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Javascript 对数组中的时间码求和,仅使用reduce_Javascript - Fatal编程技术网

Javascript 对数组中的时间码求和,仅使用reduce

Javascript 对数组中的时间码求和,仅使用reduce,javascript,Javascript,我有一个带有时间码的数组,它看起来是这样的:[“5:43”,“2:33”,“3:45”,“0:47”,“5:21”,“6:56”,“3:46”,“5:25”,“3:14”,“3:31”,“5:59”,“3:07”,“11:29”,“8:57”,“5:49”,“5:52”,“5:50”,“9:13”,“11:51”,“7:58”,“4:40”,“4:45”,“6:46”,“7:24”,“7:12”,“5:23”,“3:34”,“8:22”,“5:17”,“3:10”,“4:43”,“0:47”,“

我有一个带有时间码的数组,它看起来是这样的:
[“5:43”,“2:33”,“3:45”,“0:47”,“5:21”,“6:56”,“3:46”,“5:25”,“3:14”,“3:31”,“5:59”,“3:07”,“11:29”,“8:57”,“5:49”,“5:52”,“5:50”,“9:13”,“11:51”,“7:58”,“4:40”,“4:45”,“6:46”,“7:24”,“7:12”,“5:23”,“3:34”,“8:22”,“5:17”,“3:10”,“4:43”,“0:47”,“3:59”,“2:43”,“4:17”,“6:56”,“3:05”,“2:06”,“1:59”,“1:49”,“3:36”,“7:10”,“3:44”,“4:36”,“3:16”,“1:10”,“6:10”,“2:14”,“3:44”,“5:05”,“6:03”,“12:39”,“1:56”,“4:04”]
。 我需要用一个reduce方法对数组中的秒数求和

Var timer为我提供了正确的数字列表,我需要求和。出于某种原因,它不想自己求和。如果我将计时器变量设置为一个数组,并将其放入其中,那么它们没有索引,因此无法减少它们

time.reduce((sum, cur) => {
  const [mins, secs] = cur.split(':');
  let timer;
  timer = (parseFloat(mins * 60) + parseFloat(secs));
  console.log(timer);
}); 

您就快到了,但问题是您没有从reduce回调返回任何内容。该回调返回的值将用作下一次迭代的
sum

const-arr=[“5:43”,“2:33”,“3:45”,“0:47”,“5:21”,“6:56”,“3:46”,“5:25”,“3:14”,“3:31”,“5:59”,“3:07”,“11:29”,“8:57”,“5:49”,“5:52”,“5:50”,“9:13”,“11:51”,“7:58”,“4:40”,“4:45”,“6:46”,“7:24”,“7:12”,“5:23”,“3:34”,“8:22”,“5:17”,“3:10”,“4:43”,“19:43”,“0:47”,“0:47”,“3:47”,“3:40”,“4:45”,“5:05”,“5:56”, "1:59", "1:49", "3:36", "7:10", "3:44", "3:44", "4:36", "3:16", "1:10", "6:10", "2:14", "3:44", "5:05", "6:03", "12:39", "1:56", "4:04"]
const total=arr.reduce((ac,a)=>{
设[min,secs]=a.split(':').map(Number);
返回ac+秒+(最小*60)
}, 0)
log(总计)
您已经使用了映射方法)没关系,我已经直接向VAR添加了数字。非常感谢你!