[SOLVED] Another while() problem

[Rocketaka]

I'm building a check out system for a project and everything in my code works just fine. The only problem is that when I press the button to check something out all the buttons only check out the very last entry in the table. I'm not exactly sure what is causing this problem. If you could take a look at my code below and see if you could figure it out I'd appreciate it. Thank you!

 

<?php
include("check.php");
include("include.php");
if($_SESSION['auth']) {
$query = mysql_query("SELECT id, title, category, status FROM software");
echo "<table align='center' cellpadding='2' cellspacing='2' width='100%'>
	<tr bgcolor='#1469A2'>
	<td width='50%'><b>Software Title</b></td>
	<td width='20%'><b>Category</b></td>
	<td width='15%'><b>Status</b></td>
	<td width='15%'><b>Check Out?</b></td>
	</tr>";
while ($data = mysql_fetch_array($query)) {
	$title = $data['title'];
	$category = $data['category'];
	$status = $data['status'];
	$date = date("Y/m/d");
	$username = $_COOKIE['cis_username'];
	$sql = "UPDATE `software` SET `status` = 'Checked Out', `by` = '$username', `date` = '$date' WHERE `title` = '$title'";
	echo "<tr bgcolor='#003C64'>
		<td width='50%'>".$title."</td>
		<td width='20%'>".$category."</td>
		<td width='15%'>".$status."</td>
		<td width='15%'>";
	if ($status == "Available") {
		echo "<form method='POST' style='margin:0px;'>
			<input type='submit' name='checkout' style='border:1px #1469A2 solid;' value='Check Out'>
			</form>";
	}
	else {
		echo "<i>Not Available</i>";
	}
	echo "</td></tr>";
}
echo "</table>";
if(isset($_POST['checkout'])) {
	mysql_query("$sql") or die (mysql_error());
	echo "You have checked out <b>".$title."</b> successfully";
}
}
else {
echo "You need to login to view this page! <a href='index.php?page=login'>Login here</a>";
}
?>

[corbin]

Try doing echo mysql_num_rows($query); and make sure that it is more than 1...

 

Also I think mysql_query("$sql") should be mysql_query($sql)

[hitman6003]

Change:

 

$query = mysql_query("SELECT id, title, category, status FROM software");

 

to:

 

$query = mysql_query("SELECT id, title, category, status FROM software") or die(mysql_error());

 

Then see what error you are getting.

[Rocketaka]

I did what both of you suggestion and so far nothing. It's still the same. I'm receiving no errors. It' just not working like it should. I did the mysql_num_rows and it displayed the correct amount of rows in my table so I have no clue why every button I push will only check out the last entry in the table. It may have to do with the while() loop but I don't know.

[corbin]

try

<?php
include("check.php");
include("include.php");
if($_SESSION['auth']) {
$query = mysql_query("SELECT id, title, category, status FROM software");
echo "<table align='center' cellpadding='2' cellspacing='2' width='100%'>
	<tr bgcolor='#1469A2'>
	<td width='50%'><b>Software Title</b></td>
	<td width='20%'><b>Category</b></td>
	<td width='15%'><b>Status</b></td>
	<td width='15%'><b>Check Out?</b></td>
	</tr>";
while ($data = mysql_fetch_array($query)) {
	$title = $data['title'];
	$category = $data['category'];
	$status = $data['status'];
	$date = date("Y/m/d");
	$username = $_COOKIE['cis_username'];
	$sql = "UPDATE `software` SET `status` = 'Checked Out', `by` = '$username', `date` = '$date' WHERE `title` = '$title'";
	if(isset($_POST['checkout'])) { mysql_query($sql) or die (mysql_error());}
	echo "<tr bgcolor='#003C64'>
		<td width='50%'>".$title."</td>
		<td width='20%'>".$category."</td>
		<td width='15%'>".$status."</td>
		<td width='15%'>";
	if ($status == "Available") {
		echo "<form method='POST' style='margin:0px;'>
			<input type='submit' name='checkout' style='border:1px #1469A2 solid;' value='Check Out'>
			</form>";
	}
	else {
		echo "<i>Not Available</i>";
	}
	echo "</td></tr>";
}
echo "</table>";
if(isset($_POST['checkout'])) {
	echo "You have checked out <b>".$title."</b> successfully";
}
}
else {
echo "You need to login to view this page! <a href='index.php?page=login'>Login here</a>";
}
?>

[Rocketaka]

Yeah I tried what you suggested and instead of just the last entry in the table getting checked out it will check out everything.

[alpine]

The problem is that you are constantly overwriting the $sql variable while the loop is running

 

Try something like this:

<?php
include("check.php");
include("include.php");
if($_SESSION['auth']) {
$query = mysql_query("SELECT id, title, category, status FROM software");
echo "<table align='center' cellpadding='2' cellspacing='2' width='100%'>
	<tr bgcolor='#1469A2'>
	<td width='50%'><b>Software Title</b></td>
	<td width='20%'><b>Category</b></td>
	<td width='15%'><b>Status</b></td>
	<td width='15%'><b>Check Out?</b></td>
	</tr>";
while ($data = mysql_fetch_array($query)) {

	echo "<tr bgcolor='#003C64'>
		<td width='50%'>".$data['title']."</td>
		<td width='20%'>".$data['category']."</td>
		<td width='15%'>".$data['status']."</td>
		<td width='15%'>";
	if ($data['status'] == "Available") {
		echo "<form method='POST' style='margin:0px;'>
                                <input type='hidden' name='id' value='$data['id']'>
			<input type='submit' name='checkout' style='border:1px #1469A2 solid;' value='Check Out'>
			</form>";
	}
	else {
		echo "<i>Not Available</i>";
	}
	echo "</td></tr>";
}
echo "</table>";
if(isset($_POST['checkout'])) {
        $date = date("Y/m/d");
        $username = $_COOKIE['cis_username'];
        $id = $_POST['id'];
        $sql = "UPDATE `software` SET `status` = 'Checked Out', `by` = '$username', `date` = '$date' WHERE `id` = '$id'";
mysql_query("$sql") or die (mysql_error());
echo "You have checked out <b>".$title."</b> successfully";
}
}
else {
echo "You need to login to view this page! <a href='index.php?page=login'>Login here</a>";
}
?>

[Rocketaka]

I tried your code and it got an error because for the hidden input field the value was $date['id'] so i made that into a variable called id and it ended up working. Thank you very much!