How to get different buttons to submit different mySQL queries - php

I'm trying to create a button in a HTML form that adds a number to the corresponding mySQL location. Basically, a voting option for a post. Right now, the buttons all do the same thing: they submit the original post query from the bottom. How do I make them act individually?
PHP:
if (!$thepipeline) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
if (!empty ($_POST)) {
$squery = "INSERT INTO votes (votes) VALUE (1)";
print $squery;
$votes_result = mysqli_query ($thepipeline, $squery);
}
$query = "SELECT p.post_id, p.member_id, p.content, p.date_created,
SUM(v.votes) AS votes_total
FROM posts p LEFT JOIN
votes v
ON p.post_id = v.post_id
GROUP BY post_id
ORDER BY votes_total;";
echo "<table>\n";
echo '<tr>
<td> USER ID</td>
<td> PROPOSAL</td>
<td> DATE POSTED</td>
<td> SCORE</td>
</tr>';
if ($amorphous_cloud = mysqli_query ($thepipeline,$query)) {
while ($row = mysqli_fetch_assoc ($amorphous_cloud)){
echo '<tr>
<td> ' . $row['member_id'].'</td>
<td> ' . $row['content'].'</td>
<td> ' . $row['date_created'].'</td>
<td> ' . $row['votes_total'].'
<br><form method="POST"><input type="button" name="vote" value="Vote" /></form></td>
</tr>';
}
}
else {
print ":(";
}
echo "</table>\n";
?>
<br>
</div>
<div id="addpost">
<h3>Enter Proposal</h3>
<p>Off-topic posts will be deleted without warning.</p>
<?php
$member_id = $_GET['member_id'];
$content = $_GET['content'];
$link = mysqli_connect ('localhost', '******', '************,', '*************');
if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
if (!empty ($_GET)) {
$query = "INSERT INTO posts (member_id, content) VALUES ('".$member_id."', '".$content."')";
print $query;
$member_result = mysqli_query ($link, $query);
}
$user_query = "SELECT member_id, member_name FROM members";
$user_result = mysqli_query ($link, $user_query);
$user_html = "";
while ($row = mysqli_fetch_assoc ($user_result)) {
$user_html = $user_html . "<option value='".$row['member_id']."'>".$row['member_name']."</option>\n";
}
?>
<form method="GET">
<!-- member -->
<p>Member</br>
<select name="member_id" >
<?php echo $user_html; ?>
</select>
</p>
<!-- date -->
<p>Content
</p>
<textarea name="content" cols="40" rows="5"></textarea>
</br>
<input type="submit" />
</form>
</div>
</div>

Related

How to insert selected value from dropdown list into another MySQL table

I'm trying to make a dropdown list that allows users to select a parts that they need, so after selecting all they need and submit it should go to MySQL database. But after selecting submit nothing is inserting into my database.
My code and connection:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db = "userregistration";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $db);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo 'good connection';
?>
<form action="trying.php" method=" POST">
<table border=" 1">
<thead>
<tr>
<th>Component</th>
<th>Item Name</th>
<th>Price </th>
</tr>
</thead>
<tbody>
<tr>
<td>CPU</td>
<td>
<?php
//Retrieving CPU table
$query = $conn->query("SELECT * FROM cpu");
echo '<select name="cpu" class="cpu" onChange = $("#cpuprice").val($(this).find("option:selected").attr("cpuprice"))>';
while ($obj = mysqli_fetch_assoc($query)) {
echo '<option cpuprice = ' . $obj['price'] . ' cpuname=' . $obj['cpuname'] . ' >' . $obj['cpuname'] . '</option> /n';
}
echo '</select>';
?>
</td>
<td>
<output id="cpuprice" disabled value="">
</td>
</tr>
</tbody>
<tbody>
<tr>
<td>GPU</td>
<td>
<?php
//Retrieving GPU table
$query = $conn->query("SELECT * FROM gpu");
echo '<select name="gpu" class ="gpu" onChange = $("#gpuprice").val($(this).find("option:selected").attr("gpuprice"))>';
while ($obj = mysqli_fetch_assoc($query)) {
echo '<option gpuprice = "' . $obj['price'] . '" gpuname = "' . $obj['gpuname'] . '">' . $obj['gpuname'] . '</option>';
}
echo '</select>';
?>
</td>
<td>
<output class="form-control prc" id="gpuprice" disabled value="">
</td>
</tr>
</tbody>
</table>
<input class="submit" type="submit" />
</form>
I tried this but it doesn't work, after cliking submit it echo adding error and nothing inserted into my database
<?php
if (!empty($_POST["cpu"]) && !empty($_POST["gpu"])) {
$cpu = isset($_POST["cpu"]);
$gpu = isset($_POST["gpu"]);
$qstr = "INSERT INTO trycombuild(cpuname, gpuname) VALUES ('$cpu' , '$gpu')";
$query = mysqli_query($conn, $qstr);
} else
echo 'adding error';
?>
I tried to echo the $cpu and $gpu and it says undefined variable
I also tried this:
if (!empty($_POST['cpu']) && !empty($_POST['gpu'])) {
$cpu = $_POST['cpu'];
$gpu = $_POST['gpu'];
$qstr = $conn->prepare("INSERT INTO trycombuild (cpuname, gpuname) VALUES (?, ?)");
$qstr->bind_param("ss", $cpu, $gpu);
$qstr->execute();
$sqtr->close();
}
Your html is not correct. remove unnecessary space and try
<tbody>
<tr>
<td>CPU</td>
<td>
<?php
//Retrieving CPU table
$query = $conn->query("SELECT * FROM cpu");
echo '<select name="cpu" class="cpu" onChange = $("#cpuprice").val($(this).find("option:selected").attr("cpuprice"))>';
echo "<option>---select your CPU---</option>/n";
while ($obj = mysqli_fetch_assoc($query)) {
echo '<option cpuprice = ' . $obj['price'] . ' cpuname=' . $obj['cpuname'] . ' >' . $obj['cpuname'] . '</option> /n';
}
echo '</select>';
?>
</td>
<td>
<output id="cpuprice" disabled value="">
</td>
</tr>
</tbody>
<tbody>
<tr>
<td>GPU</td>
<td>
<?php
//Retrieving GPU table
$query = $conn->query("SELECT * FROM gpu");
echo '<select name="gpu" class ="gpu" onChange = $("#tpuprice").val($(this).find("option:selected").attr("gpuprice"))>';
echo "<option>---select your GPU---</option>";
while ($obj = mysqli_fetch_assoc($query)) {
echo '<option gpuprice = "' . $obj['price'] . '" gpuname = "' . $obj['gpuname'] . '">' . $obj['gpuname'] . '</option>';
}
echo '</select>';
?>
</td>
<td>
<output class="form-control prc" id="tpuprice" disabled value="">
</td>
</tr>
</tbody>

I Want to move data from one table to another

PHP Code in codecancel.php:
<?php
if(isset($_POST['codecancel_btn']))
{
$con = new mysqli('localhost', 'root', '', 'halcondentalclinic');
if($con->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
$id=$_POST['codecancel_id'];
$sql = "INSERT INTO cancelled_table SELECT * FROM bookings WHERE booking_id='$id';";
if ($con->query($sql) === TRUE) {
$query = "DELETE FROM bookings WHERE booking_id='$id' ";
if ($con->query($query) === TRUE) {
$message = "Success!";
} else {
echo "Error: " . $query . "<br>" . $con->error;
}
} else {
echo "Error: " . $sql . "<br>" . $con->error;
}
}
?>
For my Table appointment_calendar.php
<?php
if(mysqli_num_rows($query_run)>0)
{
while($row= mysqli_fetch_assoc($query_run))
{
?>
<tr>
<td><?php echo $row['booking_id']; ?> </td>
<td> <?php echo $row['name']; ?></td>
<td><?php echo $row['email'];?></td>
<td><?php echo $row['timeslot'];?> </td>
<td><?php echo $row['date'];?> </td>
<td>
<form action="codecancel.php" method="post">
<input type="hidden" name="codecancel_id" value="<?php echo $row['booking_id']; ?> ">
<button type="submit" name="codecancel_btn" class="btn btn-danger"> CANCEL</button>
</form>
</td>
</tr>
<?php
}
}
else {
echo" No Appointment Found";
}
?>
I want to transfer/move data if the patient cancels his appointment and the cancel appointment move to cancelled_table in the database then the canceled appointment will be removed to the bookings table in the database this is my table I want to cancel.
if I clicked the cancel button the error is :
Sorry guys I am Newbie in PHP AND MYSQL :)
INSERT INTO cancel_table(Booking_ID, Name ,Email,Date)
SELECT id as Booking_id, Name ,Email,Date
FROM bookings
WHERE id = 1
try This And Take care To standard of col Name in DB

UPDATE single column in database: PHP&MYSQL

So, I am trying to figure out how do this this and it boggling me. THIS WILL NOT BE USED ONLINE LIVE SO SQL INJECTION I DONT' CARE ABOUT. What am I doing wrong/right?
<?php
$db = mysql_connect("localhost", "root", "root");
if (!$db) {
die("Database connect failed: " . mysql_error());
}
$db_select = mysql_select_db("UNii", $db);
if (!$db_select) {
die("Database selection failed: " . mysql_error());
}
$comment = $_GET['comment'];
$id = $_GET['id'];
$sql = "UPDATE Dbsaved SET comment = '{$comment}' WHERE id = $id";
$comment1 = mysql_query($sql);
if (!$comment1) {
die("did not save comment: " . mysql_error());
}
echo $sql;
The main problem is with the statement itself, the connection is fine. I am trying to read $comment, and then update that into a MYSQL table and then have it read back in a different file.
EDIT: Mark up for the form I'm taking $comment from.
<!DOCTYPE html>
<html lang="en">
<LINK href="stylesheet.css" rel="stylesheet" type="text/css">
<script src ="js/validateform.js"></script>
<head>
<meta charset="UTF-8">
<title>UniHelp Home</title>
</head>
<body>
<div id="headeruni">
<h1>Welcome <?php echo $_GET["name"]; ?> to UniHelp!</h1>
</div>
<div id ="infouni">
<h3>Welcome to UniHelp. The social Network getting you connected to other people all over the University for any help you require!</h3>
</div>
<div id ="nameandemail">
<form action="formsend.php" method="post">
First name: <br> <input type="text" name="name"><br>
Email: <br> <input type="text" name="email"><br>
Comment: <br> <input type="text" name="message"><br>
<input type="submit" name="submit">
</form>`enter code here`
</div>
<div id="grabphpdiv">
<?php
$db = mysql_connect("localhost", "root", "root");
if (!$db) {
die("Database connect failed: " . mysql_error());
}
$db_select = mysql_select_db("UNii", $db);
if (!$db_select) {
die("Database selection failed: " . mysql_error());
}
$result = mysql_query("SELECT * FROM Dbsaved", $db);
if (!$result) {
die ("Database query failed: " . mysql_error());
}
$comment = $_POST['$comment'];
while ($row = mysql_fetch_array($result)) {
echo "<div id='posts'>";;
echo "<h2>";
echo $row[1] . "";
echo "</h2>";
echo "<p>";
//echo $timestamp = date('d-m-y G:i:s ');
echo "<br>";
echo "<br>";
echo $row[2] . "";
echo "</p>";
echo "<p>";
echo $row[3] . "";
echo "</p>";
echo 'Delete';
echo "<br>";
echo "<br>";
echo 'Comment: <br>
<input type=text name=comment><br>
<a href=addcomment.php?id=' . $row[0]. '&comment='. $row['$comment'].'>Comment</a>';
echo "<p>";
echo $row['comment'] . "";
echo "</p>";
echo "</div>";
echo "<br>";
}
?>
</div>
</body>
<div id="footer">Copyright &copy James Taylor 2016</div>
</html>
I just ran this code:
$comment = "Hello World!";
$id = 1;
$sql = "UPDATE Dbsaved SET comment = '{$comment}' WHERE id = {$id}";
echo $sql;
and saw:
UPDATE Dbsaved SET comment = 'Hello World!' WHERE id = 1
which is a correct SQL statement, so if it is not working, you might want to play with SQL directly to get something working. Hope that helps!
SOLUTION:
$comment = $_GET['$comment'];
$id = $_GET['$id'];
while ($row = mysql_fetch_array($result)) {
echo "<div id='posts'>";;
echo "<h2>";
echo $row[1] . "";
echo "</h2>";
echo "<p>";
//echo $timestamp = date('d-m-y G:i:s ');
echo "<br>";
echo "<br>";
echo $row[2] . "";
echo "</p>";
echo "<p>";
echo $row[3] . "";
echo "</p>";
echo 'Delete';
echo "<br>";
echo "<br>";
echo $row[4] . "";
echo "<br>";
echo 'Comment: <br>
<form action="addcomment.php?id=' . $row[0]. '" method="post">
<input type=text name=comment><br>
<input type=submit name="submit">
</form>';
echo "<p>";
echo $row['comment'] . "";
echo "</p>";
echo "</div>";
echo "<br>";
}
?>
and:
<?php
$db = mysql_connect("localhost", "root", "root");
if (!$db) {
die("Database connect failed: " . mysql_error());
}
$db_select = mysql_select_db("UNii", $db);
if (!$db_select) {
die("Database selection failed: " . mysql_error());
}
$comment = $_POST['comment'];
$id = $_GET['id'];
$sql = "UPDATE Dbsaved SET comment = '$comment' WHERE id = $id ";
$comment1 = mysql_query($sql);
echo $sql;
if (!$comment1) {
die("did not save comment: " . mysql_error());
}
else {
header("location: UniHelpindex.php");
}
It was to do with mainly needing to get the id which was used in $row[0]' in the form created in the while loop. And actually using the correct syntax for the update Dbsaved... bit.

Why can't my users delete their comment?

I have created a website with a comments page for users to delete the comments that they upload. However the delete comment button appears however it doesn't seem to work. Can anyone shed some light on this for me please?
<?php
require_once("checklog.php");
include_once("nihome_start_logged.php");
require_once("nifunctions.php");?>
<div id="navigation">
<ul class="container">
<li><a href='nihome.php'>Home</a></li>
<li> Search for your service</li>
<li><a href='nisalons.php' class='button'>Salons and Reviews</a></li>
<li><a href='nichangepassword.php' class='button'>Change Password</a></li>
<li><a href='nilogout.php' class='button'>Logout</a></li>
</ul>
</div>
<?php
if (!$db_server){
die("Unable to connect to MySQL: " . mysqli_connect_error($db_server));
$db_status = "not connected";
}else{
//Capture form data, if anything was submitted
if (isset($_GET['salonid']) and ($_GET['salonid'] != '')){
$salonid = clean_string($db_server, $_GET['salonid']);
//code to delete comments
if($db_server){
mysqli_select_db($db_server, $db_database);
mysqli_query($db_server, "DELETE FROM comments WHERE username = '$username' and salonid = '$salonid' ");
$message= "<p> Comment deleted </p>";
//If connected, get Salons from database and write out
mysqli_select_db($db_server, $db_database);
$query = "SELECT ID, salon_name, address, postcode, telephone, email, website FROM salon WHERE ID=$salonid";
$result = mysqli_query($db_server, $query);
if (!$result) die("Query failed: " . mysqli_error($db_server));
while($row = mysqli_fetch_array($result)){
$str_result .= "<h2>" . $row[ 'salon_name'] . "</h2>";
$str_result .= "<p>" . $row['address'] . "</p>";
$str_result .= "<p>" . $row['postcode'] . "</p>";
$str_result .= "<p>" . $row['telephone'] . "</p>";
$str_result .= "<p>" . $row['email'] . "</p>";
$str_result .= "<p>" . $row['website'] . "</p>";
}
}
mysqli_free_result($result);
}else{
$str_result = "<h2>No salon selected</h2>";
}
}
echo $str_result;
?>
<?php
if(trim($_POST['submit']) == "Submit comment"){
//Get any submitted comments and insert
$comment = clean_string($db_server, $_POST['comment']);
if ($comment != '') {
$name=$_FILES['photo']['name'];
if ($name == "") $error .= "<p class='error'>You must write a review and upload an image!</p>";
$originalname=$_FILES['photo']['name'];
$type=$_FILES['photo']['type'];
if ($type=="image/jpeg") $type=".jpeg"; //if true change
else if ($type=="image/jpg") $type=".jpg";// if not true check this one
else if ($type=="image/png") $type=".png";
$name=uniqid() . $type;
$path="images/" . $name;
$tempname=$_FILES['photo']['tmp_name'];
$size=$_FILES['photo']['size'];
//Error checking
if ($size >1000000) $error .= "<p class='error'>Your image file is to big, it have to be less than 200 mb</p>";
if ($error=="") {
if (move_uploaded_file($tempname, $path)){
$uploadquery="INSERT INTO comments (comment, imagename, salonID, userID) VALUES ('$comment', '$path', $salonid, ". $_SESSION['userID'].")";
mysqli_query($db_server,$uploadquery) or die ("Insert failed " . mysqli_error($db_server) . " " . $uploadquery);
$message= "<h2>Thanks for your comment!</h2><p>Your upload was succesful</p>";
}
}
}
}
//Print out existing comment
$query = "SELECT * FROM comments JOIN users ON comments.userID = users.ID WHERE salonID=$salonid";
$result = mysqli_query($db_server, $query);
if (!$result) die("Database access failed: " . mysqli_error($db_server));
while ($row = mysqli_fetch_array($result)){
if ($_SESSION['username'] == $row['username']){
$deletecomment = "<input class='delete comment' type='submit' id='submit' name='submit' value='Delete comment'/>";
}else{
$deletecomment = " ";
}
$str_comments .= "<p><span class='comments'>" . $row['Username'] ." : " . $row['comment'] . "</span></p>";
$str_comments .="<p><img src='" . $row['imagename'] ."' /></p>";
$str_comments .= $deletecomment ;
}
mysqli_free_result($result);
?>
<div id="form">
<table><form id='review' action='salonpage.php?salonid=<?php echo $salonid; ?>' method='post' enctype='multipart/form-data'>
<th><h2> Do you want to review the service you recieved?</h2></th>
<tr><td><textarea name="comment" rows="6" cols="40">Write something here!</textarea></td></tr>
<tr><td><input type='file' name='photo' accept='image/jpg, image/jpeg, image/png'/></td></tr>
<br/>
<tr><td><input type='submit' id='submit' name='submit' value='Submit comment' /></td></tr>
</form></table>
<?php echo $error;
echo $message;?></div>
<h2> Reviews and comments </h2>
<?php echo $str_comments; ?>
<?php mysqli_close($db_server); ?>
<div id='footer'>
Privacy Statement
Accessibility Statement
</div>
<?php include_once("nihome_end.php"); ?>
From you code, you used form-submit-button to delete the record -- and it's stored in $str_comments.
You need
<form .....>
<?php echo $str_comments ?>
</form>
Because the submit-button needs form to live in.
#1 I suggest you start using classes.
#2 if you're using mysqli(i stands for improved) why are you doing things the "old way"?
#3 a classfull example
$drop = new CLASS_NAME_GOES_HERE;
$drop->drop_comment($id,$un);
class CLASS_NAME_GOES_HERE {
private $con; // only access from this class and its children and dont need $ anymore
function __construct() { // constructor function
$this->con = new mysqli(DB,DB_USER,DB_PASS,DB_NAME) or
die('Cannot connect.');
}
function drop_comment($id,$un) {
$sql= "DELETE FROM upload WHERE id = ? AND username = ?";
if($try = $this->con->prepare($sql)) {
$try->bind_param('ss', $id, $un);
if($try->execute()) return true;
}
}//END FUNCTION
}//end class

How do I reduce the number of variables and connections for php connecting to mysql server?

I am trying to reduce the number of connections that this page makes. Everything I read says that only one connection should be enough however if I remove the additional connections the page doesn't connect to my server and provide me the results I am looking for. Also one of the queries I am using twice but if I call the original query the second time it also does not work.
What am I doing wrong?
<?php
$dbaddress="localhost";
$dbuser="testuser";
$dbpass="testpass";
$dbname="testdb";
$dbtable="elo";
$query="SELECT Sequnce, mcacctname FROM `elo`;";
$con=mysqli_connect($dbaddress,$dbuser,$dbpass,$dbname);
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysql_connect($dbaddress, $dbuser, $dbpass);
mysql_select_db($dbname);
$sql = "SELECT Sequence, mcacctname FROM `elo`;";
$result = mysql_query($sql);
mysql_connect($dbaddress, $dbuser, $dbpass);
mysql_select_db($dbname);
$sql2 = "SELECT Sequence, mcacctname FROM `elo`;";
$result2 = mysql_query($sql2);
$sqlstart = "SELECT mcacctname, elo FROM `elo`;";
$q = mysql_query($sqlstart);
?>
<form name="player1" method="post" action="predictions.php">
<label for="Select Player 1">Select Player 1:
<?php
echo "<select name='elouser1'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['Sequence'] . "'>" . $row['mcacctname'] . "</option>";
}
echo "</select>";
?>
</label>
<input type="submit" value="Player 1 Wins">
</form>
<br>
<form name="player2" method="post" action="predictions.php">
<label for="Select Player 2">Select Player 2:
<?php
echo "<select name='elouser2'>";
while ($row2 = mysql_fetch_array($result2)) {
echo "<option value='" . $row2['Sequence'] . "'>" . $row2['mcacctname'] . "</option>";
}
echo "</select>";
?>
</label>
<input type="submit" value="Player 2 Wins">
</form>
<table>
<tr>
<?
echo '<div class="container">';
while($res = mysql_fetch_array($q)){
echo '<tr><td><div class="item">'. $res['mcacctname'] . '</td><td>' . $res ['elo'] . '</div></td></tr>';
}
echo '</div>';
mysqli_close($con);
?>
</tr>
</table>
Once a connection is open in a page you don't need to re-open it for each query. You can simply remove all the statements that create a connection and select the database, except the first one.
Note you're using mysqli to open the connection, but mysql to query it. The two sets of functions are not interchangeable: use mysqli as 'mysql' is deprecated and support will be removed in the future.
Try this:
$dbaddress="localhost";
$dbuser="testuser";
$dbpass="testpass";
$dbname="testdb";
$dbtable="elo";
$query="SELECT Sequence, mcacctname FROM `elo`;";
$con=mysqli_connect($dbaddress,$dbuser,$dbpass,$dbname);
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT Sequence, mcacctname FROM `elo`;";
$result = mysqli_query($con,$sql) or die(mysqli_error($con));
$sql2 = "SELECT Sequence, mcacctname FROM `elo`;";
$result2 = mysqli_query($con,$sql2) or die(mysqli_error($con));
$sqlstart = "SELECT mcacctname, elo FROM `elo`;";
$q = mysqli_query($con, $sqlstart) or die(mysqli_error($con));
?>
<form name="player1" method="post" action="predictions.php">
<label for="Select Player 1">Select Player 1:
<?php
echo "<select name='elouser1'>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['Sequence'] . "'>" . $row['mcacctname'] . "</option>";
}
echo "</select>";
?>
</label>
<input type="submit" value="Player 1 Wins">
</form>
<br>
<form name="player2" method="post" action="predictions.php">
<label for="Select Player 2">Select Player 2:
<?php
echo "<select name='elouser2'>";
while ($row2 = mysqli_fetch_array($result2)) {
echo "<option value='" . $row2['Sequence'] . "'>" . $row2['mcacctname'] . "</option>";
}
echo "</select>";
?>
</label>
<input type="submit" value="Player 2 Wins">
</form>
<table>
<tr>
<?
echo '<div class="container">';
while($res = mysqli_fetch_array($q)){
echo '<tr><td><div class="item">'. $res['mcacctname'] . '</td><td>' . $res ['elo'] . '</div></td></tr>';
}
echo '</div>';
mysqli_close($con);
?>
</tr>
</table>

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