I update your code.
<?php
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("database") or die(mysql_error());
?>
<html>
<head>
<title>Student list</title>
<link href="https://stackoverflow.com/questions/11200978/stylesheets/public.css" media="all" rel="stylesheet" type="text/css"/>
</head>
<body id="background">
<table >
<tr>
<td><img src=" images/Picture3.png" width="1300" height="150"/></td>
</tr>
</table>
<table>
<tr>
<td id="structure">
<?
// check for post data
if( isset($_POST['name']) && isset($_POST['email']) && isset($_POST['shift']) && isset($_POST['class']) && isset($_POST['id']) )
{
$name=$_POST['name'];
$email=$_POST['email'];
$shift=$_POST['shift'];
$class=$_POST['class'];
$id=$_POST['id'];
$username = $_POST['name']; // you must escape any input. Remember.
$query = "SELECT * FROM `data` WHERE `name` = '".$username."'";
$result = mysql_query($query);
// check for duplicate
if ( mysql_num_rows ( $result ) > 1 )
{
echo 'Username already exists';
}
else
{
// insert new record
mysql_query("INSERT INTO `data`(name,email,shift,class) VALUES ('".$name."', '".$email."', '".$shift."','".$class."')");
print "Your information has been successfully added to the database.";
}
}
// list data
$data = mysql_query("SELECT * FROM data") or die(mysql_error());
print "<table border cellpadding=5>";
while($info = mysql_fetch_array( $data ))
{
print "<tr>";
print "<th>Id:</th><td>".$info['id']."</td>";
print "<th>Name:</th> <td>".$info['name'] . "</td> ";
print "<th>Email:</th> <td>".$info['email'] . " </td>";
print "<th>shift:</th> <td>".$info['shift'] . " </td>";
print "<th>class:</th> <td>".$info['class'] . " </td>";
print "<tr><td><a href="update.php?id={$info["id']}'>EDIT</a></td></tr>";
print "<tr><td><a href="delete.php?id={$info["id']}'>DELETE</a></td></tr>";
}
print "</table>";
?>
</td>
</tr>
</table>
</body>
</html>
3
solved Preventing duplicates in the database [closed]