如何使元素的可见度.hide(), .show(),或.toggle()?
如果一个元素是visible或hidden?
当前回答
此选项可以检查标签是否可见
// using a pure CSS selector
if ($('p:visible')) {
alert('Paragraphs are visible (checked using a CSS selector) !');
};
// using jQuery's is() method
if ($('p').is(':visible')) {
alert('Paragraphs are visible (checked using is() method)!');
};
// using jQuery's filter() method
if ($('p').filter(':visible')) {
alert('Paragraphs are visible (checked using filter() method)!');
};
// you can use :hidden instead of :visible to reverse the logic and check if an element is hidden
// if ($('p:hidden')) {
// do something
// }; 其他回答
$(document).ready(function() {
if ($("#checkme:hidden").length) {
console.log('Hidden');
}
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="checkme" class="product" style="display:none">
<span class="itemlist"><!-- Shows Results for Fish --></span> Category:Fish
<br>Product: Salmon Atlantic
<br>Specie: Salmo salar
<br>Form: Steaks
</div>有两种方法可以检查元素的可见度。
解决方案 # 1
if($('.selector').is(':visible')){
// element is visible
}else{
// element is hidden
}
解决方案 # 2
if($('.selector:visible')){
// element is visible
}else{
// element is hidden
}
用于检查元素是否可见、未显示或甚至不透明水平的扩展功能
它返回时返回false如果元素不可见。
function checkVisible(e) {
if (!(e instanceof Element)) throw Error('not an Element');
const elementStyle = getComputedStyle(e);
if (elementStyle.display === 'none' || elementStyle.visibility !== 'visible' || elementStyle.opacity < 0.1) return false;
if (e.offsetWidth + e.offsetHeight + e.getBoundingClientRect().height +
e.getBoundingClientRect().width === 0) {
return false;
}
const elemCenter = {
x: e.getBoundingClientRect().left + e.offsetWidth / 2,
y: e.getBoundingClientRect().top + e.offsetHeight / 2
};
if (elemCenter.x < 0 || elemCenter.y < 0) return false;
if (elemCenter.x > (document.documentElement.clientWidth || window.innerWidth)) return false;
if (elemCenter.y > (document.documentElement.clientHeight || window.innerHeight)) return false;
let pointContainer = document.elementFromPoint(elemCenter.x, elemCenter.y);
do {
if (pointContainer === e) return true;
} while (pointContainer = pointContainer.parentNode);
return false;
}
你可以试试这个
$(document).ready(function() {
var view = $(this).is(':visible');
if(view) {
alert("view");
// Code
}
else
{
alert("hidden");
}
});
您可以使用这个:
$(element).is(':visible');
示例代码
$(document).ready(function()
{
$("#toggle").click(function()
{
$("#content").toggle();
});
$("#visiblity").click(function()
{
if( $('#content').is(':visible') )
{
alert("visible"); // Put your code for visibility
}
else
{
alert("hidden");
}
});
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>
<p id="content">This is a Content</p>
<button id="toggle">Toggle Content Visibility</button>
<button id="visibility">Check Visibility</button>