如何计算24小时格式的两次之间的差异?

时间:2012-06-14 17:23:40

标签: javascript jquery html

在JavaScript中,如何计算两次24小时格式之间的差异?

示例:获取从08:00:0023:00:00已过去的小时数。

下面我从两个下拉菜单中获取两个时间值并尝试计算两次之间的小时差异。我的结果错了......

工作示例:http://jsfiddle.net/VnwF7/1/

脚本:

$(document).ready(function() {
    function calculateTime() {
             //get values
        var valuestart = $("select[name='timestart']").val();
        var valuestop = $("select[name='timestop']").val();

             //create date format       
        var timeStart = new Date("01/01/2007 " + valuestart);
        var timeEnd = new Date("01/01/2007 " + valuestop);

        var difference = timeEnd - timeStart;            
        var diff_result = new Date(difference);    

        var hourDiff = diff_result.getHours();

        $("p").html("<b>Total Hours:</b> " + hourDiff )          
    }
    $("select").change(calculateTime);
    calculateTime();
});

HTML:

<select name="timestart">
<option value="00:00:00">12:00 am</option>
<option value="01:00:00">1:00 am</option>
<option value="02:00:00">2:00 am</option>
<option value="03:00:00">3:00 am</option>
<option value="04:00:00">4:00 am</option>
<option value="05:00:00">5:00 am</option>
<option value="06:00:00">6:00 am</option>
<option value="07:00:00">7:00 am</option>
<option value="08:00:00">8:00 am</option>
<option value="09:00:00">9:00 am</option>
<option value="10:00:00">10:00 am</option>
<option value="11:00:00">11:00 am</option>
<option value="12:00:00">12:00 pm</option>
<option value="13:00:00">1:00 pm</option>
<option value="14:00:00">2:00 pm</option>
<option value="15:00:00">3:00 pm</option>
<option value="16:00:00">4:00 pm</option>
<option value="17:00:00">5:00 pm</option>
<option value="18:00:00">6:00 pm</option>
<option value="19:00:00">7:00 pm</option>
<option value="20:00:00">8:00 pm</option>
<option value="21:00:00">9:00 pm</option>
<option value="22:00:00">10:00 pm</option>
<option value="23:00:00">11:00 pm</option>
</select>

<select name="timestop">
<option value="00:00:00">12:00 am</option>
<option value="01:00:00">1:00 am</option>
<option value="02:00:00">2:00 am</option>
<option value="03:00:00">3:00 am</option>
<option value="04:00:00">4:00 am</option>
<option value="05:00:00">5:00 am</option>
<option value="06:00:00">6:00 am</option>
<option value="07:00:00">7:00 am</option>
<option value="08:00:00">8:00 am</option>
<option value="09:00:00">9:00 am</option>
<option value="10:00:00">10:00 am</option>
<option value="11:00:00">11:00 am</option>
<option value="12:00:00">12:00 pm</option>
<option value="13:00:00">1:00 pm</option>
<option value="14:00:00">2:00 pm</option>
<option value="15:00:00">3:00 pm</option>
<option value="16:00:00">4:00 pm</option>
<option value="17:00:00">5:00 pm</option>
<option value="18:00:00">6:00 pm</option>
<option value="19:00:00">7:00 pm</option>
<option value="20:00:00">8:00 pm</option>
<option value="21:00:00">9:00 pm</option>
<option value="22:00:00">10:00 pm</option>
<option value="23:00:00">11:00 pm</option>
</select>

<p></p>

5 个答案:

答案 0 :(得分:20)

你可以马上以这种方式减去小时数

var valuestart = $("select[name='timestart']").val();
var valuestop = $("select[name='timestop']").val();

//create date format          
var timeStart = new Date("01/01/2007 " + valuestart).getHours();
var timeEnd = new Date("01/01/2007 " + valuestop).getHours();

var hourDiff = timeEnd - timeStart;             

这是工作小提琴http://jsfiddle.net/VnwF7/4/

答案 1 :(得分:4)

我将代码更改为只使用差异而不必创建另一个日期对象:

$(document).ready(function() {    
function calculateTime() {
        //get values
        var valuestart = $("select[name='timestart']").val();
        var valuestop = $("select[name='timestop']").val();

         //create date format          
         var timeStart = new Date("01/01/2007 " + valuestart);
         var timeEnd = new Date("01/01/2007 " + valuestop);

         var difference = timeEnd - timeStart;             

         difference = difference / 60 / 60 / 1000;


    $("p").html("<b>Hour Difference:</b> " + difference)             

}
$("select").change(calculateTime);
calculateTime();
});​

答案 2 :(得分:3)

您不需要将它们转换为此特定情况的日期。做这个

$(document).ready(function() {    
    function calculateTime() { 
        var hourDiff = 
        parseInt($("select[name='timestart']").val().split(':')[0],10) -         
        parseInt($("select[name='timestop']").val().split(':')[0],10);

        $("p").html("<b>Hour Difference:</b> " + hourDiff )         
    }
    $("select").change(calculateTime);
    calculateTime();
});

Working fiddle

答案 3 :(得分:2)

这不是Jquery,而是所有与JavaScript中的时间和日期相关的内容......

这可能会有所帮助:datejs http://code.google.com/p/datejs/

以下是文档中的日期和时间计算

 Date.today().set({ day: 15 })          // Sets the day to the 15th of the current month and year. Other object values include year|month|day|hour|minute|second.

        Date.today().set({ year: 2007, month: 1, day: 20 })


Date.today().add({ days: 2 })          // Adds 2 days to the Date. Other object values include year|month|day|hour|minute|second.

        Date.today().add({ years: -1, months: 6, hours: 3 })


Date.today().addYears(1)               // Add 1 year.
Date.today().addMonths(-2)             // Subtract 2 months.
Date.today().addWeeks(1)               // Add 1 week
Date.today().addHours(6)               // Add 6 hours.
Date.today().addMinutes(-30)           // Subtract 30 minutes
Date.today().addSeconds(15)            // Add 15 seconds.
Date.today().addMilliseconds(200)      // Add 200 milliseconds.

Date.today().moveToFirstDayOfMonth()   // Returns the first day of the current month.
Date.today().moveToLastDayOfMonth()    // Returns the last day of the current month.

new Date().clearTime()                 // Sets the time to 00:00 (start of the day).
Date.today().setTimeToNow()            // Resets the time to the current time (now). The functional opposite of .clearTime()
编辑:由于这是一个相当陈旧的答案,我最近也使用moment.js进行日期相关操作也非常有用。 http://momentjs.com/ https://github.com/moment/moment/

答案 4 :(得分:2)

尝试这种方式

var time_start = new Date();
var time_end = new Date();
var value_start = "06:00:00".split(':');
var value_end = "23:00:00".split(':');

time_start.setHours(value_start[0], value_start[1], value_start[2], 0)
time_end.setHours(value_end[0], value_end[1], value_end[2], 0)

time_end - time_start // millisecond