充分利用嵌套數組的整個數組由MongoDB中

Question

給數組名想這是我收集

db.collection.insert([ 
    { 
     "5496":[ 
      { 
       "Asset Name":"HiScanSQ", 
       "Chiller Temperature (deg)":"2", 
       "Trays Used (nos)":"12", 
       "Power (volts)":"123", 
       "Pressure (psi)":"15", 
       "Usage (hours)":"140", 
       "Volume (ml)":"150", 
       "Viscosity":"1.8", 
       "Cartridge Revolution (rpm)":"1000" 
      } 
     ] 
    }, 
    { 
     "4963":[ 
      { 
       "Asset Name":"Genome Analyzer", 
       "Chiller Temperature (deg)":"451", 
       "Trays Used (nos)":"3", 
       "Power (volts)":"82", 
       "Pressure (psi)":"55", 
       "Usage (hours)":"280", 
       "Volume (ml)":"1000", 
       "Viscosity":"1.2", 
       "Cartridge Revolution (rpm)":"976" 
      } 
     ] 
    } 
]); 

我如何獲取整個陣列？說的「5496」全值。我需要獲得與特定資產相關的所有值。所以如果我提供資產id，mongo應該返回所有的資產細節。

Answer 1

試試這個：

db.collection.find({}, { fields: { "5496": 1 }, function(err ,result) { 
    ... 
}); 

Answer 2

db.collection.find({ "5496": { $exists: true } }, { "5496": 1, _id: 0 }).pretty() 
{ 
     "5496" : [ 
       { 
         "Asset Name" : "HiScanSQ", 
         "Chiller Temperature (deg)" : "2", 
         "Trays Used (nos)" : "12", 
         "Power (volts)" : "123", 
         "Pressure (psi)" : "15", 
         "Usage (hours)" : "140", 
         "Volume (ml)" : "150", 
         "Viscosity" : "1.8", 
         "Cartridge Revolution (rpm)" : "1000" 
       } 
     ] 
}