javascript 使用时间选择器选择的 jQuery 验证开始时间和结束时间

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/14052432/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-10-26 20:35:13  来源:igfitidea点击:

jQuery-validating start time and end time selected using time picker

javascriptjqueryjquery-uidatepicker

提问by Midhun

I want to validate start time and end time selected using timepickerwhich is in 12 hr format. The end time should be greater than the start time. I used an ifstatement but when the test is using values such as 8:00 AM as start time and 1:00 PM as end time, it is not working. What can I do to solve this problem. Someone please help me with this. I am stuck with this since yesterday. I want just time,i don't need date.

我想验证使用12 小时格式的timepicker选择的开始时间和结束时间。结束时间应大于开始时间。我使用了一个if语句,但是当测试使用诸如 8:00 AM 作为开始时间和 1:00 PM 作为结束时间之类的值时,它不起作用。我能做些什么来解决这个问题。有人请帮我解决这个问题。从昨天开始我就坚持这个。我只想要时间,我不需要日期

$("#dateTimeAddButton").click(function () 
{       
    var Date=$('#myDatePickerId').val()
    var startTime = $('#i').val();   
    var endTime   = $('#i1').val();    
    if (startTime > endTime) 
    {
        alert('End time always greater then start time.');
    }
});

回答by Shiplu Mokaddim

First convert to lowest denominational (minute here). Then compare it.

首先转换为最低面额(此处为分钟)。然后比较一下。

st = minFromMidnight(startTime);
et = minFromMidnight(endTime);
if(st>et){
    alert("End time must be greater than start time");
}

function minFromMidnight(tm){
 var ampm= tm.substr(-2)
 var clk = tm.substr(0, 5);
 var m  = parseInt(clk.match(/\d+$/)[0], 10);
 var h  = parseInt(clk.match(/^\d+/)[0], 10);
 h += (ampm.match(/pm/i))? 12: 0;
 return h*60+m;
}

回答by Mukesh Salaria

please try that below code in case you have length of time more than 6 char like 10:00am instead of 9:00am will throw error above code.

请尝试下面的代码,以防您的时间长度超过 6 个字符,例如 10:00 am 而不是 9:00 am 会在代码上方抛出错误。

I did minor changes in above code and it worked like charm for me.

我对上面的代码做了一些小改动,它对我来说很有魅力。

$('#btnSubmit').on('click', function () {
           
            var startTime = $('#txtJobStartTime').val();
            var endTime = $('#txtJobEndTime').val();
           
            var st = minFromMidnight(startTime);
            var et = minFromMidnight(endTime);
            if (st > et) {
                alert('End time always greater then start time.');
                return false;
            }

            function minFromMidnight(tm) {
                var ampm = tm.substr(-2);
                var clk;
                if (tm.length <= 6) {
                    clk = tm.substr(0, 4);
                } else {
                    clk = tm.substr(0, 5);
                }
                var m = parseInt(clk.match(/\d+$/)[0], 10);
                var h = parseInt(clk.match(/^\d+/)[0], 10);
                h += (ampm.match(/pm/i)) ? 12 : 0;
                return h * 60 + m;
            }
        });