Question

我有这个阵列......

var dates = [{"date_play":"2016-08-22 00:00:00"},{"date_play":"2016-08-23 00:00:00"},{"date_play":"2016-08-24 00:00:00"}];

如何格式化该日期以生成如下：

var newDates = ['08-22-2016','08-23-2016','08-24-2016'];

Answer 1

鉴于您已有文字，您可以操纵文字。

＆＃13;

var dates = [{"date_play":"2016-08-22 00:00:00"},{"date_play":"2016-08-23 00:00:00"},{"date_play":"2016-08-24 00:00:00"}];

var newDates = [];
$.each(dates,function(i,obj){
    newDates.push(obj['date_play'].replace(/(\d{4})-(\d\d)-(\d\d).*$/,'$2-$3-$1'));
});
console.log(newDates);

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

或者：

＆＃13;

var dates = [{"date_play":"2016-08-22 00:00:00"},{"date_play":"2016-08-23 00:00:00"},{"date_play":"2016-08-24 00:00:00"}];

var newDates = [];
dates.forEach(function(obj){
    newDates.push(obj['date_play'].replace(/(\d{4})-(\d\d)-(\d\d).*$/,'$2-$3-$1'));
});
console.log(newDates);

＆＃13;

或者：

＆＃13;

var dates = [{"date_play":"2016-08-22 00:00:00"},{"date_play":"2016-08-23 00:00:00"},{"date_play":"2016-08-24 00:00:00"}];

var newDates = dates.map(function(obj){
    return obj['date_play'].replace(/(\d{4})-(\d\d)-(\d\d).*$/,'$2-$3-$1');
});
console.log(newDates);

＆＃13;

或者：

＆＃13;

var dates = [{"date_play":"2016-08-22 00:00:00"},{"date_play":"2016-08-23 00:00:00"},{"date_play":"2016-08-24 00:00:00"}];

var newDates = dates.map(function(obj){
    var tmp = obj['date_play'].split(/[- ]/,3);
    return tmp[1] + '-' + tmp[2] + '-' + tmp[0];
});
console.log(newDates);

＆＃13;

解释replace()正则表达式：
在这种情况下，replace()与RegExp（正则表达式）一起使用，它使用三个capture groups来暂时保存月，日和年，然后用刚刚替换字符串捕获组的内容由-个字符分隔，但顺序不同（月 - 日 - 年）。

replace(/(\d{4})-(\d\d)-(\d\d).*$/,'$2-$3-$1')
         │ ││ ││ │ │ ││ │ │ │││││    │  │  └───Replace with capture group #1 (year)
         │ │┕┯┙│ │ │ ││ │ │ │││││    │  └──────Replace with capture group #3 (day)
         │ │ │ │ │ │ ││ │ │ │││││    └─────────Replace with capture group #2 (month)
         │ │ │ │ │ │ ││ │ │ ││││└────────────── $   = The end of the string (not really needed here)
         │ │ │ │ │ │ ││ │ │ │││└─────────────── *   = Repeat the previous character, or group, 0 up to as many times as possible ( .* matches the rest of the string)
         │ │ │ │ │ │ ││ │ │ ││└──────────────── .   = Any character
         │ │ │ │ │ │ ││ │ │ │└───────────────── )   = End capture group #3 (which is 2 digits, the day)
         │ │ │ │ │ │ ││ │ │ └────────────────── \d  = A digit
         │ │ │ │ │ │ ││ │ └──────────────────── \d  = A digit
         │ │ │ │ │ │ ││ └────────────────────── (   = Begin capture group #3
         │ │ │ │ │ │ │└──────────────────────── )   = End capture group #2 (which is 2 digits, the month)
         │ │ │ │ │ │ └───────────────────────── \d  = A digit
         │ │ │ │ │ └─────────────────────────── \d  = A digit
         │ │ │ │ └───────────────────────────── (   = Begin capture group #3
         │ │ │ └─────────────────────────────── )   = End capture group #1 (which is 4 digits, the year)
         │ │ └───────────────────────────────── {4} = Repeat the previous character, or group, exactly 4 times (4 digits, the year)
         │ └─────────────────────────────────── \d  = A digit
         └───────────────────────────────────── (   = Begin capture group #1

All of the '-' characters are just '-' characters: part of the old string, or part
of the replacement.

Answer 2

试试这个：

var newDates = [];
    $.each(data, function(i, item) {
       var date = new Date(item.date_play);
       newDates.push((date.getMonth() + 1) + '/' + date.getDate() + '/' +  date.getFullYear());
    })

Answer 3

您不需要jQuery或Date解析器;你所需要的只是分解字符串。

重要的是不要将它转换为日期对象，因为当你这样做时，它会强制执行时区，并且该对象甚至可能歪曲你的一天（例如，2016-08-22可能变成8-21-2016）时区转移。

let dates = [
  {"date_play":"2016-08-22 00:00:00"},
  {"date_play":"2016-08-23 00:00:00"},
  {"date_play":"2016-08-24 00:00:00"}
];
let newDates = dates.map(obj=>{
  let dt = obj.date_play.split(' ')[0].split('-');
  return [dt[1],dt[2],dt[0]].join('-');
});

因为看起来您的数据输入是统一的，所以要么调整来源，要么重新排列字符串内容。

Javascript格式化日期数组

3 个答案: