neridaj Posted April 19, 2008 Share Posted April 19, 2008 Hello, I'm getting this error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in output_fns.php on line 416 Here are the two functions involved: function get_user_info($username) { // connect to db $conn = db_connect(); // query user info $result = $conn->query("select first_name, last_name, email from user where username='$username'"); if (!$result) return false; else return $result; } function display_folders() { $userinforesult = get_user_info($username); if(!$userinforesult) echo 'nothing'; else while ($row = mysql_fetch_array($result, MYSQL_NUM)) { // line 416 printf("Last Name: %s Last Name: %s Email: %s", $row[0], $row[1], $row[2]); } echo '<a href="logout.php">Logout</a>'; $userfolder = $_SESSION['valid_user'] . '/'; echo '<table align="center" border="1" cellpadding="5"><tr><th></th></tr><tr>'; $dir = 'members/' . $userfolder; $files = scandir($dir); foreach($files as $value) { if(!in_array($value, array('.', '..'))) { // check for folders if(is_dir($dir.DIRECTORY_SEPARATOR.$value)) { printf('<td><a href="preview.php?pa=%s">'. '<img src="'. $dir . $value .'.jpg" width=75" height="75" />'. '<br />%s<a/></td>', $value, $value); } } } echo '</tr></table>'; } Is it because I'm not using mysql_query? I tried that and I just got connection errors so I reverted back to: $result = $conn->query(), which at least connects and returns something. Is there an alternative to mysql_fetch_array() that I should be using instead? Thanks for any help, Jason Quote Link to comment Share on other sites More sharing options...
phpretard Posted April 19, 2008 Share Posted April 19, 2008 I would look more into the connection error. Quote Link to comment Share on other sites More sharing options...
joinx Posted April 29, 2008 Share Posted April 29, 2008 me too i'm getting same error...here is my code <?php mysql_select_db($database_JaceyConn, $JaceyConn); $sql = "SELECT product_name, product_price, product_status from FROM product WHERE product_id = $productId;"; $result = mysql_query($sql, $JaceyConn); while ($row_result = mysql_fetch_array($result)) { echo $row_result['product_name']; } ?> where is the error? Quote Link to comment Share on other sites More sharing options...
moselkady Posted April 29, 2008 Share Posted April 29, 2008 Try to print mysql_error() after your mysql_query(..) statement to check for errors in it Quote Link to comment Share on other sites More sharing options...
joinx Posted April 29, 2008 Share Posted April 29, 2008 ok thnx...there was an unknow field... problem solved.. Quote Link to comment Share on other sites More sharing options...
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