Question

I want to convert time data to the format HH:mm:ss in JavaScript. I ve got a problem in my code (see comments inside the code):

function parseTime(timeString){

    var timeString = timeString.toLowerCase();
    timeString = $.trim(timeString);

    var regEx = /^([0-9]|1[0-9]|2[0-3])$/;
    var regEx2 = /^([0-9]|1[0-9]|2[0-3]).?([0-5][0-9])$/;
    var regEx3 = /^([0-9]|1[0-2])(a|p|am|pm)$/;
    var regEx4 = /^([1-9]|10|11|12).?([0-5][0-9])(a|p|am|pm)$/;

    if(regEx.test(timeString)){
        var hours = timeString;
        if(hours.length == 1){
            hours =  0  + hours;
        }
        return hours +  :00:00 ;
    }
    else if(regEx2.test(timeString)){
        var hoursEndIndex, minutesStartIndex;
        if(timeString.indexOf( . )){
            hoursEndIndex = timeString.indexOf( . );
            minutesStartIndex = timeString.indexOf( . ) + 1;
        }else if(timeString.length == 3){//Problem here timeString.length returns 3 but the code below isn t executed?
            hoursEndIndex = 1;
            minutesStartIndex = 1;
        }else if(timeString.length == 4){//Same thing here?
            hoursEndIndex = 2;
            minutesStartIndex = 2;
            return timeString.length;
        }
        var hours = timeString.substring(0, hoursEndIndex);
        if(hours.length == 1){
            hours =  0  + hours;
        }
        var minutes = timeString.substr(minutesStartIndex, 2);
        return hours +  :  + minutes +  :00 ;
    }

Answer 1

我认为,你正在使用。这里不正确:

if(timeString.indexOf( . )){

文件:

回归第一个指数,该指数在阵列中可找到某一要素,如果不存在,则回归指数为-1。

可以说:

if(timeString.indexOf( . ) > -1) {

With your code the expression in the first if statement will be true even if the string does not contain a dot. This means that the else if statement will never be executed.

Answer 2

我想将几乎任何形式的时间格式改为HHH:mm:javacript

http://www.datejs.com/“rel=“nofollow”http://www.datejs.com/。

没有理由重新发明轮.。

然而,如果要求你执行这一紧急情况,那么我认为马克解决办法将有助于

Answer 3

如果,请重新使用else>,这就要求前面的所有有条件的区块都等同于伪造。



Try this:

    if(timeString.indexOf( . )){
        hoursEndIndex = timeString.indexOf( . );
        minutesStartIndex = timeString.indexOf( . ) + 1;
    }
    if(timeString.length == 3){
        hoursEndIndex = 1;
        minutesStartIndex = 1;
    } else if(timeString.length == 4){
        hoursEndIndex = 2;
        minutesStartIndex = 2;
        return timeString.length;
    }

Answer 4

或许,你应该使用被俘虏的群体,而不是再 par:

var groups = regEx2.exec(timeString);
if(groups){
    var hours = groups[0];
    if(hours.length == 1){
        hours =  0  + hours;
    }
    var minutes = groups[1];
    return hours + ":" + minutes + ":00";
}

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