Parse error: syntax error, unexpected $end in C:\xampp\htdocs\xampp\2014\select.php on line 46

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<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<table width="500" align="center" border="0" cellspacing="0">
<tr> 
	<td> List of Student </td>
</tr> 
	<tr bgcolor="#FF0033">
    	<td>Name </td>
        <td>Father Name </td>
        <td>Gender </td>
        <td>Qualification </td>
    </tr>
    	$SERVERNAME='localhost';
	$USERNAME= 'ROOT';
	$PASSWAROD= '';
	$DBNAME='test';
		$conn = new mysqli($SERVERNAME,$USERNAME,$PASSWAROD,$DBNAME);
	if ($conn -> connect_error)
	echo "connection error";
	$sql = "SELECT * FROM  studentinfo";
	$result= $conn -> query($sql);
	if ($result -> num_rows > 0)
	 {
	 	while ($row = $result -> fetch_assoc())
			{
				$name = $row['name'];
			
	?>
    <tr>
    	<td> </td>
	
	</tr>
    	}	
     }
</table>
</body>
</html>

Posted 9-Oct-15 6:21am

Member 12046582

Updated 9-Oct-15 7:07am

Patrice T

v2

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Richard Deeming 9-Oct-15 13:27pm

Which line of that code is line 46? The code block only seems to contain 42 lines.

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