如何从输出nodejs mongodb数据库中删除数据之一

Question

实际上，我想更改结果中的一些数据，例如添加一行并从返回 json 中删除一行

这里是代码

const Ledgers = db.Ledgers;
const LedgersSync = db.LedgersSync;

      const ledgerList=await Ledgers.find(queryObj)
    
        queryObj["isTransfer"] = 0
        const ledgerListSync=await LedgersSync.find(queryObj)
        
        console.log('ledgerList',ledgerList)
        console.log('ledgerListSync',ledgerListSync)
        
        Array.prototype.push.apply(ledgerList,ledgerListSync);
        return ledgerList 

我得到的结果

{
    "_id": "608bb0725cdd5f0c93d96da0",
    "companyId": 294989,
    "company_id": "60741775b9808d1d885e66d5",
    "Name": "ABC Company",
    "reservedName": "",
    "uid": "4eb25490-8288-4841-a925-ed2512ad6d7b-000000c4",

}

我想要的结果

{
    "_id": "608bb0725cdd5f0c93d96da0",
     "new":test
}

添加新行并删除现有行...

Answer 1

  const ledgerList=await Ledgers.find(queryObj)

    queryObj["isTransfer"] = 0
    const ledgerListSync=await LedgersSync.find(queryObj)
    
    console.log('ledgerList',ledgerList)
    console.log('ledgerListSync',ledgerListSync)
    
    Array.prototype.push.apply(ledgerList,ledgerListSync);
    return {"_id":ledgerList ["_id"],  "new":ledgerList["new"]}

只需返回这两个字段！