node.js - 多字段传递获取查询失败

Question

我正在尝试根据不同的查询查找用户列表。就像我传递 firstname 或 lastname 或 location 我想要基于其中一个或所有数据的数据，但我没有得到细节。还有 2 个嵌套字段我想通过检查。这些在此处 "otherDetails.nationality": "Indian" 和 "professionalDetails.type": "Cook"。每当我向它们传递一个查询检查时，我都会得到所需的数据，但是每当我将它们全部放入 $or 聚合方法中时，它就会失败并为我提供数据库中的所有数据。在这里卡了2天没解决。帮助将不胜感激

这里是数据库数据的一瞥。不要使用相同的一条记录，我复制了多次，我还删除了一些我让它干净的字段

[{
        "_id": "604670a613a06f0017702d88",
        "firstname": "Babulal ",
        "lastname": "Gadri",
        "mobile": "00000000000",
        "otherDetails": {
            "nationality": "Indian"
        },
        "professionalDetails": {
            "type": "Cook"
        }
}

{
        "_id": "604670a613a06f0017702d8a",
        "firstname": "Babulal ",
        "lastname": "Gadri",
        "mobile": "00000000000",
        "otherDetails": {
            "nationality": "Indian"
        },
        "professionalDetails": {
            "type": "Cook"
        }
}
{
        "_id": "604670a613a06f0017702d8b",
        "firstname": "Babulal ",
        "lastname": "Gadri",
        "mobile": "00000000000",
        "otherDetails": {
            "nationality": "Indian"
        },
        "professionalDetails": {
            "type": "Cook"
        }
}
{
        "_id": "604670a613a06f0017702d88",
        "firstname": "Babulal ",
        "lastname": "Gadri",
        "mobile": "00000000000",
        "otherDetails": {
            "nationality": "Indian"
        },
        "professionalDetails": {
            "type": "Cook"
        }
}
{
        "_id": "604670a613a06f0017702d8c",
        "firstname": "Babulal ",
        "lastname": "Gadri",
        "mobile": "00000000000",
        "otherDetails": {
            "nationality": "Indian"
        },
        "professionalDetails": {
            "type": "Chef"
        }
}

]

这是我正在做的功能

searchCookForNotification: async function (req, res) {
    let { type, nationality, firstname, lastname, location } = req.query;

    var db = Person.getDatastore().manager;

    var cooks = await db
      .collection(Person.tableName)
      .find({
        $or: [
          { firstname },
          { lastname },
          { location },
          { "professionalDetails.type": type },
          { "otherDetails.nationality": nationality },
        ],
      })
      .toArray();

    console.log("Length: ", cooks.length);

    return res.successResponse(
      cooks,
      200,
      null,
      true,
      "Cooks found successfully"
    );
  },
};

Answer 1

这只是因为代码本身不稳定，如果其中一个参数不存在于“查询”中，例如缺少“位置”，那么您在 $or 查询中创建的条件将执行以下查询:

{ location: undefined }

您提供的示例与所有文档匹配。我想您只是想根据给定的输入动态构建查询:

let { type, nationality, firstname, lastname, location } = req.query;
var db = Person.getDatastore().manager;

const orCondArray = [];
if (firstname) {
    orCondArray.push({ firstname })
}
if (lastname) {
    orCondArray.push({ lastname })
}
if (location) {
    orCondArray.push({ location })
}
if (type) {
    orCondArray.push({ "professionalDetails.type": type })
}
if (nationality) {
    orCondArray.push({ "otherDetails.nationality": nationality })
}

if (orCondArray.length) {
  var cooks = await db
    .collection(Person.tableName)
    .find({$or: orCondArray})
    .toArray();
}