mongodb中的倍數組聚合

Question

可以說我有這樣的書籍收藏：

{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 "}

我想做類似的查詢：
SELECT author,category, count(*) FROM books GROUP BY category, author
==>結果：

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

我最近去過的解決方案是這樣的：

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

==>結果

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

但是我不明白如何執行帶有類別的第二個“分組依據”。
做這個的最好方式是什么 ？

謝謝

Answer 1

您可以在$group使用的_id包含多個字段，以提供多字段分組：

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

結果：

{
    "_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
}