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下面显示的是我的connection.php文件。jQuery datatable不能在php中工作
我在控制台
遇到这些错误我怎样才能摆脱这些错误的？
<?php
$con=mysqli_connect("localhost", "root", "");
mysqli_select_db($con,"college");
if($con)
{
//echo "Database Connected";
//echo "<br>";
}
else
{
//echo "Database Not Connected";
}
?>
这是我的fetch.php文件。我打电话给我所有的CSS和JS文件，但我不知道为什么我的数据表无法正常工作。
<html>
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1" >
<link rel="stylesheet" href="datatable/media/css/dataTables.jquery.dataTables.css" />
<link rel="stylesheet" href="datatable/media/css/dataTables.bootstrap.css" />
<script src="datatable/media/js/dataTables.jquery.dataTables.js"></script>
<script src="datatable/media/js/dataTables.bootstrap.js"></script>
</head>
<body>
<?php
include("conn.php");
$sql="select * from student";
$res=mysqli_query($con,$sql);
?>
<table border="1" id="datatable">
<thead>
<tr>
<td>USER NAME</td>
<td>FIRST NAME</td>
<td>LAST NAME</td>
<td>EMAIL ADDRESS</td>
<td>PASSWORD</td>
</tr>
</thead>
<tbody>
<?php
while ($result=mysqli_fetch_array($res))
{
?>
<tr>
<td><?php echo $result['Student_Uname']; ?></td>
<td><?php echo $result['Student_Fname']; ?></td>
<td><?php echo $result['Student_Lname']; ?></td>
<td><?php echo $result['Email_Id']; ?></td>
<td><?php echo $result['Password']; ?></td>
</tr>
<?php } ?>
</tbody>
</table>
</body>
</html>
<script>
$(document).ready(function() {
$('#datatable').DataTable();
});
</script>
这是我的index.php文件。这是我的学生信息页面。
<!DOCTYPE html>
<html lang="en">
<head>
<title>Form Page</title>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1" >
<link rel="stylesheet" href="style/css/bootstrap.css" />
<link rel="stylesheet" href="style/css/bootstrap-theme.css" />
</head>
<body>
<div class="container">
<h2 align="center">STUDENT INFORMATION FORM</h2>
<form method="POST" action="insert.php" class="form-horizontal" autocomplete="off">
<div class="form-group">
<label for="Uname" class="control-label col-md-2">User Name:</label>
<div class="col-md-8">
<input type="text" name="Uname" class="form-control" placeholder="User Name" required >
</div>
</div>
<div class="form-group">
<label for="Fname" class="control-label col-md-2">First Name:</label>
<div class="col-md-8">
<input type="text" name="Fname" class="form-control" placeholder="First Name" required >
</div>
</div>
<div class="form-group">
<label for="Uname" class="control-label col-md-2">Last Name:</label>
<div class="col-md-8">
<input type="text" name="Lname" class="form-control" placeholder="Last Name" required >
</div>
</div>
<div class="form-group">
<label for="email" class="control-label col-md-2">Email Address:</label>
<div class="col-md-8">
<input type="email" name="email" class="form-control" placeholder="Email Address" required >
</div>
</div>
<div class="form-group">
<label for="password" class="control-label col-md-2">Password:</label>
<div class="col-md-8">
<input type="password" name="password" class="form-control" placeholder="Password" autocomplete="new-password" required >
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2">
<button type="submit" name="submit" class="btn btn-info">Submit</button>
</div>
</div>
</form>
</div>
<script src="style/js/jquery.js"></script>
<script src="style/js/bootstrap.js"></script>
</body>
</html>
这是我的insert.php文件，我已正确插入数据库。
<?php[enter image description here][1]
include("conn.php");
if (isset($_POST['submit']))
{
$uname=$_POST['Uname'];
$fname=$_POST['Fname'];
$lname=$_POST['Lname'];
$email=$_POST['email'];
$pass=$_POST['password'];
$q="insert into student(Student_Uname,Student_Fname,Student_Lname,Email_Id,Password) values ('".$uname."', '".$fname."', '".$lname."', '".$email."', '".$pass."')";
$res=mysqli_query($con,$q);
}
?>
<h3 align="center">YOU ARE MOVING VIEW PAGE</h3>
<?php
header("refresh:1; url=fetch.php");
?>
[1]: https://i.stack.imgur.com/k8JdX.jpg
这是我抓取的数据表输出页面，所以通过这张图片你可以了解更多。我满足了jQuery数据表所需的所有要求。错误在哪里？
检查css和js文件是否在浏览器中直接使用而不是从php页面加载 –