MongoDB:找到数组中的最小元素并将其删除 [英] MongoDB: Find the minimum element in array and delete it

问题描述

我在MongoDB中有一个文档，其中一个看起来像这样:

I have a documents in MongoDB, one of them looks like this:

{
"_id" : 100,
"name" : "Something",
"items" : [
    {
        "item" : 47,
        "color" : "red"
    },
    {
        "item" : 44,
        "color" : "green"
    },
    {
        "item" : 39,
        "color" : "blue"
    }
]
}

在每个文档中，我需要找到最小的项目并将其删除.所以应该是这样的:

In every document I need to find the minimum item and delete it. So it should be like this:

{
"_id" : 100,
"name" : "Something",
"items" : [
    {
        "item" : 47,
        "color" : "red"
    },
    {
        "item" : 44,
        "color" : "green"
    }
]
}

似乎应该在这里使用findAndModify函数，但是我不能再继续了.

It looks like findAndModify function should be used here but I can't go any further.

如何找到数组中的最小元素并将其删除?

How to find the minimum element in array and delete it?

我正在使用MongoDB和Pymongo驱动程序.

I'm using MongoDB and Pymongo driver.

推荐答案

如果您不局限于将查询放在一个步骤中，则可以尝试:

If you are not restricted to having the query be in one single step, you could try:

步骤1)将聚合函数与$ unwind和$ group运算符配合使用，以查找每个文档的最小项目

step 1) use the aggregate function with the $unwind and $group operators to find the minimum item for each document

myresults = db.megas.aggregate( [ { "$unwind": "$items" },  
    {"$group": { '_id':'$_id' , 'minitem': {'$min': "$items.item" } } } ] )

第2步)遍历结果并从数组中拉出元素

step 2) the loop through the results and $pull the element from the array

for result in myresults['result']:
    db.megas.update( { '_id': result['_id'] }, 
        { '$pull': { 'items': { 'item': result['minitem'] } } } )

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