Question

考虑一下：

{
  "movies": [
    {
      "title": "Star Wars",
      "year": 1977,
      "director": "George Lucas"
    },
    {
      "title": "The Empire Strikes Back",
      "year": 1980,
      "director": "Irvin Kershner"
    },
    {
      "title": "Return of the Jedi",
      "year": 1983,
      "director": "Richard Marquand"
    },
    {
      "title": "The Phantom Menace",
      "year": 1999,
      "director": "George Lucas"
    },
    {
      "title": "Attack of the Clones",
      "year": 2002,
      "director": "George Lucas"
    },
    {
      "title": "Revenge of the Sith",
      "year": 2005,
      "director": "George Lucas"
    },
    {
      "title": "The Force Awakens",
      "year": 2015,
      "director": "J.J. Abrams"
    }
  ]
}

我正试图推出Geroge Lucas导演的所有电影。这是我尝试过的，它会返回所有项目：

db.movies.find( {"movies.director" : "George Lucas"} ).pretty()

这也会导致错误：

db.movies.find({"$pull" : {"movies.director" : "George Lucas"}}).pretty()

请让我知道如何查询数据库以仅检索导演键具有“George Lucas”值的电影。

Answer 1

您需要使用聚合框架：

db.movies.aggregate([
{
    $unwind : "$movies"
},
{
    $match : {
        "movies.director" : "George Lucas"
    }
},
{
    $project : {
        _id : 0,
        movies : 1
    }
}
])

从MongoDB中的JSON对象数组中提取值

1 个答案: