Hey guys, I need your help. It took me hours to try and fix the problem as I am still trying to get the data from the mysql database to print into the php page, but I keep getting the error message of the array which it said "data is not found". <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'username'); define('DB_PASSWORD', 'password'); define('DB_DATABASE', 'databasename'); $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)); } $datastrings = clean($_GET['strings1']); if($datastrings== '') { $errmsg_arr[] = 'data is not found'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $query = "SELECT data_strings FROM table1"; $result=mysql_query($query) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result)) { echo "<p id='data1'>"; echo $row['data1'] . "</p>"; } } ?> PHP: I am sure that the column in the database are valid. The name of the column is data_strings and the table name is called table1. Any idea why I keep getting the error message of the array?
First, are you passing the string1 variable in the URL, as in: http://site.com/page.php?strings1=stringyouwant Second, your database query is static, never changes. Is that how you want it or do you actually want to use the variable $datastrings in place of the static field data_strings as in: $query = "SELECT $datastrings FROM table1"; PHP: