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Accepted
Rishabh Jain
Asked: 2020-01-28 22:11:59 +0800 CST

如何仅获取那些在猫鼬中具有所有匹配字段的文档

  • 772

假设我有以下名为的集合survey：

{ _id: 1, results: [ { product: "abc", active: true}, { product: "xyz", active: true} ] }
{ _id: 2, results: [ { product: "abc", active: false }, { product: "xyz", active: true } ] }
{ _id: 3, results: [ { product: "abc", active: false }, { product: "xyz", active: false } ] }

通过做：

db.survey.find(
   { results: { $elemMatch: { active:  false } }
)

我将获得数组中至少一个对象为假的所有文档，即

{ _id: 2, results: [ { product: "abc", active: false }, { product: "xyz", active: true } ] }

{ _id: 3, results: [ { product: "abc", active: false }, { product: "xyz", active: false } ] }

但是我怎样才能只获得那些拥有所有关键的文档，active: false 即：

{ _id: 3, results: [ { product: "abc", active: false }, { product: "xyz", active: false } ] }

蒂亚..

mongodb mongodb-4.0

1 个回答

  1. Best Answer

    所以在这里我得到了一个解决方案，它非常适合我的需要：

    db.getCollection('survey').aggregate([
    { "$match": { "results.active": false } },
    { "$redact": {
        "$cond": {
            "if": {
                "$allElementsTrue": {
                    "$map": {
                        "input": "$results",
                        "as": "result",
                        "in": { "$eq": [ "$$result.active", false ] }
                    }
                }               
            },
            "then": "$$KEEP",
            "else": "$$PRUNE"
        }
    }}    
])
    • 0

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