拆分日期字符串到变量

Question

我有以下的日期 -

201409290000 

我想这个使用下面的分裂 -

var timePeriodDay = timePeriod.substring(6, Math.min(timePeriod.length, 2)); 
var timePeriodMonth = timePeriod.substring(4, Math.min(timePeriod.length, 2)); 
var timePeriodYear = timePeriod.substring(0, Math.min(timePeriod.length, 4)); 
var timePeriodTime = timePeriod.substring(8, Math.min(timePeriod.length, 4)); 
var timePeriodTime = [timePeriodTime.slice(0, 2), ":", timePeriodTime.slice(2)].join(''); 

但我的逻辑似乎是掉，其中所需的输出应该是 -

var timePeriodDay = 29 
var timePeriodMonth = 09 
var timePeriodYear = 2014 
var timePeriodTime = 00:00 

我该如何解决这个问题？

Answer 1

尝试使用.substr而不是.substring，因为您的第二个参数是要提取的字符数。

var timePeriod = '201409290000'; 
 
var timePeriodDay = timePeriod.substr(6, Math.min(timePeriod.length, 2)); 
 
var timePeriodMonth = timePeriod.substr(4, Math.min(timePeriod.length, 2)); 
 
var timePeriodYear = timePeriod.substr(0, Math.min(timePeriod.length, 4)); 
 
var timePeriodTime = timePeriod.substr(8, Math.min(timePeriod.length, 4)); 
 
var timePeriodTime = [timePeriodTime.slice(0, 2), ":", timePeriodTime.slice(2)].join(''); 
 
alert(timePeriodDay + ' ' + timePeriodMonth + ' ' + timePeriodYear + ' ' + timePeriodTime);

Answer 2

您可以使用此

timePerio = '201409290000'; 


var timePeriodYear = timePerio.substr(0,4); 
var timePeriodMonth =timePerio.substr(4,2); 
var timePeriodDay = timePerio.substr(6,2); 
var timePeriodTime = timePerio.substr(8,2)+':'+timePerio.substr(10,2);