mark103 Posted May 1, 2012 Share Posted May 1, 2012 Hi guys, I need your help. I am checking on a database as I want to see if I have the same value in the url and in the database. <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbtable'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); $test = clean($_GET['test']); $public = clean($_GET['public']); if (isset($_GET['user']) && (isset($_GET['pass']))) { if($username == '' || $password == '') { $errmsg_arr[] = 'username or password are missing'; $errflag = true; } } elseif (isset($_GET['user']) || (isset($_GET['test'])) || (isset($_GET['public']))) { if($username == '' || $test == '' || $public == '') { $errmsg_arr[] = 'user or others are missing'; $errflag = true; } } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'"; $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if ($username && $password) { if(mysql_num_rows($result) > 0) { $qrytable1="SELECT images, id, test, links, Public FROM user_list WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo "<p id='test'>"; echo $row['test'] . "</p>"; echo '<p id="images"> <a href="images.php?test=test&id='.$row['id'].'">Images</a></td> | <a href="http://' . $row["links"] . '">Link</a> </td> | <a href="delete.php?test=test&id='.$row['id'].'">Delete</a> </td> | <span id="test">'.$row['Public'].'</td>'; } } else { echo "user not found"; } } elseif($username && $test && $public) { $qry="SELECT * FROM members WHERE username='$username'"; $result1=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result1) > 0) { $qrytable1="SELECT Public FROM user_list WHERE username='$username' && test='$test'"; $result2=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result2) > 0) { $row = mysql_fetch_row($result2); mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'"); echo "update!"; } else { echo "already updated!"; } } else { echo "user not found"; } } } ?> This is the function I use to check the value in the database: if (mysql_affected_rows($result2) > 0) { mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'"); echo "you have update it!"; } else if (mysql_affected_rows($result2) < 0) { echo "it is not on the database"; } else { echo "you have already updated!"; } When i input the different value in a url bar while the records are not the same as the value in the url and in the database, i can't get passed and I am keep getting "you have already updated!!" when the value in a database are different than I have input in a url. Do you know how i can get pass it when I have input the different value in the url while it is not the same in the database? Any advice would be much appreicated. Thanks, Mark Link to comment https://forums.phpfreaks.com/topic/261914-how-to-check-the-value-in-mysql-database/ Share on other sites More sharing options...
McMaster Posted May 2, 2012 Share Posted May 2, 2012 What query is running through the $result2 variable? Link to comment https://forums.phpfreaks.com/topic/261914-how-to-check-the-value-in-mysql-database/#findComment-1342304 Share on other sites More sharing options...
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