我在.php中有这段代码可以检索联系表:
echo '<div class="hsk-column4 hsk-agency-inquiry-form">';
echo hsk_agency_enquiry_form();
echo '</div>';
当用id="btn-contact"
按下按钮时,如何使弹出式窗口显示在联系表单中?
谢谢。
答案 0 :(得分:2)
也许您在谈论模态?应该有帮助...
答案 1 :(得分:1)
您可以使用jQuery
PHP:
<button id="btn-contact">Open Contact</button>
<?php
echo '<div class="hsk-column4 hsk-agency-inquiry-form">';
echo hsk_agency_enquiry_form();
echo '</div>';
?>
jQuery:
jQuery(document).ready(function( $ ) {
$("body").on("click", function(){
$(".hsk-column4.hsk-agency-inquiry-form").removeClass("active-form");
});
$("#btn-contact").on("click", function(event){
$(".hsk-column4.hsk-agency-inquiry-form").addClass("active-form");
event.stopPropagation();
});
});
CSS:
.hsk-column4.hsk-agency-inquiry-form {
display: none;
}
.active-form {
display: block;
}