MongoDB where子句(MongoDB where clause)

 我有JSON格式的mongoDB“用户”集合，我想返回所有隐私真实的数据。 我该怎么做 ？ 
 
{

    "name" : "Maria Kari",
    "social" : [
        {
            "facebook" : "www.fb.com/maria.mongodb",
            "privacy" : true
        },
        {
            "twitter" : "www.tw.com/mongodb",
            "privacy" : false
        }
    ],
    "personal" : [
        {
            "cell_no" : "+1-99082198414",
            "privacy" : true
        },
        {
            "email" : "maria@mongodb.com",
            "privacy" : false
        }
    ]
}
 
 在这里，我想返回隐私真实的数据。 例如，facebook，它的隐私等于真。 如何为此构建查询？ 
 
 谢谢。 ：'） 

I have mongoDB "users" collection in JSON format and I want to return all the data having privacy is true. How can I do it ?
 
{

    "name" : "Maria Kari",
    "social" : [
        {
            "facebook" : "www.fb.com/maria.mongodb",
            "privacy" : true
        },
        {
            "twitter" : "www.tw.com/mongodb",
            "privacy" : false
        }
    ],
    "personal" : [
        {
            "cell_no" : "+1-99082198414",
            "privacy" : true
        },
        {
            "email" : "maria@mongodb.com",
            "privacy" : false
        }
    ]
}
 
Here, I want to return the data having privacy is true. For example, facebook, it has privacy is equal to true. How to build query for this ?
 
Thank You. :')

原文：https://stackoverflow.com/questions/37769629