You need to tell mysqli to throw exceptions: mysqli_report(MYSQLI_REPORT_STRICT); try { $connection = new mysqli(‘localhost’, ‘my_user’, ‘my_password’, ‘my_db’) ; } catch (Exception $e ) { echo “Service unavailable”; echo “message: ” . $e->message; // not in live code obviously… exit; } Now you will catch the exception and you can take it from there.
I just experienced this same problem, calling bind_param via call_user_func_array and passing an array of parameters. The solution is to modify the values in the array to be referenced. It’s not elegant but it works. call_user_func_array(array($stmt, ‘bind_param’), makeValuesReferenced($passArray)); function makeValuesReferenced($arr){ $refs = array(); foreach($arr as $key => $value) $refs[$key] = &$arr[$key]; return $refs; }
It’s best if you used an alias for your SUM: SELECT SUM(`field`) as `sum` FROM `table_name` And then, you’ll be able to fetch the result normally by accessing the first result row’s $row[‘sum’].
You cannot do this in mysqli: $stmt->bind_param(“ii”,0,$victimiid); The 0 needs to be a variable. Try this: $zero = 0; $stmt->bind_param(“ii”,$zero,$victimiid);
File sample.html <form action=”sample.php” method=”POST”> <input name=”name” type=”text”> <input name=”text” type=”text”> <input name=”submit” type=”submit” value=”Submit”> </form> File sample.php <?php if (isset($_POST[‘submit’])) { mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); $mysqli = new mysqli(‘localhost’, ‘user’, ‘password’, ‘mysampledb’); // replace every piece of data in SQL with question mark $sql = “INSERT INTO SampleTable (name, text) VALUES (?,?)” // 2 question … Read more
It seems you are using some outdated tutorial, that offers incorrect approach at reporting errors: a code that tries to detect the connection error manually and output something on its own. But your code shouldn’t be doing anything like that. Mysqli can report errors automatically and database errors are no different from other errors, and … Read more
You can use the num_rows on the dataset to check the number of rows returned. Example: $results = $mysqli->query(“SELECT ANNOUNCE_NUMBER,ANNOUNCEMENTS,ANNOUNCE_TYPE,POST_DATE FROM home ORDER BY ANNOUNCE_NUMBER DESC”); if ($results->num_rows === 0) { echo ‘No results’; } else { while ($obj = $results->fetch_object()) { //output results from database } }
Should be something like this: // enable error reporting for mysqli mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); // create mysqli object $mysqli = new mysqli(/* fill in your connection info here */); $email = $_POST[’email’]; // might want to validate and sanitize this first before passing to database… // set query $query = “SELECT COUNT(*) FROM users WHERE … Read more
Okay, here is a way to do it: Edited, to fix bug when fetching multiple rows $sql = “SELECT `first_name`,`last_name` FROM `users` WHERE `country` =? AND `state`=?”; $params = array(‘Australia’,’Victoria’); /* In my real app the below code is wrapped up in a class But this is just for example’s sake. You could easily throw … Read more
As some users have suggested (and is the best way), return the mysqli instance function getConnected($host,$user,$pass,$db) { $mysqli = new mysqli($host, $user, $pass, $db); if($mysqli->connect_error) die(‘Connect Error (‘ . mysqli_connect_errno() . ‘) ‘. mysqli_connect_error()); return $mysqli; } Example: $mysqli = getConnected(‘localhost’,’user’,’password’,’database’);