在mongoDB中进行更新聚合 [英] Aggregation with update in mongoDB

问题描述

我有一个包含许多相似结构化文档的集合，其中两个文档看起来像

I've a collection with many similar structured document, two of the document looks like

输入:

{ 
    "_id": ObjectId("525c22348771ebd7b179add8"), 
    "cust_id": "A1234", 
    "score": 500, 
    "status": "A"
    "clear": "No"
}

{ 
    "_id": ObjectId("525c22348771ebd7b179add9"), 
    "cust_id": "A1234", 
    "score": 1600, 
    "status": "B"
    "clear": "No"
}

默认情况下，所有文档的clear为"No"

By default the clear for all document is "No",

要求:我必须添加具有相同cust_id的所有文档的得分，前提是它们属于status "A"和status "B".如果score超过2000，那么对于具有相同cust_id的所有文档，我必须将clear属性更新为"Yes".

Req: I have to add the score of all documents with same cust_id, provided they belong to status "A" and status "B". If the score exceeds 2000 then I have to update the clear attribute to "Yes" for all of the document with the same cust_id.

预期输出:

{ 
    "_id": ObjectId("525c22348771ebd7b179add8"), 
    "cust_id": "A1234", 
    "score": 500, 
    "status": "A"
    "clear": "Yes"
}

{
    "_id": ObjectId("525c22348771ebd7b179add9"), 
    "cust_id": "A1234", 
    "score": 1600, 
    "status": "B"
    "clear": "Yes"
}

是，因为1600 + 500 = 2100，而2100> 2000.

Yes because 1600+500 = 2100, and 2100 > 2000.

我的方法: 我只能通过聚合函数获取总和，但无法更新

My Approach: I was only able to get the sum by aggregate function but failed at updating

db.aggregation.aggregate([
    {$match: {
        $or: [
            {status: 'A'},
            {status: 'B'}
        ]
    }},
    {$group: {
        _id: '$cust_id',
        total: {$sum: '$score'}
    }},
    {$match: {
        total: {$gt: 2000}
    }}
])

请建议我如何进行.

推荐答案

在经历了很多麻烦之后，尝试使用mongo shell，我终于为我的问题找到了解决方案.

After a lot of trouble, experimenting mongo shell I've finally got a solution to my question.

伪代码:

# To get the list of customer whose score is greater than 2000
cust_to_clear=db.col.aggregate(
    {$match:{$or:[{status:'A'},{status:'B'}]}},
    {$group:{_id:'$cust_id',total:{$sum:'$score'}}},
    {$match:{total:{$gt:500}}})

# To loop through the result fetched from above code and update the clear
cust_to_clear.result.forEach
(
   function(x)
   { 
     db.col.update({cust_id:x._id},{$set:{clear:'Yes'}},{multi:true}); 
   }
)

如果您对同一问题有不同的解决方案，请发表评论.

Please comment, if you have any different solution for the same question.

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