我在javascript中制作一个简单的时间计算器。为简单起见,我已将时间转换为12小时而不是24小时,但是我用于计算上午/下午的代码总是显示am。有什么理由会发生这种情况吗?
这是我的代码:
function solveTime(x) {
var suffixSolve = (utcHours + x) % 24;
var suffix = "am";
if (utcHours > 12) {
var suffix = "pm";
}
if (utcMinutes == 0) {
utcMinutesLead = "00";
}
if (utcMinutes < 10) {
utcMinutesLead = "0" + utcMinutes;
}
var timeSolve = (((utcHours + x) + 11) % 12 + 1);
var timeTotal = timeSolve + ":" + utcMinutesLead + " " + suffix;
var utcMod = x;
if (utcMod > 0) {
utcMod = "+" + utcMod;
}
document.getElementById(x).innerHTML = "(UTC" + utcMod + ") " + timeTotal;
}
这是utcHours背后的代码
var masterTimeUTC = new Date();
var utcHours = masterTimeUTC.getUTCHours();
var utcMinutes = masterTimeUTC.getUTCMinutes();
var utcSeconds = masterTimeUTC.getUTCSeconds();
var utcMinutesLead = masterTimeUTC.getUTCMinutes();
答案 0 :(得分:0)
问题是您应该检查suffixSolve
是否大于12而不是utcHours
,因为utcHours
由于x
的值而没有变化。既然你可以向前和向后移动小时,我创建了一个变量shift
来处理它。
function solveTime(x) {
if (x < 0) {
var shift = 24 + x;
} else {
var shift = x;
}
var suffixSolve = (utcHours + shift) % 24;
var suffix = "am";
if (suffixSolve > 12) {
suffix = "pm";
}
if (utcMinutes == 0) {
utcMinutesLead = "00";
}
if (utcMinutes < 10) {
utcMinutesLead = "0" + utcMinutes;
}
var timeSolve = (((utcHours + x) + 11) % 12 + 1);
var timeTotal = timeSolve + ":" + utcMinutesLead + " " + suffix;
var utcMod = x;
if (utcMod > 0) {
utcMod = "+" + utcMod;
}
document.getElementById(x).innerHTML = "(UTC" + utcMod + ") " + timeTotal;
}
var masterTimeUTC = new Date();
var utcHours = masterTimeUTC.getUTCHours();
var utcMinutes = masterTimeUTC.getUTCMinutes();
var utcSeconds = masterTimeUTC.getUTCSeconds();
var utcMinutesLead = masterTimeUTC.getUTCMinutes();
solveTime(4);
solveTime(0);
solveTime(-8);
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