如何從MongoDB獲取匹配的子文件？

Question

我有這樣

{ 
    "_id" : ObjectId("545dad3562fa028fb48832f0"), 
    "key" : "123456", 
    "members" : [ 
      { 
        "name" : "Shiva", 
        "country" : "India", 
        "profession" : "Software" 
      }, 
      { 
        "name" : "Neil", 
        "country" : "Australia", 
        "profession" : "Software" 
      }, 
      { 
        "name" : "anil", 
        "country" : "India", 
        "profession" : "Software" 
      } 

    ] 
} 

集合現在我想檢索與國印度的記錄。 當我試圖這樣

db.user_details.find({ 'key' :'123456','members':{$elemMatch:{'country': "India"}}}, {'members.$': 1}).pretty() 

在執行上面的代碼我得到只有第一次出現，如何從文件

Answer 1

的$elemMatch運營商處獲得所有匹配的子文檔（如name:Shiva和name:Anil）沿對方positional $運營商目前只會匹配符合指定條件的第一個元素。

爲了得到「多於一個的」匹配，最好的方法是使用聚合框架：

db.user_details.aggregate([ 
    # Always match first to at least find the "documents" with matching elements 
    { "$match": { 
     "members.country": "India" 
    }}, 

    # Use $unwind to "de-normalize" the array as individual documents 
    { "$unwind": "$members" }, 

    # Then match again to "filter" the content down to matches 
    { "$match": { 
     "members.country": "India" 
    }}, 

    # Group back to keep an array 
    { "$group": { 
     "_id": "$_id", 
     "key": { "$first": "$key" }, 
     "members": { "$push": "$members" } 
    }} 
]) 

即用於從數組「過濾」多於一個的匹配的基本過程。基本投影目前不能做到。

它返回這樣的：

{ 
    "_id" : ObjectId("545dad3562fa028fb48832f0"), 
    "key" : "123456", 
    "members" : [ 
      { 
        "name" : "Shiva", 
        "country" : "India", 
        "profession" : "Software" 
      }, 
      { 
        "name" : "anil", 
        "country" : "India", 
        "profession" : "Software" 
      } 
    ] 
}