在你的情况下,这样的事情应该有效:

app.Users.find(projection={"username": 1, "stats": 1})

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