Answer the question
In order to leave comments, you need to log in
How to integrate tabs in select?
I want to make a choice of tabs through select, the design <option VALUE="#" data-tab="1">Скриншоты</option>
does not work.
There is a select form from Bootstrap
<select class="form-select form-select-sm" aria-label=".form-select-sm example" onchange="window.location.href=this.options[this.selectedIndex].value">
<option selected>2022</option>
<option VALUE="#" data-tab="1">Скриншоты</option>
<option VALUE="#" data-tab="2">2024</option>
</select>
<div class="nav-tabs1">
<a href="#" data-tab="1" class="nav-tab active">Скриншоты</a>
<a href="#" data-tab="2" class="nav-tab">Видео</a>
</div>
<div class="tab-content ">
<div data-tab-content="1" class="tab-pane active">
<h5 class="fw-bold mb-3 my-1">Скриншоты</h5>
Контент
</div>
<div data-tab-content="2" class="tab-pane">
<h5 class="fw-bold mb-3 my-1">Видео</h5>
контент
</div>
</div>
var tabNavs = document.querySelectorAll(".nav-tab");
var tabPanes = document.querySelectorAll(".tab-pane");
for (var i = 0; i < tabNavs.length; i++) {
tabNavs[i].addEventListener("click", function(e){
e.preventDefault();
var activeTabAttr = e.target.getAttribute("data-tab");
for (var j = 0; j < tabNavs.length; j++) {
var contentAttr = tabPanes[j].getAttribute("data-tab-content");
if (activeTabAttr === contentAttr) {
tabNavs[j].classList.add("active");
tabPanes[j].classList.add("active");
} else {
tabNavs[j].classList.remove("active");
tabPanes[j].classList.remove("active");
}
};
});
}
.nav-tabs1{
display: flex;
margin-bottom:-17px !important;
}
.nav-tab{
margin-right: 20px;
text-decoration: none;
}
.nav-tab.active{
border-bottom: 3px solid #0d6efd;
padding-bottom: 12px;
cursor: default;
font-weight: 700 !important;
}
.tab-pane{
display: none;
}
Answer the question
In order to leave comments, you need to log in
Why is this all here:
tabNavs[j].classList.add("active");
tabPanes[j].classList.add("active");
Didn't find what you were looking for?
Ask your questionAsk a Question
731 491 924 answers to any question