我有这个代码,我无法第二次正确格式化:
setInterval(function() {
var local = new Date();
var localdatetime = local.getHours() + ":" + local.getMinutes() + ":" + local.getSeconds();
var remote = new Date();
var remotedatetime = remote.getHours() + ":" + remote.getMinutes() + ":" + remote.getSeconds();
var remoteoffset = remote.setHours(local.getHours() - 5);
$('#local-time').html(localdatetime);
$('#remote-time').html(remoteoffset);
}, 1000);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
My Time:
<div id="local-time"></div>
Their time:
<div id="remote-time"></div>
local-time
非常完美,并显示“hh:mm:ss”
remote-time
只显示一个随机数列表。
我怎样才能remote-time
“hh:mm:ss”?
答案 0 :(得分:6)
在获取字符串表示后,您正在调整remote
,这样做对您没有好处。
然后你显示setHours()
的结果(自1970年1月1日以来的毫秒数)而不是字符串。
这就是我认为你的目标:
setInterval(function() {
var local = new Date();
var localdatetime = local.getHours() + ":" + pad(local.getMinutes()) + ":" + pad(local.getSeconds());
var remote = new Date();
remote.setHours(local.getHours() - 5);
var remotedatetime = remote.getHours() + ":" + pad(remote.getMinutes()) + ":" + pad(remote.getSeconds());
$('#local-time').html(localdatetime);
$('#remote-time').html(remotedatetime);
}, 1000);
function pad(t) {
var st = "" + t;
while (st.length < 2)
st = "0" + st;
return st;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
My Time:
<div id="local-time"></div>
Their time:
<div id="remote-time"></div>
答案 1 :(得分:0)
setInterval(function() {
var local = new Date();
var localdatetime = local.getHours() + ":" + local.getMinutes() + ":" + local.getSeconds();
var remote = new Date();
remote.setHours(local.getHours() - 5);
var remotedatetime = remote.getHours() + ":" + remote.getMinutes() + ":" + remote.getSeconds();
$('#local-time').html(localdatetime);
$('#remote-time').html(remotedatetime);
},1000);