MongoDB中的倍数组聚合

Question

假设我收藏了这样的书籍：

{author:"john",  category:"action", title:"foobar200"},  
{author:"peter", category:"scifi" , title:"42test"},  
{author:"peter", category:"novel",  title:"whatever_t"},  
{author:"jane",  category:"novel",  title:"the return"},  
{author:"john",  category:"action", title:"extreme test"},  
{author:"peter", category:"scifi",  title:"such title"},  
{author:"jane",  category:"action", title:"super book "}

我想做一个类似的查询：
SELECT author,category, count(*) FROM books GROUP BY category, author
==> 结果：

john -> action -> 2  
john -> novel  -> 0  
john -> scifi  -> 0  
jane -> action -> 1
etc...

我最接近解决方案的是：

 db.books.aggregate(
       { 
         $match: {category:"action"} 
        }, 
       {
         $group: { _id: '$author', result: { $sum: 1 } } 
      }
);

==> 结果

{ "_id" : "jane",  "result" : 1 }
{ "_id" : "john",  "result" : 2 }
{ "_id" : "peter", "result" : 0 }

但我不明白如何使用类别执行第二个“分组依据”。
最好的方法是什么？

谢谢

Answer 1

您可以在$group 使用的_id 中包含多个字段以提供多字段分组：

db.books.aggregate([
    {$group: {
        _id: {category: '$category', author: '$author'}, 
        result: {$sum: 1}
    }}
])

结果：

{
    "_id" : {
        "category" : "action",
        "author" : "jane"
    },
    "result" : 1
}, 
{
    "_id" : {
        "category" : "novel",
        "author" : "jane"
    },
    "result" : 1
}, 
{
    "_id" : {
        "category" : "novel",
        "author" : "peter"
    },
    "result" : 1
}, 
{
    "_id" : {
        "category" : "scifi",
        "author" : "peter"
    },
    "result" : 2
}, 
{
    "_id" : {
        "category" : "action",
        "author" : "john"
    },
    "result" : 2
}