Question

我有这个mongodb文件。我正在使用pymongo，我想将所有 exercise-names 字段提取为字典，然后使用它在网页上以列表的形式查看它们。

{
"_id" : ObjectId("5b28bfed9ebcea299d08976e"),
"username" : "cybr1998@gmail.com",
"workouts" : [
    {
        "workout-name" : "Treadmill",
        "workout-duration" : "1",
        "workout-frequency" : "Everyday",
        "_id" : "td"
    },
    {
        "workout-name" : "Squats",
        "workout-duration" : "1",
        "workout-frequency" : "Everyday",
        "_id" : "sq"
    },
    {
        "workout-name" : "Running",
        "workout-duration" : "0.5",
        "workout-frequency" : "Everyday",
        "_id" : "rn"
    }
    ]
}

我希望得到所有锻炼名称，如下所示：

{"Treadmill", "Pushups", "Squats"}

我尝试过这个查询：

workouts.find_one({"username":session['username']}, {"workouts.workout-name" : , "_id" : 0})

我收回了这个结果：

{'workouts': [{'workout-name': 'Treadmill'}, {'workout-name': 'Squats'}, {'workout-name': 'Running'}]}

如何改进查询以获得所需的结果？

Answer 1

此代码似乎有效：

list_workouts = []
available_workouts = workouts.find_one({"username": session['username']}, {"workouts.workout-name": 1, "_id": 0})
list_len = len(available_workouts['workouts'])
for i in range(0, list_len):
    list_workouts.append(available_workouts['workouts'][i]['workout-name'])

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1 个答案: