Question

我在JSON数组中有完整的日期，但是根据我的需要，我只希望日期而不是月份和年份，就像完整日期是2018-12-01一样，我只想要01。

因为这是我的要求，所以我不知道如何将其连接起来

const raw = [
  [
    "JAYANAGAR",
    "2018-09-01",
    "476426"
  ],
  [
    "MALLESHWARAM",
    "2018-09-01",
    "92141"
  ],
  [
    "KOLAR",
    "2018-09-01",
    "115313"
  ],
  [
    "JAYANAGAR",
    "2018-09-02",
    "511153"
  ],
  [
    "MALLESHWARAM",
    "2018-09-02",
    "115704"
  ],
  [
    "KOLAR",
    "2018-09-02",
    "83597"
  ],
  [
    "JAYANAGAR",
    "2018-09-03",
    "167421"
  ],
  [
    "KOLAR",
    "2018-09-03",
    "53775"
  ]
]
let types = new Set();
const rawObj = raw.reduce((memo, [type, date, value]) => {
  date = date.split('-').reverse().join('-');
  memo[date] = memo[date] || {};
  memo[date][type] = parseInt(value);
  types.add(type);
  return memo;
}, {});
types = [...types];

const data = Object.entries(rawObj).reduce((memo, [date, value]) => {
  memo.push([date, ...types.map(type => value[type] || 0)]);
  return memo;
}, [
  ['Billdate', ...types.map(type => `${type[0]}${type.substr(1).toLowerCase()}`)]
]);

console.log(data)

在我的代码中，我正在格式化我的JSON数据，因为我现在只想要日期而不是月份和年份。

任何人请我指导或建议我如何实现这一目标

Answer 1

var date = new Date('2019-01-02');
alert(date.getDate())

.getDate()是从日期返回日期的函数。

如果日期是字符串做-

console.log(date.slice(-2));

Answer 2

您可以先将旅行日期字符串转换为日期对象，然后使用getDate()获取日期。这将保证支持各种日期时间格式。

var date = new Date('2018-09-01')
console.log(date.getDate());

根据您的数组raw数据，假设您的日期始终为1索引，则可以尝试

const raw = [
  [
    "JAYANAGAR",
    "2018-09-01",
    "476426"
  ],
  [
    "MALLESHWARAM",
    "2018-09-01",
    "92141"
  ],
  [
    "KOLAR",
    "2018-09-01",
    "115313"
  ],
  [
    "JAYANAGAR",
    "2018-09-02",
    "511153"
  ],
  [
    "MALLESHWARAM",
    "2018-09-02",
    "115704"
  ],
  [
    "KOLAR",
    "2018-09-02",
    "83597"
  ],
  [
    "JAYANAGAR",
    "2018-09-03",
    "167421"
  ],
  [
    "KOLAR",
    "2018-09-03",
    "53775"
  ]
]

var dateList = raw.map(data => {
  return new Date(data[1]).getDate();
})

console.log(dateList)

Answer 3

您可以将日期字符串更改为Date对象，并使用getDate函数获取日期部分

var d = new Date('2018-09-03');
console.log(d.getDate())

Answer 4

尝试添加此内容：

const parent = this.props.navigation.dangerouslyGetParent();
const isDrawerOpen = parent && parent.state && parent.state.isDrawerOpen;

Answer 5

您可以通过一点正则表达式魔术来完成此操作：

var regex = /^\d{4}-\d\d-(\d\d)$/;
var dates = []

for (let i = 0; i < raw.length; i++) {
    var match = regex.exec(raw[i][1]);
    dates.push(match[1])
}

正则表达式匹配您的日期字符串（4位数字“-” 2位数字“-” 2位数字），并在将日期推送到数组之前捕获日期。

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions

Answer 6

我将告诉您在第一条语句中以所需的形式获取原始数组，如下所示。

注意：然后在此之后继续前进。使用new Date()语句创建新日期没有意义。

const raw = [
   [
     "JAYANAGAR",
     "2018-09-01",
     "476426"
   ],
   [
     "MALLESHWARAM",
     "2018-09-01",
     "92141"
   ],
   [
     "KOLAR",
     "2018-09-01",
     "115313"
   ],
   [
     "JAYANAGAR",
     "2018-09-02",
     "511153"
   ],
   [
     "MALLESHWARAM",
     "2018-09-02",
     "115704"
   ],
   [
     "KOLAR",
     "2018-09-02",
     "83597"
   ],
   [
     "JAYANAGAR",
     "2018-09-03",
     "167421"
   ],
   [
     "KOLAR",
     "2018-09-03",
     "53775"
   ]
 ].map((l) => { l[1] = l[1].split('-')[2]; return l})

console.log(raw)
/*
  [ [ 'JAYANAGAR', '01', '476426' ],
    [ 'MALLESHWARAM', '01', '92141' ],
    [ 'KOLAR', '01', '115313' ],
    [ 'JAYANAGAR', '02', '511153' ],
    [ 'MALLESHWARAM', '02', '115704' ],
    [ 'KOLAR', '02', '83597' ],
    [ 'JAYANAGAR', '03', '167421' ],
    [ 'KOLAR', '03', '53775' ] ]
*/

Answer 7

尝试此代码

var weekday = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"];
var d = new Date('08/11/2015');
alert(weekday[d.getDay()]);

Answer 8

您可以使用简单易用的moment.js：

<script src="https://momentjs.com/downloads/moment.js"/>

var day = moment(new Date('2018-09-03')).format('DD');

如何仅打印日期而不是完整日期？

8 个答案: