MongoDB:查找字段名称开头为的对象 [英] MongoDB: Find objects with field names starting with

问题描述

查询MongoDB:从给定的集合中(请参见下面的示例)，我只需要列出包含字段的对象，其中字段名称以"need_"开头.

Query for a MongoDB: From a given collection (see example below) I need only objects listed that comprise fields, where the field name starts with "need_".

具有三个对象的集合的示例

Example of a collection with three objects

/* 1 */
{
    "_id" : 1,
    "need_some" : "A",
    "need_more" : 1,
    "website_id" : "123456789"
}

/* 2 */
{
    "_id" : 2,
    "need_more" : 2,
    "website_id" : "123456789"
}

/* 3 */
{
    "_id" : 3,
    "website_id" : "123456789"
}

所需的输出:

/* 1 */
{
    "_id" : 1,
    "need_some" : "A",
    "need_more" : 1,
    "website_id" : "123456789"
}

/* 2 */
{
    "_id" : 2,
    "need_more" : 2,
    "website_id" : "123456789"
}

查询可能类似于

db.getCollection('nameCollection').find({ "need_.*"  : { "$exists" : true }})

推荐答案

您可以使用 $objectToArray 在mongodb 3.4 及更高版本中

You can use below aggregation using $objectToArray in mongodb 3.4 and above

db.collection.aggregate([
  { "$addFields": {
    "field": { "$objectToArray": "$$ROOT" }
  }},
  { "$match": { "field.k": { "$regex": "need_" }}},
  { "$project": { "field": 0 }}
])

将为您提供 输出

Will give you output

[
  {
    "_id": 1,
    "need_more": 1,
    "need_some": "A",
    "website_id": "123456789"
  },
  {
    "_id": 2,
    "need_more": 2,
    "website_id": "123456789"
  }
]

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