nickholt1972 Posted October 6, 2006 Share Posted October 6, 2006 I had a shopping page which was working fine only I didn't like it. So in trying to fix what wasn't broken ... guess what? I broke it.I introduced sessions and shortly afterwards made the short trip to errormessagesvilleMy products.php page starts with the following code[code]<?phpsession_start();if (!isset($_SESSION['cart'])) { $_SESSION['cart'] = array();}if (isset($_GET['buy'])) { // Add item to the end of the $_SESSION['cart'] array $_SESSION['cart'][] = $_GET['buy']; header('location: ' . $_SERVER['PHP_SELF'] . '?' . SID); exit();}?>[/code]then it does some other SELECT FROM related shenanigans, before giving it some of this.. (and its this line which brings about the error)[code]while ($row = mysql_fetch_array($result)) {[/code]within my while statement, I display my product details and include the following link, which is SUPPOSED to add the product id to my array of purchased products.[code]<a href="' . $_SERVER['PHP_SELF'] .'?buy=' . $row['id'] . '">Buy</a>[/code]When I run the page, the number of items in my cart goes up, which is fine. However, I get the following message:[i]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/.sites/123/site117/web/display2.inc.php on line 3[/i](line 3 is [b]while ($row = mysql_fetch_array($result)) {[/b])What I think i happening is the [b]while ($row...[/b] line is losing track of the preceding SELECT line so there's effectively no data but this only happened since I started piddling about with these pesky sessions.can any one suggest a solution? Quote Link to comment Share on other sites More sharing options...
nickholt1972 Posted October 6, 2006 Author Share Posted October 6, 2006 Further to my original post, I think I might not have explained myself too well so i'm posting the complete script (more or less, slightly simplified so you've less to trawl through)[code]<?phpsession_start();if (!isset($_SESSION['cart'])) { $_SESSION['cart'] = array();}if (isset($_GET['buy'])) { // Add item to the end of the $_SESSION['cart'] array $_SESSION['cart'][] = $_GET['buy']; header('location: ' . $_SERVER['PHP_SELF'] . '?' . SID); exit();}?><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"><html><head><title>apple tree nappies online shop</title></head><body><?phpinclude_once 'dbconnect.inc.php';if ($choice == "allnappies"){$result = @mysql_query('SELECT * FROM stocks WHERE alphatype = "NAPPY" AND display = "yes" ORDER BY id ASC'); }elseif ($choice == "onesize"){$result = @mysql_query('SELECT * FROM stocks WHERE sizetype = "One Size" AND display = "yes" ORDER BY id ASC');}elseif ($choice == "multisize"){$result = @mysql_query('SELECT * FROM stocks WHERE sizetype = "Multi Size" AND display = "yes" ORDER BY id ASC');} // Display the title of each entry with a 'buy' link in a table // THIS FOLLOWING LINE IS THE ONE GIVING ME AN ERROR //while ($row = mysql_fetch_array($result)) {<table class="products"><tr><td>' . $row['alphaname'] . '</td></td><td><a href="' . $_SERVER['PHP_SELF'] .'?buy=' . $row['id'] . '">Buy</a></td></tr></table>'; }?><?php echo count($_SESSION['cart']); ?><p><a href="cart.php">View Cart</a></p></body></html>[/code]I'd really appreciate your input.Thanks. Quote Link to comment Share on other sites More sharing options...
redbullmarky Posted October 7, 2006 Share Posted October 7, 2006 $choice does not appear to get a value from anywhere, effectively meaning non of these:[code]if ($choice == "allnappies"){$result = @mysql_query('SELECT * FROM stocks WHERE alphatype = "NAPPY" AND display = "yes" ORDER BY id ASC'); }elseif ($choice == "onesize"){$result = @mysql_query('SELECT * FROM stocks WHERE sizetype = "One Size" AND display = "yes" ORDER BY id ASC');}elseif ($choice == "multisize"){$result = @mysql_query('SELECT * FROM stocks WHERE sizetype = "Multi Size" AND display = "yes" ORDER BY id ASC');}[/code]will get executed and meaning that $result, which you expect to be the result of a query, will not get set,resulting in the error you've gotten here. it's worth noting that in many cases, using the error suppressor (@) can make debugging pretty tricky. as the error coming back from mysql_query will only be stuff like not being able to connect to the database, or invalid SQL, you'd be better doing a proper check elsewhere - i.e, let the mysql_query's throw errors if they have to, so you can fix them. something like:[code]<?php// first make sure we actually have a choiceif ($choice){ $query = ''; // using switch to simplify all the ifs/elses etc switch ($choice) { case 'allnappies': $query = 'SELECT * FROM stocks WHERE alphatype = "NAPPY" AND display = "yes" ORDER BY id ASC'; break; case 'onesize': $query = 'SELECT * FROM stocks WHERE sizetype = "One Size" AND display = "yes" ORDER BY id ASC'; break; case 'multisize': $query = 'SELECT * FROM stocks WHERE sizetype = "Multi Size" AND display = "yes" ORDER BY id ASC'; break; } // if we have a query, run it. if ($query) { $result = mysql_query($query) or die('Problem with '.$query); // rest of code here... } else { echo 'Invalid choice'; die; }}else{ echo 'No choice specified'; die;}?>[/code] Quote Link to comment Share on other sites More sharing options...
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