如何迭代列表中的对象以及对象中的列表和列表中的对象

时间：2019-12-23 16:06:16

标签： javascript node.js 

var timeslots = [
    {
        "details": [{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/16"
        },
        {
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/26"
        }],
        "time": "10:30 AM",
        "status": "Booked"
    },
    {
        "details": [{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/26"
        },{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/26"
        }],
        "time": "11:30 AM",
        "status": "Booked"
    },
    {
        "details": [{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/16"
        },{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/23"
        }],
        "time": "12:30 PM",
        "status": "Booked"
    },
    {
        "details": [{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/16"
        },{
            "name":"naveen",
            "email":"navee@gmail.com",
            "date":"2019/12/16"
        }],
        "time": "02:30 PM",
        "status": "Booked"
    },
    {
        "details": [{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/16"
        },{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/16"
        }],
        "time": "03:30 PM",
        "status": "Booked"
    },
    {
        "details": [{
            "name":"naveen",
            "email":"naveeni@gmail.com",
            "date":"2019/12/16"
        },{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/16"
        }],
        "time": "04:30 PM",
        "status": "Booked"
    },
    {
        "details": [{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/16"
        },{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/16"
        }],
        "time": "05:30 PM",
        "status": "Booked"
    }
]

基于今天的日期，我有09:00 am --09：00 pm之类的时间段，我需要迭代对象列表，并且需要分别在nodejs，javascript中删除以前的日期对象

  

所以请帮助我

     

预期产量

var date ="23/12/2019"
var timeslots = [
    {
        "details": [
        {
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/26"
        }],
        "time": "10:30 AM",
        "status": "Booked"
    },
    {
        "details": [{
            "name":"naveen",
            "email":"naveeni@gmail.com",
            "date":"2019/12/26"
        },{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/26"
        }],
        "time": "11:30 AM",
        "status": "Booked"
    },
    {
        "details": [{
            "name":"naveen",
            "email":"naveen@gmail.com",
            "date":"2019/12/23"
        }],
        "time": "12:30 PM",
        "status": "Booked"
    },
    {
        "details": [],
        "time": "02:30 PM",
        "status": "Booked"
    },
    {
        "details": [],
        "time": "03:30 PM",
        "status": "Booked"
    },
    {
        "details": [],
        "time": "04:30 PM",
        "status": "Booked"
    },
    {
        "details": [],
        "time": "05:30 PM",
        "status": "Booked"
    }
]
  

我该如何迭代

基于今天的日期，我有09:00 am –09：00 pm之类的时间段，我需要迭代对象列表，并且需要分别在nodejs，javascript中删除以前的日期对象

  

所以请帮助我

来自下面的评论

timeslots.time.forEach(function(value) {
  console.log(value.time);
  value.details.forEach(function(model) {
    console.log(value.details);
    value.date.forEach(function(date) {
      console.log(value.date)
    })
  });
});

for (var i = 0; i < timeslots.length; i++) {
  if (timeslots[i].details[i].date < today) {
    list.push(timeslots[i].details[i].date)
  }
}
slots.reduceRight(function(acc, obj, idx) {
  if (list.indexOf(obj.date) > -1) slots.splice(idx, 1);
}, 0);

2 个答案:

答案 0 :(得分：0)

我仔细检查了您的代码，发现有几处不正确的地方可以改进。

var date ="23/12/2019" // string needs to be "2019/12/23" to be parsed by 'Date'
for (var i = 0; i < timeslots.length; i++) { // can be replaced with map
  if (timeslots[i].details[i].date < today) { // this is comparing strings, not dates
    list.push(timeslots[i].details[i].date) // can be replaced by filter on details
  }
}

我已经使用所看到的东西转换了您的代码：

const today = "2019/12/23";

const timeslots = [...]

const answer = timeslots.map(timeSlot => {
    return {
        ...timeSlot, // this is called the spread operator
        details: timeSlot.details.filter(detail => {
            return new Date(detail.date).getTime() >= new Date(today).getTime()
        })
    }
})

我们map timeslots到一个新对象，使用散布运算符复制现有属性。

然后，我们通过将对象的详细信息置于对象声明的下面，使用数组的过滤版本来覆盖其详细信息。

我们正在基于已解析日期字符串的getTime结果进行过滤。 getTime返回Date对象的纪元时间戳。这使我们的比较更加容易，因为它们只是数字。

答案 1 :(得分：0)

如果我的理解正确，您可以这样考虑您的问题：（1）您要删除所有早于特定日期和（2）那么您要删除所有没有任何详细信息的时隙。

考虑学习object-scan。一旦您将数据包起来，它对于数据处理就非常强大。这是解决问题的方法

const objectScan = require('object-scan');

const prune = (date, input) => {
  const jsDate = new Date(date);
  const logic = {
    '[*].details': (value) => value.length === 0,
    '[*].details[*].date': (value) => jsDate > new Date(value)
  };
  return objectScan(Object.keys(logic), {
    rtn: 'bool',
    filterFn: ({ matchedBy, key, value, parents }) => {
      if (matchedBy.some((n) => logic[n](value))) {
        parents[1].splice(key[key.length - 2], 1);
        return true;
      }
      return false;
    }
  })(input);
};

const timeslots = [{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/16"},{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/26"}],"time":"10:30 AM","status":"Booked"},{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/26"},{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/26"}],"time":"11:30 AM","status":"Booked"},{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/16"},{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/23"}],"time":"12:30 PM","status":"Booked"},{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/16"},{"name":"naveen","email":"navee@gmail.com","date":"2019/12/16"}],"time":"02:30 PM","status":"Booked"},{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/16"},{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/16"}],"time":"03:30 PM","status":"Booked"},{"details":[{"name":"naveen","email":"naveeni@gmail.com","date":"2019/12/16"},{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/16"}],"time":"04:30 PM","status":"Booked"},{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/16"},{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/16"}],"time":"05:30 PM","status":"Booked"}];

console.log(prune('2019/12/23', timeslots)); // return true iff changes made
// => true
console.log(JSON.stringify(timeslots));
// => [{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/26"}],"time":"10:30 AM","status":"Booked"},{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/26"},{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/26"}],"time":"11:30 AM","status":"Booked"},{"details":[{"name":"naveen","email":"naveen@gmail.com","date":"2019/12/23"}],"time":"12:30 PM","status":"Booked"}]
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