仅获取mongodb中数组属性的第一项

Question

鉴于集合中的以下布局......

{
  vehicle_id: 1
  ,// bunch of properties I don't want
  ,vehicle: {
    mfg_year: 1928
    ,mfg_make: "Ford"
    ,mfg_model: "Model A"
    ,mfg_trim: "T-Bucket"
    ,// bunch of properties I don't want
    ,images: [
      {url:'...',...}
      ,...
    ]
  }
}

如何只返回上述字段的结果,而只返回图像下的第一个结果？我不介意结果是否在一个展平对象中,只有图像是嵌套对象.

我查看了聚合框架,它似乎与我正在寻找的不匹配.我知道我可以在结果集上做map/reduce,或者在listing_id上做一个组,我只是希望有一个更简单的查询结构,而不需要诉诸组或减少.

如果目前无法通过聚合框架实现这一点,那么工作组或map-reduce将是可接受的答案.

---

编辑:在最终结果中有大约50多个我不想要的属性..使用$ slice指令,似乎我不能只指定我想要的字段.

Answer 1

怎么样:

db.vehicles.find({"vehicle_id":1}, {images:{$slice: 1}})

来源:http: //www.mongodb.org/display/DOCS/Retrieving+a+Subset+of+Fields#RetrievingaSubsetofFields-RetrievingAsubrangeofArrayElements

例

db.vehicles.insert({"vehicle_id": 1, "mfg_year": "1928", "mfg_make": "Ford", "mfg_model": "Model A", "images": [{"url":"www.a.com"}, {"url":"www.b.com"}, {"url":"www.c.com"}]})

db.vehicles.insert({"vehicle_id": 2, "mfg_year": "1999", "mfg_make": "BMW", "mfg_model": "Model B", "images": [{"url":"www.a.com"}, {"url":"www.b.com"}, {"url":"www.c.com"}]})

db.vehicles.insert({"vehicle_id": 3, "mfg_year": "1998", "mfg_make": "FMerc", "mfg_model": "Model C", "images": [{"url":"www.a.com"}, {"url":"www.b.com"}, {"url":"www.c.com"}]})


//now the query
db.vehicles.find({"vehicle_id":1}, {images:{$slice: 1}})

输出:

{
        "vehicle_id" : 1,
        "mfg_year" : "1928",
        "mfg_make" : "Ford",
        "mfg_model" : "Model A",
        "images" : [
                {
                        "url" : "www.a.com"
                }
        ]
}

编辑

您可以指定只是要返回这样的字段:

db.vehicles.find({"vehicle_id":1}, {"mfg_make":1, images:{$slice: 1}})

所以,在这种情况下,只有mfg_make和images返回.

另一个....

db.vehicles.find({"vehicle_id":1}, {"mfg_make":1, "some_other_field":1, images:{$slice: 1}})

如果这是一个RDBMS,则此查询等效于:

SELECT mfg_make, some_other_field FROM tblVehicles WHERE vehicle_id = 1