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Question

这是一个很简单的问题，我找遍了所有，但could'nt找到答案

我只有

{ 
"_id" : 1, 
    "name" : { 
     "first" : "John", 
     "last" : "Backus" 
    }, 
"awards" : [ 
     { 
      "award" : "W.W. McDowell Award", 
      **"year" : 1967,** 
      "by" : "IEEE Computer Society" 
     }, 
     { 
      "award" : "Draper Prize", 
      **"year" : 1967,** 
      "by" : "National Academy of Engineering" 
     }, 
     { 
      "award" : "National Medal of Science", 
      "year" : 1975, 
      "by" : "National Science Foundation" 
     } 
    ] 
}, 
{ 
"_id" : 2, 
    "name" : { 
     "first" : "John", 
     "last" : "McCarthy" 
    }, 
"awards" : [ 
     { 
      "award" : "Turing Award", 
      "year" : 1971, 
      "by" : "ACM" 
     }, 
     { 
      "award" : "Kyoto Prize", 
      **"year" : 1967,** 
      "by" : "Inamori Foundation" 
     }, 
     { 
      "award" : "National Medal of Science", 
      "year" : 1990, 
      "by" : "National Science Foundation" 
     } 
    ] 
}, 

所有我需要的是，一直是所有奖项在1967年赢得了！

，我试图

，我只得到

"_id" : 1, 
"awards" : [ 
     { 
      "award" : "W.W. McDowell Award", 
      "year" : 1967, 
      "by" : "IEEE Computer Society" 
     }, 

], 

，但我需要在所有的文件，找到（在firsdt一个有两个这样totalyy 3样

{ 
"_id" : 1, 
    "awards" : [ 
      { 
       "award" : "W.W. McDowell Award", 
       "year" : 1967, 
       "by" : "IEEE Computer Society" 
      }, 
      { 
       "award" : "Draper Prize", 
       **"year" : 1967,** 
       "by" : "National Academy of Engineering" 
      } 
    ] 
}, 
{ 
    "_id" : 2, 
    "award":[{ 
       "award" : "Kyoto Prize", 
       "year" : 1967, 
       "by" : "Inamori Foundation" 
      } 
    ] 
} 

我尝试使用$all，但无法得到结果。我希望我的明确 是的，我正在使用BIOS的例子。 这一定很简单，我正在做短暂的动作 请帮忙。

Answer 1

您必须使用aggregation framework，因为$仅指第一个匹配项。在你的情况，以下可能的工作：

db.bios.aggregate([ 
    { $match : {"awards.year":1967} }, 
    { $unwind : "$awards"}, 
    { $match : {"awards.year" :1967} } 
]); 

屈服

{ 
     "_id" : 1, 
     "name" : { 
       "first" : "John", 
       "last" : "Backus" 
     }, 
     "awards" : { 
       "award" : "W.W. McDowell Award", 
       "year" : 1967, 
       "by" : "IEEE Computer Society" 
     } 
} 
{ 
     "_id" : 1, 
     "name" : { 
       "first" : "John", 
       "last" : "Backus" 
     }, 
     "awards" : { 
       "award" : "Draper Prize", 
       "year" : 1967, 
       "by" : "National Academy of Engineering" 
     } 
} 
{ 
     "_id" : 2, 
     "name" : { 
       "first" : "John", 
       "last" : "McCarthy" 
     }, 
     "awards" : { 
       "award" : "Kyoto Prize", 
       "year" : 1967, 
       "by" : "Inamori Foundation" 
     } 
} 

正如你所看到的，将会复制该文件（所以约翰·巴库斯的文档返回两次）的其余部分，但你可以通过添加一个$group来改变，例如

db.bios.aggregate([ { $match : {"awards.year":1967} },  
    { $unwind : "$awards"},  
    { $match : {"awards.year" :1967} }, 
    { $group : { _id : "$_id", "awards" : { $push : "$awards" } } } 
]).pretty(); 

但是有很多方法可以做到这一点，所以我建议您深入研究文档以获取所需的投影。