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Accepted
Abhijit Mondal Abhi
Asked: 2023-08-20 05:29:46 +0800 CST

如何检查mongodb查询中的数组元素是否相同？

  • 772

我有以下收藏：

[
  {
    "_id": 1,
    "favorite": {
      "color": "red",
      "foods": {
        "fruits": "banana",
        "fastfood": [
          "burger",
          "sandwich"
        ]
      }
    }
  },
  {
    "_id": 2,
    "favorite": {
      "color": "green",
      "foods": {
        "noodles": "ramen",
        "fastfood": [
          "fries",
          "burger",
          "corn dog"
        ]
      }
    }
  },
  {
    "_id": 3,
    "favorite": {
      "color": "red",
      "foods": {
        "soup": "cream soup"
      }
    }
  }
]

我必须进行以下查询：

db.collection.find({
  "favorite.foods.fruits": "banana",
  "favorite.foods.fastfood": {
    "$all": [
      "burger"
    ]
  }
})

当前结果：

[
  {
    "_id": 1,
    "favorite": {
      "color": "red",
      "foods": {
        "fastfood": [
          "burger",
          "sandwich"
        ],
        "fruits": "banana"
      }
    }
  }
]

预期结果： 但这不是我预期的结果。我需要为快餐数组匹配完整的favorite.foods.fastfood对象。如果快餐与我的查询不同，则查询应返回 null/false/empty。

另外，使用 mongoTemplate 的查询结果对我来说也很好。

mongodb

2 个回答

  1. Best Answer

    只需用于$setEquals执行数组比较。

    db.collection.find({
  $expr: {
    $let: {
      vars: {
        fruits_input: "banana",
        fastfood_input: [
          "burger"
        ]
      },
      in: {
        $and: [
          {
            $eq: [
              "$$fruits_input",
              "$favorite.foods.fruits"
            ]
          },
          {
            "$setEquals": [
              "$$fastfood_input",
              "$favorite.foods.fastfood"
            ]
          }
        ]
      }
    }
  }
})

    蒙戈游乐场

    • 1

  2. 如果为了保持相同而顺序很重要，那么只需删除$all运算符即可：

    db.collection.find({
  "favorite.foods.fruits": "banana",
  "favorite.foods.fastfood": [
    "burger"
  ]
})

    游乐场演示

    • 0

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