Mongodb聚合。我需要过滤在多个数组中匹配的结果

Question

我有一个查询需要 $in 参数不同的数组。第一个是 openBehaviourIds，第二个是 closeBehaviours。我想避免重复的代码并使用聚合进行一次查询以获得结果 {openBehaviours：[{}，{}...]，closeBehaviours：[{}，{}，{}...]} 代替使用：

const openBehaviours = await behavioursCollection.aggregate([...])
const closeBehaviours = await behavioursCollection.aggregate([...])

打开和关闭 behaviorId 是字符串数组

获取以下两种行为的完整查询（几乎相同）：

const openBehaviours = await behavioursCollection.aggregate([
            { $match: { 'behaviourId': { $in: openBehaviourIds } } }, {
                $lookup: {
                    from: 'colors',
                    localField: 'domainId',
                    foreignField: 'domainId',
                    as: 'color'
                }
            }, { $unwind: '$color' }
        ], { session }).toArray()

const closeBehaviours = await behavioursCollection.aggregate([
            { $match: { 'behaviourId': { $in: closeBehaviours } } }, {
                $lookup: {
                    from: 'colors',
                    localField: 'domainId',
                    foreignField: 'domainId',
                    as: 'color'
                }
            }, { $unwind: '$color' }
        ], { session }).toArray()

如何将其合并到一个查询中以使其 crystal 清晰？感谢帮助

Answer 1

您可以使用 $facet 运算符来实现。

behavioursCollection.aggregate([
{$facet:{
    "openBehaviours":[
        { $match: { 'behaviourId': { $in: openBehaviourIds } } },
        {
            $lookup: {
                from: 'colors',
                localField: 'domainId',
                foreignField: 'domainId',
                as: 'color'
            }
        },
        { $unwind: '$color' }
    ],
    "closeBehaviours":[
        { $match: { 'behaviourId': { $in: closeBehaviours  } } },
        {
            $lookup: {
                from: 'colors',
                localField: 'domainId',
                foreignField: 'domainId',
                as: 'color'
            }
        },
        { $unwind: '$color' }   
    ]
}}]);

Mongodb聚合。我需要过滤在多个数组中匹配的结果

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