我正在尝试清理我的代码。只是想知道是否有办法更好地编写这段代码?我在下面添加了一段代码:
$('.popUp-block').on("click",function() {
$('.protect-field-container').removeClass('is-displayed');
$(this).closest('.protect-field-container').addClass('is-displayed');
});
$('.popUp-block').hover(function() {
$('.protect-field-container').removeClass('is-displayed');
$(this).closest('.protect-field-container').addClass('is-displayed');
});
$('.popUp-block').focusout(function() {
$('.protect-field-container').removeClass('is-displayed');
});
$('.popUp-block').on("mouseout", function() {
$('.protect-field-container').removeClass('is-displayed');
});
最佳答案
简单来说,你可以用 .on()
来做到这一点:
$('.popUp-block').on("click mouseover",function() {
$('.protect-field-container').removeClass('is-displayed');
$(this).closest('.protect-field-container').addClass('is-displayed');
});
$('.popUp-block').on("mouseout focusout", function() {
$('.protect-field-container').removeClass('is-displayed');
});
来自docs :
One or more space-separated event types and optional namespaces, such as
"click"
or"keydown.myPlugin"
.
关于javascript - 使用具有相同功能的多个事件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37052762/