问题
I have selected a specific class via .querySelectorAll
:
var hit3 = document.querySelectorAll(".lattern.hit-3 .circle");
I am now trying to target and adjust a .style.visibility
attribut on this element, by doing the following:
hit3.style.visibility = "visible";
This however results in an error:
Uncaught TypeError: Cannot set property 'visibility' of undefined
How do I target a specific .style
with the above .querySelectorAll
selector?
Fiddle
回答1:
querySelectorAll
return a node list so you should specify the index of element you want to change :
hit3[0].style.visibility = "visible";
If you want to change the css of all the elements returned you should loop through them, see Johny's answer.
Hope this helps.
回答2:
querySelectorAll
returns an array like structure(NodeList) which does not have the style
attribute.
But I think what you need is slightly different, I assume want to display the circle for the clicked element then
var latternElement = document.querySelectorAll('.lattern');
function toggleElement(el) {
el.querySelector('.circle').classList.add('visible'); //also minor tweaks, use css rules
}
for (var i = 0; i < latternElement.length; i++) {
latternElement[i].addEventListener('click', function(event) {
if (this.classList.contains("hit-3")) { //minor tweaks - only supported in modern browsers
toggleElement(this);
}
});
}
.lattern {
position: relative;
width: 100px;
height: 50px;
background-color: red;
margin: 0 0 10px 0;
cursor: pointer;
}
.circle {
position: relative;
top: 20px;
left: 20px;
border-radius: 50% 50%;
width: 16px;
height: 16px;
background-color: green;
visibility: hidden;
}
.circle.default,
.circle.visible {
visibility: visible;
}
<div class="lattern hit-1">
<div class="circle"></div>
</div>
<div class="lattern hit-2">
<div class="circle default"></div>
</div>
<div class="lattern hit-3">
<div class="circle"></div>
click me
</div>
来源:https://stackoverflow.com/questions/35601756/how-to-target-style-attribut-with-queryselectorall-selector