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Hi, I am back after almost 6 months to continue on this part of the project, using the advice I had from you, I have reached this point where I am still getting all records inserted into the table instead of the selected ones: $checkbox = $_POST['checkbox']; $class_id = trim($_POST['class_id']); foreach ($checkbox as $key => $value) { echo "{$key} => {$value} "; print_r($checkbox); echo '<pre>'; print_r($checkbox); echo '</pre>'; $sql = "INSERT INTO Class_List (user_id,class_id) VALUES (?, ?)"; if($stmt = mysqli_prepare($link, $sql)){ mysqli_stmt_bind_param($stmt, "ii", $value, $class_id); if(mysqli_stmt_execute($stmt)){ echo ""; } else{ echo "ERROR in inserting records: $sql. " . mysqli_error($link); } } else{ echo "ERROR: Could not insert records: $sql. " . mysqli_error($link); } } // Close statement mysqli_stmt_close($stmt); // Close connection mysqli_close($link); echo "The selected students have been added"; ?>

Thank you, When I select three records with user IDs 6, 7 , 8, I get this output and assigns user_id (8) to all the selected records. <td><input type="checkbox" name="checkbox[<?php echo $row['user_id']; ?>]"></td> I am really stuck as I have no worked example to work from.

OK, so I have progressed a little, I am now using Prepared statements and now I can insert the number of rows based on the selections, and it will insert the class ID, however, only the last student's userID gets inserted in each row even when it is not selected. <?php ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL); ?> <?php mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT); ?> <?php $servername = "xxxxxt"; $username = "webxxxx"; $password = "xxxx!"; $dbname = "xxx"; // Create connection $link = mysqli_connect($servername, $username, $password, $dbname); // Check connection if (!$link) { die("Connection failed: " . mysqli_connect_error()); } $checkbox = $_POST['checkbox']; $user_id = $_POST['user_id']; $class_id = $_POST['class_id']; $count = count($checkbox); for ($i = 0; $i < $count; $i++) { $sql = "INSERT INTO Class_List (user_id,class_id) VALUES (?, ?)"; if($stmt = mysqli_prepare($link, $sql)){ mysqli_stmt_bind_param($stmt, "ss", $user_id, $class_id); if(mysqli_stmt_execute($stmt)){ echo "Success!"; } else{ echo "ERROR in inserting records: $sql. " . mysqli_error($link); } } else{ echo "ERROR: Could not insert records: $sql. " . mysqli_error($link); } } // Close statement mysqli_stmt_close($stmt); // Close connection mysqli_close($link); ?>

My first attempt is in post number 1 My second attempt is is post number 7 after trying to act upon your suggestions with my little PHP coding experience. My third attempt was making the checkbox field an array, with the array index being the user_id value I am not a coder, but I can make some sense from worked examples/tutorials. So now my code is at this stage and really need some guidance: $class_id = $_POST['class_id']; // This value is from a dropdown $checkbox = $_POST['checkbox']; // This is a checkbox next to each record $user_id = $_POST['user_id']; // this is a hidden value output from the database (in a textbox) if (isset($_POST['submit'])) { if(!empty($_POST['checkbox'])) { foreach($_POST['checkbox'] as $i) { $query="INSERT INTO Class_List(user_id, class_id)VALUES('$user_id','$class_id')"; if (mysqli_query($conn, $query)) { echo "The selected records have been added successfully !"; } else { echo "Error: " . $sql . ":-" . mysqli_error($conn); } mysqli_close($conn); } } } ?>

Step 1 completed: Thanks I now need guidance with: - if any of the checkboxes are checked, you will get an array with the indexes being the user_ids (i would use array_keys() to get all of them as an array to loop over). you would then use a prepared query, prepared once before the start of any looping, to insert each row of data.

Hi, Thank you. I am trying to follow what you advised, I am not a programmer and new to PHP, I did some further research and tried to apply to what you said: HTML: <?php $outcome = ''; $count=1; $sel_query="SELECT * FROM User_Accounts_"; $stmt = $conn->prepare($sel_query); $stmt->execute(); $result = $stmt->get_result(); // get the mysqli result if($result->num_rows === 0) exit('No Users'); $i = 0; while($row = $result->fetch_assoc()) { ?> <td><input name="checkbox[]" type="checkbox" value="Yes" value=<?php echo $i++;?>></td> <td><input name="user_id" type="text" size="2" value="<?php echo $row["user_id"]; ?>" /></td> <td><?php echo $row["username"]; ?></td> <td><?php echo $row["firstname"]; ?></td> <td><?php echo $row["lastname"]; ?></td> <td><?php echo $row["School"]; ?></td> <td><?php echo $row["Year"]; ?></td> <td><?php echo $row["Form"]; ?></td> PHP <?php ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL); ?> <?php mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT); ?> <?php $servername = "xxx"; $username = "xxx"; $password = "xxx"; $dbname = "xxx"; // Create connection $conn = mysqli_connect($servername, $username, $password, $dbname); // Check connection if (!$conn) { die("Connection failed: " . mysqli_connect_error()); } $class_id = $_POST['class_id']; // This value is from a dropdown $checkbox = $_POST['checkbox']; // This is a checkbox next to each record $user_id = $_POST['user_id']; // this is a hidden value output from the database (in a textbox) $i=0 if (isset($_POST['submit'])) { foreach($_POST['checkbox'] as $i) { $query="INSERT INTO Class_List(user_id, class_id)VALUES('{$_POST['user_id'][$i]}', '{$_POST['class_id'][$i]}')"; if (mysqli_query($conn, $query)) { echo "The selected records have been added successfully !"; } else { echo "Error: " . $sql . ":-" . mysqli_error($conn); } mysqli_close($conn); } } ?> Now, I am getting an error: Notice: Uninitialized string offset: 1 (at the line foreach($_POST['checkbox'] as $i) { )

<td><input name="checkbox" type="checkbox" value="Yes"></td> <td><input name="user_id" type="text" size="2" value="<?php echo $row["user_id"]; ?>" /></td> <td><?php echo $row["username"]; ?></td> <td><?php echo $row["firstname"]; ?></td> <td><?php echo $row["lastname"]; ?></td> <td><?php echo $row["School"]; ?></td> <td><?php echo $row["Year"]; ?></td> <td><?php echo $row["Form"]; ?></td>

OK great, I found that at the EOF there was a curly bracket missing. So now, the append is working but it only adds the last row in the table, even if the others are selected and the last row isn't.

<?php $servername = "xxx"; $username = "xxxx"; $password = "xxxx"; $dbname = "web216-admin-6d5"; // Create connection $conn = mysqli_connect($servername, $username, $password, $dbname); // Check connection if (!$conn) { die("Connection failed: " . mysqli_connect_error()); } $class_id = $_POST['class_id']; // This value is from a dropdown $checkbox = $_POST['checkbox']; // This is a checkbox next to each record $user_id = $_POST['user_id']; // this is a hidden value output from the database (in a textbox) if(isset($_POST['submit'])) { for($i=0;$i<count($checkbox);$i++){ $query="INSERT INTO Class_List(user_id,class_id)VALUES('".$user_id[$i]."','".$class_id."')"; if (mysqli_query($conn, $query)) { echo "The selected records have been added successfully !"; echo "<meta http-equiv=\"refresh\" content=\"2;url=search_courses.php\"/>"; } else { echo "Error: " . $sql . ":-" . mysqli_error($conn); } mysqli_close($conn); } ?> Hi, I am trying to figure out how to insert the following values into another table if they have a checked box: $class_id = $_POST['class_id']; // This value is from a dropdown on the top of the page $checkbox = $_POST['checkbox']; // This is a checkbox next to each row $user_id = $_POST['user_id']; // this is a hidden value output from the database (in a textbox) in each row basically, each row in the table has a checkbox If the checkbox is checked then it should insert the user_id field from the checked rows and the value from a drop down (class_id) at the top of a page into another table. I have managed the code up to here but I need a tutorial or guidance to do the rest. At the moment I am getting error HTTP 500 I understand that the code is subject to sql injection, but will sort that out later

Thank you! (A silly mistake 🙉) Yes, on login, the user level is checked and the user is re-directed to the relevant page! so non admin will not see the admin options.

Hi, I am trying to get this script to execute as an administrator of an online system. If a user has forgotten their password, I enter their username and enter a new password which they can update later. I am not sure why this is not updating the password for the username entered? <?php // Initialize the session session_start(); // Check if the user is logged in, if not then redirect to login page if(!isset($_SESSION["loggedin"]) || $_SESSION["loggedin"] !== true){ header("location: login.php"); exit; } // Include config file require_once "config.php"; // Define variables and initialize with empty values $new_password = $confirm_password = ""; $new_password_err = $confirm_password_err = ""; // Processing form data when form is submitted if($_SERVER["REQUEST_METHOD"] == "POST"){ // Validate new password if(empty(trim($_POST["new_password"]))){ $new_password_err = "Please enter the new password."; } elseif(strlen(trim($_POST["new_password"])) < 6){ $new_password_err = "Password must have atleast 6 characters."; } else{ $new_password = trim($_POST["new_password"]); } // Validate confirm password if(empty(trim($_POST["confirm_password"]))){ $confirm_password_err = "Please confirm the password."; } else{ $confirm_password = trim($_POST["confirm_password"]); if(empty($new_password_err) && ($new_password != $confirm_password)){ $confirm_password_err = "Password did not match."; } } // Check input errors before updating the database if(empty($new_password_err) && empty($confirm_password_err)){ // Prepare an update statement $sql = "UPDATE User_Accounts_ SET password = ? WHERE username = ?"; if($stmt = mysqli_prepare($link, $sql)){ // Bind variables to the prepared statement as parameters mysqli_stmt_bind_param($stmt, "ss", $param_password, $username); // Set parameters $param_password = password_hash($new_password, PASSWORD_DEFAULT); $username = $_POST['username']; // Attempt to execute the prepared statement if(mysqli_stmt_execute($stmt)){ // Password updated successfully. Destroy the session, and redirect to login page session_destroy(); header("location: login.php"); exit(); } else{ echo "Oops! Something went wrong. Please try again later."; } // Close statement mysqli_stmt_close($stmt); } } // Close connection mysqli_close($link); } ?> <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <title>Reset Password</title> <style type="text/css"> body{ font: 14px sans-serif; } .wrapper{ width: 350px; padding: 20px; } </style> </head> <body> <div class="wrapper"> <h2>Reset Password</h2> <p>Please fill out this form to reset your password.</p> <p><strong>Username</strong> <input type="text" name="username" class="form-control"> </p> <p>&nbsp;</p> <form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post"> <div class="form-group <?php echo (!empty($new_password_err)) ? 'has-error' : ''; ?>"> <label>New Password</label> <input type="password" name="new_password" class="form-control" value="<?php echo $new_password; ?>"> <span class="help-block"><?php echo $new_password_err; ?></span> </div> <div class="form-group <?php echo (!empty($confirm_password_err)) ? 'has-error' : ''; ?>"> <label>Confirm Password</label> <input type="password" name="confirm_password" class="form-control"> <span class="help-block"><?php echo $confirm_password_err; ?></span> </div> <div class="form-group"> <input type="submit" class="btn btn-primary" value="Submit"> <a class="btn btn-link" href="welcome.php">Cancel</a> </div> </form> </div> </body> </html>

I didn't seem to locate the missing curly brace

Hi ! I am trying to translate my mysqli count query that works perfectly into prepared statements. Unfortunately, after playing around and using my knowledge of PS, I have come up with this script which fails to execute and returns a http 500 error. I may have missed something very silly, I require some guidance on fixing the error. <?php $conn = mysqli_connect("xxxx", "xxxx", "xxxx", "xxx"); $sel_query = "SELECT S1, B1 COUNT(IF(S1 = ?, 1, NULL)) 'Accepted', COUNT(IF(S1 = ?, 1, NULL)) 'Rejected', COUNT(IF(S1 = ?, 1, NULL)) 'Under_Review' FROM Enrol"; $stmt = $conn->prepare($sel_query); $Accepted="Accepted"; $Rejected="Rejected"; $Under_Review="Under Review"; $stmt->bind_param("sss",$Accepted, $Rejected, $Under_Review); $stmt->execute(); $result = $stmt->get_result(); // get the mysqli result if($result->num_rows === 0) exit('No records found!'); while($row = $result->fetch_assoc()) { ?> <tr> <td><?php echo $row["Accepted"]; ?></td> <td><?php echo $row["Rejected"]; ?></td> <td><?php echo $row["Under_Review"]; ?></td> </tr> </table>

I have but not sure which section can help me. Guidance needed please.

Thank you so much, with your support, I am learning as well as becoming a step closer to the end result. So, after looking at the PHP manual, I have altered the script , but now it is producing the error ""Notice: Undefined index for the following lines: <td><?php echo $row['Form_Group']; ?></td> <td><?php echo $row['Present']; ?></td> <td><?php echo $row['Absent']; ?></td> Here is the entire code <?php ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL); ?> <?php mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT); ?> <?php //Connection for database $conn = mysqli_connect("xxx", "xxxxxxxxxx", "xxxxxxxxxx", "xxxxxxx"); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); exit(); } $sql = "SELECT Form_Group, COUNT(Presence LIKE '/') AS Present, COUNT(Presence LIKE 'N') AS Absent FROM attendance"; ?> <!doctype html> <html> <body> <h1 align="center">Summary</h1> <table border="1" align="center" style="line-height:25px;"> <tr> <th>Form</th> <th>Present</th> <th>Absent</th> </tr> <?php //Fetch Data form database if ($result = mysqli_query($conn, $sql)) { while ($row = mysqli_fetch_row($result)) { ?> <tr> <td><?php echo $row['Form_Group']; ?></td> <td><?php echo $row['Present']; ?></td> <td><?php echo $row['Absent']; ?></td> </tr> <?php } mysqli_free_result($result); } mysqli_close($conn); ?> </table> </body> </html>

I believe it is a mysql error , after removing the ' ' around the table name, I am getting: Fatal error: Call to undefined method mysqli_result::fetch_rows() in \\143FF240DC.STORAGE-1B.HOSTING.STACKCP.NET\SITES\1b\1\143ff240dc\public_html\Progress_Tracker\Pages\reports_attendance.php on line 65 After removing the single quotes, I am getting: Fatal error: Call to undefined method mysqli_result::fetch_rows() in \\143FF240DC.STORAGE-1B.HOSTING.STACKCP.NET\SITES\1b\1\143ff240dc\public_html\Progress_Tracker\Pages\reports_attendance.php on line 65

Oh ok, so I fixed some foolish errors (missing commas etc) *embarassed* There now seems to be this final error but I just cant seem to spot what it is - near attendance $sql = "SELECT Form_Group, COUNT(Presence LIKE '/') AS Present, COUNT(Presence LIKE 'N') AS Absent FROM 'attendance'"; $result = $conn->query($sql); ?>

Thank you Barand. The mysql server is showing the following error: Error 1: Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'COUNT(Presence LIKE '/') AS Present, COUNT(Presence LIKE 'N') AS Absent, F' at line 3' Error 2: Line 47: $result = $conn->query($sql); i tried to follow the MariaDB syntax but to no success: $sql = "SELECT Form_Group COUNT(DISTINCT Presence LIKE '/') AS Present, COUNT(DISTINCT Presence LIKE 'N') AS Absent, FROM 'attendance'"; $result = $conn->query($sql);

Error message is: Notice: Trying to get property of non-object in \\.........\reports_attendance.php on line ... I looked very closely at the query for a very long time and just cannot find the problem,. It worked well before using the case with just the one count, when the second count went in, it showed this error. I am new to PHP so any guidance or tutorial to help will be greatly appreciated.

I am trying to run a count query and faced with the error: Notice: Trying to get property of non-object in Not sure how to resolve this error in these lines: if($result->num_rows > 0){ while($row = $result->fetch_rows()){ Code: <?php $sql = "SELECT Form_Group COUNT(Presence LIKE '/') AS Present, COUNT(Presence LIKE 'N') AS Absent, FROM 'attendance'"; $result = $conn->query($sql); ?> <?php //Fetch Data form database if($result->num_rows > 0){ while($row = $result->fetch_rows()){ ?> <tr> <td><?php echo $row['Form_Group']; ?></td> <td><?php echo $row['Present']; ?></td> <td><?php echo $row['Absent']; ?></td> </tr> <?php } } else { ?> <tr> <th colspan="6">There's No data found!!!</th> </tr> <?php } ?>

Hello, Thank you, since learning how to turn on error reporting, I was able to resolve all the previous issues and I have used mysqli which has been new learning for me. I am now able to display filtered data using variables posted from a form and I am able to update the data from multiple rows. The problem I am having now is that after the update I need the table to display the data using the filter fields from the form that was originally submitted. But I am getting the errors: Notice: Undefined index: Att_Date on line 33 Notice: Undefined index: Form_Group on line 34 I have tried to use session but with no luck. I feel stuck at this point and need some guidance. <?php ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL); ?> <?php //Connection for database $conn = mysqli_connect("xxxx", "xxxxxxxxx", "xxxxx", "xxxxxxx"); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); exit(); } $Att_Date=$_POST['Att_Date']; $Form_Group=$_POST['Form_Group']; session_start(); /*session is started if you don't write this line can't use $_Session global variable*/ $_SESSION["newsession1"] = $Att_Date; $_SESSION["newsession2"] = $Form_Group; echo $_SESSION["newsession1"]; echo $_SESSION["newsession2"]; //Select Database $sql = "SELECT * FROM attendance WHERE Form_Group ='$Form_Group' AND Att_Date = '$Att_Date'"; $result = $conn->query($sql); ?> <?php /* Check if button name "Submit" is active, do this */ if(isset($_POST['Submit1'])) { header('Location: '.$_SERVER['REQUEST_URI']); $count=count($_POST["Student_id"]); for($i=0;$i<$count;$i++){ $sql1="UPDATE attendance SET Presence='" . $_POST['Presence'][$i] . "' WHERE Student_id='" . $_POST['Student_id'][$i] . "'"; $result = $conn->query($sql1); } } echo ""; ?> <!doctype html> <html> <body> <h1 align="center">Register</h1> <table border="1" align="center" style="line-height:25px;"> <tr> <th>Stduent ID</th> <th>Name</th> <th>Surname</th> <th>Form</th> <th>Year Group</th> <th>Date</th> <th>Presence</th> </tr> <?php //Fetch Data form database if($result->num_rows > 0){ while($row = $result->fetch_assoc()){ ?> <form method="post"> <tr> <td><input name="Student_id" type="text" value="<?php echo $row['Student_id']; ?>"></td> <td><?php echo $row['firstname']; ?></td> <td><?php echo $row['lastname']; ?></td> <td><?php echo $row['Form_Group']; ?></td> <td><?php echo $row['Year_Group']; ?></td> <td><?php echo $row['Att_Date']; ?></td> <td><input name="Presence" type="text" value="<?php echo $row['Presence']; ?>"></td> </tr> <?php } } else { ?> <tr> <th colspan="6">There's No data found!!!</th> </tr> <?php } ?> </table> <p class="auto-style1"><input name="Submit1" type="submit" value="submit"></p> </form> </body> </html>

🙈

It is supposed to execute the query after sub1 button is pressed on the form which has two fields : Form_Group and Att_Date. The query is: $sql=("SELECT * FROM $tbl_name WHERE Form_Group ='$Form_Group' AND Att_Date = '$Att_Date'");; The query displays no data matching the filter fields, although it worked with $sql="SELECT * FROM attendance";

Sorry, I should have been clear The records display without the Where clause, however, when I refer to the controls on the form to filter the data no records are displayed. Why would only class A be displayed?

Thank you guys. I got the script to display the data and also edit and refresh the page. I am now wanting to filter the data but it does not seem to work. The form with the filter fields: <body> <?php include 'Navigation.php';?> <form method="post" action="Registers_(1).php"> <table frame="box" border="0"align="center" style="width: 38%; height: 63px;"> <tr> <td class="auto-style2"><strong>Select Date</strong></td> <td class="auto-style3"><input name="Att_Date" type="date" />&nbsp;</td> </tr> <tr> <td class="auto-style2"><strong>Select Form Group</strong></td> <td class="auto-style3"><select name="Form_Group"> <option value='SELECT' selected="selected">SELECT</option> <option value='Class A'>Class A</option> <option value='Class A'>Class B</option> </select> </td> </tr> <tr> <td class="auto-style1" colspan="2"> <input name="Sub1" type="submit" value="Load Register" />&nbsp;</td> </tr> </table> </form> The php script: <?php $host="shareddb1d.hosting.stackcp.net"; // Host name $username="xxxx"; // Mysql username $password="xxx"; // Mysql password $db_name="xxxx"; // Database name $tbl_name="xxxxxx"; /* Table name */ // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); if(isset($_POST["Sub1"])){ $Att_Date=$_POST['Att_Date']; $Form_Group=$_POST['Form_Group']; } $sql=("SELECT * FROM $tbl_name WHERE Form_Group ='$Form_Group' AND Att_Date = '$Att_Date'";; $result=mysql_query($sql); /* Count table rows */ $count=mysql_num_rows($result); ?> <?php /* Check if button name "Submit" is active, do this */ if(isset($_POST['Submit'])) { header('Location: '.$_SERVER['REQUEST_URI']); $count=count($_POST["Student_id"]); for($i=0;$i<$count;$i++){ $sql1="UPDATE $tbl_name SET firstname='" . $_POST['firstname'][$i] . "', lastname='" . $_POST['lastname'][$i] . "', Presence='" . $_POST['Presence'][$i] . "' WHERE Student_id='" . $_POST['Student_id'][$i] . "'"; $result1=mysql_query($sql1); } } echo ""; mysql_close(); ?> <?php include 'Navigation.php';?> <table width="500" border="0" cellspacing="1" cellpadding="0"> <form name="form1" method="post" action=""> <tr> <td> <table width="500" border="0" cellspacing="1" cellpadding="0"> <tr> <td align="center" class="auto-style1"><strong>Id</strong></td> <td align="center" class="auto-style1"><strong>First Name</strong></td> <td align="center" class="auto-style1"><strong>Lastname</strong></td> <td align="center" class="auto-style1"><strong>Year Group</strong></td> <td align="center" class="auto-style1"><strong>Form Group</strong></td> <td align="center" class="auto-style1"><strong>Date</strong></td> <td align="center" class="auto-style3"><strong>Attendance</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td align="center"> <input name="Student_id[]" type="text" value="<?php echo $rows['Student_id']; ?>" readonly class="auto-style1"></td> <td align="center"> <input name="lastname[]" type="text" value="<?php echo $rows['lastname']; ?>" readonly class="auto-style1"><span class="auto-style1"> </span> </td> <td align="center"> <input name="firstname[]" type="text" value="<?php echo $rows['firstname']; ?>" readonly class="auto-style1"><span class="auto-style1"> </span> </td> <td align="center"> <input name="Year_Group[]" type="text" value="<?php echo $rows['Year_Group']; ?>" readonly class="auto-style1"><span class="auto-style1"> </span> </td> <td align="center"> <input name="Form_Group[]" type="text" value="<?php echo $rows['Form_Group']; ?>" readonly class="auto-style1"><span class="auto-style1"> </span> </td> <td align="center"> <input name="Att_Date[]" type="text" value="<?php echo $rows['Att_Date']; ?>" readonly class="auto-style1"><span class="auto-style1"> </span> </td> <td align="center" class="auto-style2"> <input name="Presence[]" type="text" value="<?php echo $rows['Presence']; ?>" class="auto-style1"><span class="auto-style1"> </span> </td> </tr> <?php } ?> <tr> <td colspan="7" align="right"> <input type="submit" name="Submit" value="Submit" class="auto-style1"></td> </tr> </table> </td> </tr> </form> </table>