Javascript 超时后,图像不会重新出现在其正确的位置
ch图像消失,但当它再次出现时,它会从右向左移动Javascript 超时后,图像不会重新出现在其正确的位置,javascript,html,timeout,Javascript,Html,Timeout,ch图像消失,但当它再次出现时,它会从右向左移动 function rest() { img = document.getElementById('ch'); img.style.position= 'absolute'; img.style.top = '0px'; img.style.left = '0px'; img.style.visibility='hidden'; setTimeout(function(){reappearch()},3000); } function reapp
function rest()
{
img = document.getElementById('ch');
img.style.position= 'absolute';
img.style.top = '0px';
img.style.left = '0px';
img.style.visibility='hidden';
setTimeout(function(){reappearch()},3000);
}
function reappearch(){
img = document.getElementById('ch');
img.style.visibility='visible';
}
我该怎么做呢您已经在rest()中将left设置为0px,所以它将位于左侧。如果不想将其显示在左侧,则需要在repearch()函数中添加img.style.left='px',例如img.style.left='150px'
function reappearch() {
img = document.getElementById('ch');
img.style.visibility = 'visible';
img.style.left = '120px';
}
选中此小提琴…您可以使用
img.style.display
并将其设置为“无”
,这样您就不需要重新定位图像并将位置设置为绝对:
function rest() {
img = document.getElementById('ch');
img.style.display = 'none';
setTimeout(function () {
reappearch()
}, 3000);
}
function reappearch() {
img = document.getElementById('ch');
img.style.display = 'block';
}
祝你好运