如何在 jquery 中检查 div 的显示（无/块）？
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How to check display (none/block) of a div in jquery?
提问by Mubeen
I am using this,
我正在使用这个，
$("#loginanchor1").click(function (e) {
e.preventDefault();
$("#signin_menu1").slideDown("slow");
});
$(document).mouseup(function (e) {
if ($(e.target).parent("a.loginanchor1").length == 0) {
//$(".signin").removeClass("menu-open");
$("#signin_menu1").slideUp("slow");
}
});
Everything works fine but what happens is when the signin_menu1is displayed block and i click my mouse button inside the div the div slidesup... I want mouseup function to be prevented when the signin_menu1is displayed block. So i thought of changing the condition like,
一切正常，但发生的情况是当signin_menu1显示块并且我在 div 内单击鼠标按钮 div 滑动...我希望在signin_menu1显示块时阻止 mouseup 功能。所以我想改变条件，比如，
if(($(e.target).parent("a.loginanchor1").length==0) &&( //check the display of the div)
if(($(e.target).parent("a.loginanchor1").length==0) &&( //check the display of the div)
Now how to check the display?
现在如何检查显示？
回答by Reigel
try
尝试
$(document).mouseup(function (e) {
var $parent = $(e.target).parent("a.loginanchor1");
if ($parent.length == 0 && !$("#signin_menu1").is(':visible')) {
//$(".signin").removeClass("menu-open");
$("#signin_menu1").slideUp("slow");
}
});
I'm confused with the problem, but $("#signin_menu1").is(':visible')would check if the div is visible (display:block).
我对这个问题感到困惑，但$("#signin_menu1").is(':visible')会检查 div 是否可见（显示：块）。
added notes:
补充说明：
you may try to check if the $(e.target)is the signin_menu1or the is inside signin_menu1. do it like this,
您可以尝试检查$(e.target)is thesignin_menu1或 is inside signin_menu1。像这样做，
$(document).mouseup(function (e) {
if ($(e.target).is('#signin_menu1') || $(e.target).closest('#signin_menu1').length > 0) { return ; } // do nothing on mouseup
var $parent = $(e.target).parent("a.loginanchor1");
if ($parent.length == 0) {
//$(".signin").removeClass("menu-open");
$("#signin_menu1").slideUp("slow");
}
});
