paulbray Posted April 11, 2011 Share Posted April 11, 2011 Hi, If anyone can assist I am receiving these errors Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /websites/123reg/LinuxPackage21/co/ns/ol/console-entertainment.co.uk/public_html/Test/view_products.php on line 80 Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in /websites/123reg/LinuxPackage21/co/ns/ol/console-entertainment.co.uk/public_html/Test/view_products.php on line 102 My Code is <?php # View products $page_title = 'View Inventory'; require_once ('connect.php'); $display = 35; // Number of records per page if (isset($_GET['p']) && is_numeric($_GET['p'])) { // How many pages there are $pages = $_GET['p']; } else { // Count number of records $q = "SELECT COUNT(Game_Title) FROM Products"; $r = @mysqli_query ($dbc, $q); $row = @mysqli_fetch_array ($r, MYSQLI_NUM); $records = $row[0]; // Calculate number of pages if ($records > $display) {// More than one page $pages = ceil ($records/$display); } else { $pages = 1; } } // End of p IF if (isset($_GET['s']) && is_numeric ($_GET['s'])) { $start = $_GET['s']; } else { $start = 0; } //make query $q = "SELECT SKU_ProductID, SKU_MerchSKUID, Game_Title, Platform, Genre, Weight, Supplier, Supplier_Price, Cashback, Cashback_Amount AS dr, FROM Products ORDER BY Game_Title ASC LIMIT $start, $display"; $r = @mysqli_query ($dbc,$q); echo '<TABLE ALIGN="CENTER" cellspacing="0" cellpadding="5" width="75%"> <TR> <TD align="left"><b>SKU_ProductID</b></td> <TD align="left"><b>SKU_MerchSKUID</b></td> <TD align="left"><b>Game_Title</b></td> <TD align="left"><b>Platform</b></td> <TD align="left"><b>Genre</b></td> <TD align="left"><b>Weight</b></td> <TD align="left"><b>Supplier</b></td> <TD align="left"><b>Supplier_Price</b></td> <TD align="left"><b>Cashback</b></td> <TD align="left"><b>Cashback_Amount</b></td> </TR> '; $bg = '#eeeeee'; while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) { $bg = ($bg=='#eeeeee' ? '#ffffff' : '#eeeeee'); // Switch background color echo ' <tr bgcolor="' . $bg . '"> <td align="left">' . $row['SKU_ProductID'] . '</td> <td align="left">' . $row['SKU_MerchSKUID'] . '</td> <td align="left">' . $row['Game_Title'] . '</td> <td align="left">' . $row['Platform'] . '</td> <td align="left">' . $row['Genre'] . '</td> <td align="left">' . $row['Weight'] . '</td> <td align="left">' . $row['Supplier'] . '</td> <td align="left">' . $row['Supplier_Price'] . '</td> <td align="left">' . $row['Cashback'] . '</td> <td align="left">' . $row['Cashback_Amount'] . '</td> </tr> '; } echo '</table>'; mysqli_free_result ($r); mysqli_close($dbc); if ($pages > 1) { echo '<br /><p>'; $current_page = ($start/$display) + 1; if ($current_page != 1) { echo '<a href="view_products.php"?s=' . ($start - $display) . '&p=' . $pages . ' ">Previous</a> '; for ($i = 1; $i <= $pages; $i++) { if ($i != $current_page) { echo '<a href="view_products.php"?s=' . (($display * ($i - 1))) . '&p=' . $pages . '"> ' , $i . '</a> '; } else { echo $i . ' '; } } if ($current_page != $pages) { echo '<a href="view_products.php"?s=' . ($start + $display) . '&p=' . $pages . '">Next</a>'; } echo '</p>'; }} ?> Quote Link to comment Share on other sites More sharing options...
Adam Posted April 11, 2011 Share Posted April 11, 2011 The problem is you're blindly trying to return rows from a MySQLi result resource, that isn't actually a result resource -- i.e. the query has an error. $r = @mysqli_query ($dbc,$q); mysqli_query() never actually throws a PHP error, so there's no use in the "@" to suppress errors. However you should handle the error here and throw one yourself: $r = mysqli_query($dbc, $q) or trigger_error('MySQLi Error: ' . mysqli_error()); That will tell you why the query is failing. Obviously you'll fix any errors before pushing to a production server (where display_errors should be turned off anyway), so it should never throw an error the user will see. Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted April 11, 2011 Share Posted April 11, 2011 If you search for that error message, you will find that it generally means that your query failed due to an error of some kind and you need to have some error checking and error reporting logic in your code to get it to tell you why the query failed. For troubleshooting purposes, you can echo mysqli_error($dbc) to find out why the query is failing. And I going to guess that all the @'s you have put in front of the other mysqli_ statements in your code were to hide other similar error messages. Hiding the error messages doesn't fix your code, it just means that your code doesn't work and you aren't getting any of the errors that would let you know what isn't working. DON'T put @'s in your code to hide error messages, there's no valid reason to do so. And, is there some reason this is the third thread you have started for this same code? Just because you are getting a different error with your code, doesn't mean it is a different problem. It's the same code and you should stick to one thread for it. Quote Link to comment Share on other sites More sharing options...
paulbray Posted April 11, 2011 Author Share Posted April 11, 2011 ok sorry for multiple threads, I have removed @ signs and this is the output Warning: mysqli_error() expects exactly 1 parameter, 0 given in /websites/123reg/LinuxPackage21/co/ns/ol/console-entertainment.co.uk/public_html/Test/view_products.php on line 60 Notice: MySQLi Error: in /websites/123reg/LinuxPackage21/co/ns/ol/console-entertainment.co.uk/public_html/Test/view_products.php on line 60 SKU_ProductID SKU_MerchSKUID Game_Title Platform Genre Weight Supplier Supplier_Price Cashback Cashback_Amount Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /websites/123reg/LinuxPackage21/co/ns/ol/console-entertainment.co.uk/public_html/Test/view_products.php on line 81 Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in /websites/123reg/LinuxPackage21/co/ns/ol/console-entertainment.co.uk/public_html/Test/view_products.php on line 103 Warning: mysqli_error() [function.mysqli-error]: Couldn't fetch mysqli in /websites/123reg/LinuxPackage21/co/ns/ol/console-entertainment.co.uk/public_html/Test/view_products.php on line 141 My current coding <?php # View products $page_title = 'View Inventory'; require_once ('connect.php'); $display = 35; // Number of records per page if (isset($_GET['p']) && is_numeric($_GET['p'])) { // How many pages there are $pages = $_GET['p']; } else { // Count number of records $q = "SELECT COUNT(Game_Title) FROM Products"; $r = mysqli_query ($dbc, $q); $row = mysqli_fetch_array ($r, MYSQLI_NUM); $records = $row[0]; // Calculate number of pages if ($records > $display) {// More than one page $pages = ceil ($records/$display); } else { $pages = 1; } } // End of p IF if (isset($_GET['s']) && is_numeric ($_GET['s'])) { $start = $_GET['s']; } else { $start = 0; } //make query $q = "SELECT SKU_ProductID, SKU_MerchSKUID, Game_Title, Platform, Genre, Weight, Supplier, Supplier_Price, Cashback, Cashback_Amount AS dr, FROM Products ORDER BY Game_Title ASC LIMIT $start, $display"; $r = mysqli_query($dbc, $q) or trigger_error('MySQLi Error: ' . mysqli_error()); echo '<TABLE ALIGN="CENTER" cellspacing="0" cellpadding="5" width="75%"> <TR> <TD align="left"><b>SKU_ProductID</b></td> <TD align="left"><b>SKU_MerchSKUID</b></td> <TD align="left"><b>Game_Title</b></td> <TD align="left"><b>Platform</b></td> <TD align="left"><b>Genre</b></td> <TD align="left"><b>Weight</b></td> <TD align="left"><b>Supplier</b></td> <TD align="left"><b>Supplier_Price</b></td> <TD align="left"><b>Cashback</b></td> <TD align="left"><b>Cashback_Amount</b></td> </TR> '; $bg = '#eeeeee'; while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) { $bg = ($bg=='#eeeeee' ? '#ffffff' : '#eeeeee'); // Switch background color echo ' <tr bgcolor="' . $bg . '"> <td align="left">' . $row['SKU_ProductID'] . '</td> <td align="left">' . $row['SKU_MerchSKUID'] . '</td> <td align="left">' . $row['Game_Title'] . '</td> <td align="left">' . $row['Platform'] . '</td> <td align="left">' . $row['Genre'] . '</td> <td align="left">' . $row['Weight'] . '</td> <td align="left">' . $row['Supplier'] . '</td> <td align="left">' . $row['Supplier_Price'] . '</td> <td align="left">' . $row['Cashback'] . '</td> <td align="left">' . $row['Cashback_Amount'] . '</td> </tr> '; } echo '</table>'; mysqli_free_result ($r); mysqli_close($dbc); if ($pages > 1) { echo '<br /><p>'; $current_page = ($start/$display) + 1; if ($current_page != 1) { echo '<a href="view_products.php"?s=' . ($start - $display) . '&p=' . $pages . ' ">Previous</a> '; for ($i = 1; $i <= $pages; $i++) { if ($i != $current_page) { echo '<a href="view_products.php"?s=' . (($display * ($i - 1))) . '&p=' . $pages . '"> ' , $i . '</a> '; } else { echo $i . ' '; } } if ($current_page != $pages) { echo '<a href="view_products.php"?s=' . ($start + $display) . '&p=' . $pages . '">Next</a>'; } echo '</p>'; }} ?> Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted April 11, 2011 Share Posted April 11, 2011 The mysqli_error() statement requires the connection link, other wise it produces an error it self that prevents it from showing any actual query error - mysqli_error($dbc) Quote Link to comment Share on other sites More sharing options...
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