mikebarbaro Posted February 2, 2010 Share Posted February 2, 2010 Hello, I have the following code processing a form located at tandoorpgh.com/placeanorder which declares two variables, $qty and $spice_level from the array. I try saving them as Session variables and then passing them onto the following script but nothing prints out. They echo here perfectly: $items = $_POST['items']; foreach($items as $item){ $qty = $_POST['qty_' . $item]; // now the quantity of that particular item will be stored here $spice_level = $_POST['spice_level_' . $item]; // now the spice_level of that particular item will be stored here $_SESSION['qty'] = $_POST['qty_' . $item]; $_SESSION['spice_level'] = $_POST['spice_level_' . $item]; $query = mysql_query("SELECT * FROM `menu` WHERE id = '$item'"); while($row = mysql_fetch_array($query)) { if ($qty > 0) { $total = $total + $row['price'] * $qty; echo "<p>$qty - " . $row['title'] . " - <span style=color:#000;>$" . $row['price'] . "</span> - Spice Level: $spice_level</p>"; } else { $total = $total + $row['price']; echo "<p>1 - " . $row['title'] . " - <span style=color:#000;>$" . $row['price'] . "</span> - Spice Level: $spice_level</p>"; } But when called up on the following script here within $finalitems, they display as blank. This script here is for faxing the variables: $items = $_SESSION['items']; foreach ($items as $item) { $spice_level = $_SESSION['spice_level']; $qty = $_SESSION['quantity']; $sql = mysql_query("SELECT * FROM `menu` WHERE id = '$item'"); while($row = mysql_fetch_array($sql)){ $finalitems .= $qty . ' - ' . $row['title'] . ' - Spice Level: ' . $spice_level . '<br><br />'; //changed this line }} Any help is highly appreciated. I've been working on this for days. Thanks! Link to comment https://forums.phpfreaks.com/topic/190682-session-variable-problem/ Share on other sites More sharing options...
teamatomic Posted February 2, 2010 Share Posted February 2, 2010 $_SESSION['qty']=$items HTH Teamatomic Link to comment https://forums.phpfreaks.com/topic/190682-session-variable-problem/#findComment-1005610 Share on other sites More sharing options...
mikebarbaro Posted February 2, 2010 Author Share Posted February 2, 2010 $_SESSION['qty']=$items HTH Teamatomic Now it just displays each qty and spice_level as the word Array?? First script: $items = $_POST['items']; foreach($items as $item){ $qty = $_POST['qty_' . $item]; // now the quantity of that particular item will be stored here $spice_level = $_POST['spice_level_' . $item]; // now the spice_level of that particular item will be stored here $_SESSION['qty']=$items; $_SESSION['spice_level']=$items; $query = mysql_query("SELECT * FROM `menu` WHERE id = '$item'"); while($row = mysql_fetch_array($query)) { if ($qty > 0) { $total = $total + $row['price'] * $qty; echo "<p>$qty - " . $row['title'] . " - <span style=color:#000;>$" . $row['price'] . "</span> - Spice Level: $spice_level</p>"; } else { $total = $total + $row['price']; echo "<p>1 - " . $row['title'] . " - <span style=color:#000;>$" . $row['price'] . "</span> - Spice Level: $spice_level</p>"; } Faxing script: $items = $_SESSION['items']; foreach ($items as $item) { $spice_level = $_SESSION['spice_level']; $qty = $_SESSION['qty']; $sql = mysql_query("SELECT * FROM `menu` WHERE id = '$item'"); while($row = mysql_fetch_array($sql)){ $finalitems .= $qty . ' - ' . $row['title'] . ' - Spice Level: ' . $spice_level . '<br><br />'; //changed this line }} Link to comment https://forums.phpfreaks.com/topic/190682-session-variable-problem/#findComment-1005613 Share on other sites More sharing options...
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