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<title>Transparent Login Form
	
	body
{
	margin: 0;
	padding:0;
	background:url(digital-innovation.jpg);
	background-size:cover;
	font-family:sans-serif;
}
.loginBox
{
	position:absolute;
	top:50%;
	left: 50%;
	transform: translate(-50%,-50%);
	width:350px;
	height: 420px;
	padding: 80px 40px;
	box-sizing: border-box;
	background: rgba(0,0,0,.6);
}
.user
{
	width: 100px;
	height: 100px;
	border-radius: 50%;
	overflow: hidden;
	position:absolute;
	top: calc(-100px/2);
	left:calc(50% - 50px);
}
h2
{
	margin: 0;
	padding: 0 0 20px;
	color: yellow;
	text-align: center;
}
.loginBox p
{
  margin: 0;
  padding: 0; 
  font-weight: bold;
  color:#fff;
}
.looginBox input
{
	width: 100% ;
	margin-bottom: 20px;
}
.loginBox input[type="text"],
.loginBox input[type="password"]
{
	border:none;
	border-bottom: 2px solid white;
	background: transparent;
	outline: none;
	height: 40px;
	color:#fff;
	font-size: 16px;
}
::placeholder
{
	color: rgba(255,255,255,.5);
}
.loginBox input[type="submit"]
{
	border-radius: none;
	outline: none;
	height: 40px;
	color: #fff;
	font-size: 16px;
	background:#ff267e;
	cursor: pointer;
	border-radius: 20px;
}



	<div class="loginBox">
		
		<h2>Admin LogIn</h2>
		
			<p>Username</p>
			
			<p>Password</p>
			<br><br>
			
		
	</div>


What I have tried:

I am new PHP,mysql learner.I just wanted to create a login page and connect to my sql server.Then I've faced Parse error above. i am using xampp server.
Posted
Updated 11-Feb-20 13:06pm
v2
Comments
Dave Kreskowiak 11-Feb-20 14:04pm    
And you posted HTML/CSS, not the PHP code that's throwing the error.

1 solution

 
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