我有一个非常简单的窗体示例,它被分成了一些标签,它被发布到自己的验证和处理与PHP,简化版本的代码如下(没有验证和处理步骤):用javascript提交的窗体打开标签提交
<style>
/* Style the list */
ul.tab {
list-style-type: none;
margin: 0;
padding: 0;
overflow: hidden;
border: 1px solid #ccc;
background-color: #f1f1f1;
}
/* Float the list items side by side */
ul.tab li {float: left;}
/* Style the links inside the list items */
ul.tab li a {
color: black;
display: inline-block;
font-size: 15px;
font-weight: lighter;
padding: 14px 16px;
text-align: center;
text-decoration: none;
transition: all 0.3s ease 0s;
}
/* Change background color of links on hover */
ul.tab li a:hover {background-color: #ddd;}
/* Create an active/current tablink class */
ul.tab li a:focus, .active {background-color: #ccc;}
/* Style the tab content */
.tabcontent {
display: none;
padding: 60px 10px;
border: 1px solid #ccc;
border-top: none;
margin-bottom: 20px;
}
</style>
<ul class="tab">
<li><a href="#" class="tablinks" onclick="openTab(event, 'tab_1')">tab 1</a></li>
<li><a href="#" class="tablinks" onclick="openTab(event, 'tab_2')">tab 2</a></li>
</ul>
<!-- FORM Starts -->
<form method="POST" action="">
<div id="tab_1" class="tabcontent">
Tab One Content
</div>
<div id="tab_2" class="tabcontent">
Tab Two Content
</div>
<input type="submit" name="save_settings_button" value="Save Settings" />
</form>
<script>
function openTab(evt, tabName) {
var i, tabcontent, tablinks;
tabcontent = document.getElementsByClassName("tabcontent");
for (i = 0; i < tabcontent.length; i++) {
tabcontent[i].style.display = "none";
}
tablinks = document.getElementsByClassName("tablinks");
for (i = 0; i < tabcontent.length; i++) {
tablinks[i].classList.remove("active");
}
document.getElementById(tabName).style.display = "block";
evt.currentTarget.classList.add("active");
}
var mybtn = document.getElementsByClassName("tablinks")[0];
mybtn.click();
</script>
这有助于大单分解成部分。然而如果在选项卡用户,并按下提交按钮,他们会回到标签之一,不是一个很好的用户体验!
有没有什么办法可以在提交表单时识别哪个选项卡处于焦点状态,然后在页面重新加载后重新加载该选项卡上的页面?