下面是我的代码。在按钮上单击,我需要提交表单并通过邮寄传递值。然后div#con1
应该被隐藏起来,并且应该显示div #con2
。在div中,#con2
我需要显示通过发布获取的值,但是问题出在单击后,页面会继续重新加载。投票者请提及您的评论
<?php include("../view/common/head.php"); ?>
<script>
$(document).ready(function() {
$("#myform").submit(function() {
$("#con1").hide();
$("#con2").show();
});
});
</script>
<div class="container" id="con1">
<div class="row">
<div class="col-lg-12 col-md-12 col-sm-12 col-xs-12">
<form name="myform" class="form-horizontal" id="myform" method="post">
<ul class="devices">
<li>
<div class="dev-inner">
<div class="dei-mid"><p>Computer Tower</p></div>
<div class="dei-rgt">
<input type="text" class="form-control inpt-bx-txtclr-home" name="computername" id="computerid" placeholder="000">
</div>
</div>
</li>
</ul>
<button type="submit" id="grad-btn">Calculate</button>
</div>
</form>
</div>
</div>
<div class="container" id="con2" style="display:none">
<div class="row">
<div class="col-lg-12 col-md-12 col-sm-12 col-xs-12">
<p><?php echo $_POST['computername'];?></p>
</div>
</div>
</div>
在您的代码div
中隐藏和显示工作正常,但是页面重新加载,这就是为什么它会导致问题的原因,
我建议您使用php代码而不是jquery
。
如果未提交表单,则显示con1,否则显示con2
<?php include("../view/common/head.php"); ?>
<?php if(!isset($_POST['computername'])){ ?> <!-Add condition Here->
<div class="container" id="con1">
<div class="row">
<div class="col-lg-12 col-md-12 col-sm-12 col-xs-12">
<form name="myform" class="form-horizontal" id="myform" method="post">
<ul class="devices">
<li>
<div class="dev-inner">
<div class="dei-mid"><p>Computer Tower</p></div>
<div class="dei-rgt">
<input type="text" class="form-control inpt-bx-txtclr-home" name="computername" id="computerid" placeholder="000">
</div>
</div>
</li>
</ul>
<button type="submit" id="grad-btn">Calculate</button>
</div>
</form>
</div>
</div>
<?php }else{ ?>
<div class="container" id="con2" style="display:none">
<div class="row">
<div class="col-lg-12 col-md-12 col-sm-12 col-xs-12">
<p><?php echo $_POST['computername'];?></p>
</div>
</div>
</div>
<?php } ?>
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