我想在 php 条件检查后显示此模态弹出窗口
<div id="myModal65" class="modal fade">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title">Subscribe our Newsletter</h4>
</div>
<div class="modal-body">
<p>Subscribe to our mailing list to get the latest updates straight in your inbox.</p>
<form>
<div class="form-group">
<input type="text" class="form-control" placeholder="Name">
</div>
<div class="form-group">
<input type="email" class="form-control" placeholder="Email Address">
</div>
<button type="submit" class="btn btn-primary">Subscribe</button>
</form>
</div>
</div>
</div>
</div>
在此 PHP 代码中用于显示弹出窗口。但它无法显示。
<?php
if($intwschdle==1)
{
echo "<script type='text/javascript'>$('#myModal65').modal('show');</script>";
}
?>
我认为如果您默认显示模态并将其置于 php 条件下会好得多,例如:
<?php
if($intwschdle==1){
?>
<!--full html for modal goes here showing by default - - >
<?php } ? >