Question

我有2017-07-25 09:30:49这样的日期，当我减去2017-07-25 10:30:00和2017-07-25 09:30:00时，我需要1 Hours之类的结果。

我找不到正确的搜索键来搜索我需要的内容。任何人都知道我应该在谷歌上搜索什么？或者有人知道某些功能？

PS。 Mysql或Javascript

Answer 1

在javascript中尝试使用日期对象
喜欢这个

var d1 = new Date("2017-07-25 10:30:00"); 
var d2 = new Date("2017-07-25 09:30:49") 
var diff = Math.abs(d1-d2);  // difference in milliseconds

然后将毫秒转换为小时

var hours = parseInt((diff/(1000*60*60))%24);

Answer 2

你可以通过它 Get the time difference between two datetimes

但查询不清楚您是否只想要小时差异，或者您希望差异转换为小时格式

如果2017-07-25 09:30:49和2017-07-26 10:30:00将会给出什么？ 25小时或1小时？

Answer 3

这里有一个如何操作的代码示例

＆＃13;

var date1 = new Date("2017-07-25 09:30:49"); 
var date2 = new Date("2017-07-25 10:30:00");

var datesum = new Date(date1 - date2);
var hours = datesum.getHours();
var minutes = datesum.getMinutes();
var seconds = datesum.getSeconds();

console.log(hours + " hour, " + minutes + " minutes, " + seconds + " seconds" )

＆＃13;

Answer 4

var dateString = "2017-07-25 09:30:49";
var dateString2= "2017-07-25 11:30:00";
var reggie = /(\d{4})-(\d{2})-(\d{2}) (\d{2}):(\d{2}):(\d{2})/;
var dateArray = reggie.exec(dateString); 
var dateArray2= reggie.exec(dateString2); 
var dateObject1= new Date(
(+dateArray[1]),
(+dateArray[2])-1, // Careful, month starts at 0!
(+dateArray[3]),
(+dateArray[4]),
(+dateArray[5]),
(+dateArray[6])
);
var dateObject2= new Date(
(+dateArray2[1]),
(+dateArray2[2])-1, // Careful, month starts at 0!
(+dateArray2[3]),
(+dateArray2[4]),
(+dateArray2[5]),
(+dateArray2[6])
);
var diff = Math.abs(dateObject2-dateObject1);  // difference in milliseconds
 var hours = parseInt((diff/(1000*60*60))%24);

Answer 5

尝试使用以下dateFormatter函数：

var d1 = new Date("2017-07-25 10:30:00"); 
var d2 = new Date("2017-07-25 09:30:00") 
var diff = Math.abs(d1-d2);
var d = dateFormatter(diff);
console.log(d);

function dateFormatter(t){
    var cd = 24 * 60 * 60 * 1000;
    var ch = 60 * 60 * 1000;
    var cm = 60*1000;
    var d = Math.floor(t / cd);
    var h = '0' + Math.floor( (t - d * cd) / ch);
    var m = '0' + Math.round( (t - d * cd - h * ch) / 60000);
    var s = '0' + Math.round((t - (d * cd) - (h * ch) - (m * cm))/1000);
    return d + " days, " + h.substr(-2) + " hours, " + m.substr(-2) + " minutes, " +s.substr(-2)+ " seconds";
}

如何知道减法日期的结果

5 个答案: