如何在 mongodb 中严格搜索？

Question

我有一个集合，其中有 70 个文档，如下所示：-

{
    "_id" : ObjectId("629090f9772c8f43cd264662"), 
    "songName" : "Deep End",
    "albumName": "Deep End"
},
{ 
    "_id" : ObjectId("629090f9772c8f43cd264665"), 
    "songName" : "Deep Blue", 
    "albumName": "Pages"
},
{ 
    "_id" : ObjectId("629090f9772c8f43cd264665"), 
    "songName" : "The World", 
    "albumName": "Deep End"
},
{ 
    "_id" : ObjectId("629090f9772c8f43cd264665"), 
    "songName" : "The River",
    "albumName": "A Different Kind Of Human - Step 2"
}

我想严格搜索这 70 个文档，目前我的代码如下所示：-

db.songDetails.aggregate([{$search:{index:"searchDB",text:{query:"deep end",path:{"wildcard":"*"}}}}])

当我运行这样做时，我得到的结果是：-

[
  {
    "_id" : ObjectId("629090f9772c8f43cd264662"), 
    "songName" : "Deep End",
    "albumName": "Deep End"
  },
  { 
    "_id" : ObjectId("629090f9772c8f43cd264665"), 
    "songName" : "Deep Blue",  
    "albumName": "Pages"
  },
  { 
    "_id" : ObjectId("629090f9772c8f43cd264665"), 
    "songName" : "The World",  
    "albumName": "Deep End"
  }
]

但我只想要结果

{
    "_id" : ObjectId("629090f9772c8f43cd264662"), 
    "songName" : "Deep End",
    "albumName": "Deep End"
},
{ 
    "_id" : ObjectId("629090f9772c8f43cd264665"), 
    "songName" : "The World",  
    "albumName": "Deep End"
}

我的意思是，我需要 精确搜索 以从任何字段 {可能来自“songName”或“albumName”}，我该怎么做？

谢谢:)

Answer 1

如果有人正在搜索相同的查询，这里是答案：- 正如@prasad_ 提到的，您可以使用 $or 运算符，如下所示：-

注意：-查询是case-sensitive

var search="Deep End"
db.songDetails.find({$or:[{songName:search},{albumName:search}]})

如果您想搜索 in-case 敏感的查询，那么您可以使用正则表达式，如下所示：-

db.songDetails.find({$or:[{songName:{$regex:/deep end/i}},{albumName:{$regex:/deep end/i}}]})

确保您在正则表达式中添加了 i，因为它会使您的查询 in-case 敏感，请注意，这将导致每个包含 [=21= 的条目]deep end匹配查询（deep end，deep end XYZ，deep end now）

如何在 mongodb 中严格搜索？

how to search strictly in mongodb?

mongodb

aggregation-framework