node.js - 将字符串转换为日期并获取差异

Question

我在“food”模式-mongodb中很少有这样的数据记录

{
    "_id": "5b220199acbec1409cbf84dd",
    "Name": "Mongo",
    "ID_No": "BA1233",
    "Content": "Bottle",
    "No_packages": 5,
    "No_items": 6,
    "Qty": 30,
    "Mfg": "2018-05-27",
    "Exp": "2018-06-30",
    "__v": 0
  }

我想获取日期差(Exp-today Date)小于30的数据集。我使用以下命令尝试了它。但没有成功。

db.food.find({$lt: { $subtract: [ Date() , Exp ],"30"}}  )

Answer 1

您需要先使用 $dateFromString (aggregation)将日期从字符串转换为日期格式，然后 $redact (aggregation)消除不匹配的文档

db.collection.aggregate([
  { "$addFields": {
    "date": {
      "$dateFromString": {
        "dateString": "$EXP"
      }
    }
  }},
  { "$redact": {
    "$cond": [
      { "$lt": [
        { "$divide": [
          { "$subtract": [new Date(), "$date"] },
          1000 * 60 * 60 * 24
        ]},
        30
      ]},
      "$$KEEP",
      "$$PRUNE"
    ]
  }}
])