Mongodb 子文档数组包含公共数据时返回文档的Mongo查询

我正在尝试编写一个mongo查询，以返回来自cat1和cat2源的具有完全相同电子邮件的文档

请帮忙。我正在包含一个样本集合架构，其中包含预期匹配的文档

样本采集

{
        "id" : "1",
        "name" : "John",
        "contacts" : [
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": aa@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": john@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat2",
            "detail": john@yahoo.com
                }
        ]
},
{
        "id" : "2",
        "name" : "Bell",
        "contacts" : [
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": jj@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat2",
            "detail": john@yahoo.com
                }
        ]
},

{
        "id" : "3",
        "name" : "Sam",
        "contacts" : [
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": exmaple@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat3",
            "detail": exmaple@yahoo.com
                }
        ]
}

预期结果如下，因为这是具有公共电子邮件地址的文档john@yahoo.com来自源cat1和cat2

{
        "id" : "1",
        "name" : "John",
        "contacts" : [
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": aa@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": john@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat2",
            "detail": john@yahoo.com
                }
        ]
}

> db.collection.find().forEach(function(doc) {
        var cat1Email = new Set(); 
        var cat2Email = new Set(); 

        doc.contacts.forEach(function(c){
            if (c.source === 'cat1'){
                // save the cat1 emails
                cat1Email.add(c.detail); 
                if (cat2Email.has(c.detail)){
                    // if there is same email from cat2, print this document 
                    return printjson(doc);
                }
            }else if(c.source === 'cat2'){
                // save cat2 email
                cat2Email.add(c.detail); 
                if (cat1Email.has(c.detail)){
                    // if there is same email from cat1, print this document
                    return printjson(doc)
                }
            }
        }); 
    })  

不应返回id为2的文档，因为虽然它们都有来自cat1和cat2源的电子邮件，但电子邮件地址不同

不应返回id为3的文档，因为通过电子邮件地址，它们的类别不是cat1和cat2

{
        "id" : "1",
        "name" : "John",
        "contacts" : [
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": aa@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": john@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat2",
            "detail": john@yahoo.com
                }
        ]
}

> db.collection.find().forEach(function(doc) {
        var cat1Email = new Set(); 
        var cat2Email = new Set(); 

        doc.contacts.forEach(function(c){
            if (c.source === 'cat1'){
                // save the cat1 emails
                cat1Email.add(c.detail); 
                if (cat2Email.has(c.detail)){
                    // if there is same email from cat2, print this document 
                    return printjson(doc);
                }
            }else if(c.source === 'cat2'){
                // save cat2 email
                cat2Email.add(c.detail); 
                if (cat1Email.has(c.detail)){
                    // if there is same email from cat1, print this document
                    return printjson(doc)
                }
            }
        }); 
    })  

---

请尝试使用普通JavaScript代码来过滤来自cat1和cat2的相同电子邮件

{
        "id" : "1",
        "name" : "John",
        "contacts" : [
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": aa@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat1",
            "detail": john@yahoo.com
                },
                {
                        "type" : "email",
                        "source": "cat2",
            "detail": john@yahoo.com
                }
        ]
}

> db.collection.find().forEach(function(doc) {
        var cat1Email = new Set(); 
        var cat2Email = new Set(); 

        doc.contacts.forEach(function(c){
            if (c.source === 'cat1'){
                // save the cat1 emails
                cat1Email.add(c.detail); 
                if (cat2Email.has(c.detail)){
                    // if there is same email from cat2, print this document 
                    return printjson(doc);
                }
            }else if(c.source === 'cat2'){
                // save cat2 email
                cat2Email.add(c.detail); 
                if (cat1Email.has(c.detail)){
                    // if there is same email from cat1, print this document
                    return printjson(doc)
                }
            }
        }); 
    })  

---

什么意思？两个文档都有具有相同“cat”源的数组，那么为什么只返回一个文档呢？这是否意味着代表两者的“交叉点”？唯一相交的元素不是“cat1”数据吗？您是否期望从集合中获得所有“交叉点”？或者您只是对第一个文档的“交叉点”感兴趣？欢迎来到StackOverflow。当你向同龄人提问时，你需要比这更清楚。你的问题标题似乎与你的问题内容不同。若要满足您的问题内容，请尝试使用

>db.collection.find（{$and:[{'contacts.source'：'cat1'}，{'contacts.source'：'cat2'}]}）

通过

$and

操作符进行查找。@zangw这并没有比较cat1和cat2电子邮件地址是同一电子邮件地址的事实。谢谢，伙计。实际上，我正试图找到一个解决方案，从.NET级别使用它时，它会更加健壮。我对Mongo相当陌生，但我在想是否需要使用聚合框架来实现这一点。@Raki，我试图通过聚合来解决它，但我失败了。希望有人通过汇总指出。