mysqli_query does not return your number directly. It returns a mysqli_result as you can see here. To get the row you have parsed, you should fetch it first: $row = mysqli_fetch_assoc($result) A lot of information on using mysqli can be found here. 6 solved Object of class mysqli_result could not be converted to int – … Read more
Returns an array that corresponds to the fetched row and moves the internal data pointer ahead. The internal data point is still at the end when you try to use your second while loop. You have done nothing to reset it. You can move it back to the start with mysqli_data_seek($result, 0); Update: when first … Read more
You’re generating invalid HTML. This: ‘<option value=”. $spName . ” ‘ . $spPro .’>’ ^— WITH the added space Will produce something like this: <option value=SomeName SomeProfession> The browser has no way of knowing that these two values are part of the same attribute. In short, you forgot the quotes. You want to generate something … Read more
The html code below is a snippet of how your form should look, though you have mentioned you already have this part done: page1.html <form method=”POST” action=”page2.php”> <input type=”text” name=”usernameForm”> <input type=”password” name=”passwordForm”> <input type=”submit” value=”Submit”> </form> The php code below then obtains the variables from page1.html after a user Submits their information from the … Read more
I got it to work with $result = $result->get_result(); 0 solved Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, object given
Just for a quick testing purpose, try this below which is a bare bones method. You can then slowly build up sanitizing and troubleshoot from thereon. <?php $username = $_POST[‘username’]; $password = sha1($_POST[‘password’]); $link = mysqli_connect(‘xxx’, ‘xxx’, ‘xxx’, ‘xxx’); $query = “SELECT password, id FROM users WHERE username=”$username” AND password=’$password'”; $result = mysqli_query($link, $query); if(mysqli_num_rows($result) … Read more
You need to escape the single quotes using php’s str_replace, e.g.: $exp_title = str_replace(“‘”, “\'”, $_REQUEST[‘exp_title’]); $exp_description = str_replace(“‘”, “\'”, $_REQUEST[‘exp_description’]); $exp_time = $_REQUEST[‘exp_time’]; $update=”UPDATE experience SET exp_title=””.$exp_title.”” , exp_description='”.$exp_description.”‘ , exp_time=””.$exp_time.”” WHERE expid='”.$id.”‘”; However, you should really really use preparedstatements instead of concatenating strings and escaping characters, e.g.: $exp_title = $_REQUEST[‘exp_title’]; $exp_description = $_REQUEST[‘exp_description’]; … Read more
If you are going to convert your code to mysqli_*, then it should be in a prepared statement manner. Lets re-establish your MySQL connection in db.inc.php: <?php $conn = new mysqli(“localhost”, “Datenbank-Username”, “Datenbank-Passwort”, “Datenbank-Name”); /* CHECK CONNECTION */ if (mysqli_connect_errno()) { printf(“Connect failed: %s\n”, mysqli_connect_error()); exit(); } /* CHANGE CHARACTER SET TO utf8 */ if … Read more
i probably not guess what is the issue with you and your code but i wanna suggest you to use alternet syntax this make more sense writing very long query. “INSERT INTO table_name SET column_name = $value, column_name = $value, column_name = $value, column_name = $value, column_name = $value, column_name = $value”; solved Accessing MySQL … Read more
It was pretty simple, I don’t know why do I get a vote down every time I ask question. $mysqli = new mysqli(“localhost”, “root”, “”, “database”); if ($mysqli->connect_errno) { echo “Failed to connect to MySQL: (” . $mysqli->connect_errno . “) ” . $mysqli->connect_error; } $username = $_SERVER[‘REMOTE_ADDR’]; $stmt = $mysqli->prepare(“select * from `vpb_uploads` where `username` … Read more
You get the error because you bind 7 params to the query, but don’t use any of them. Instead you insert the variables directly into the query, which leads to the data being (insecurely) inserted into the database. $insert = $this->con->db->prepare(‘UPDATE users SET firstName=”‘.$updateFirstName.'”, lastName=”‘.$updateLastName.'”, username=”‘.$updateUsername.'”, email=”‘.$updateEmail.'”, profileImage=”‘.$updateProfileImage.'”, bio=”‘.$updateBio.'” WHERE userID = ‘.$userID); should be … Read more
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘)’ at line 5 solved You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near … Read more
Try this: <?php $games_sql = “SELECT id,col1,col2,now FROM tblname ORDER BY now ASC LIMIT 9”; $sth = $conn->query($games_sql); $sth->setFetchMode(PDO::FETCH_ASSOC); $gn=0; while ($game_get = $sth->fetch()) { $id = $game_get[‘id’]; $col1 = $game_get[‘col1’]; $col2 = $game_get[‘col2’]; $now = $game_get[‘now’]; $gametimeanddate = jdate(“l d M y time G:i”,$now); $gamedate = jdate(“l d M y”,$now); $gametime = jdate(“G:i”,$now); //SAVE … Read more
You should have a database name for mysqli_connect. So, go like this: $connect=mysqli_connect(‘localhost’,’root’,”,’databse_name’); and why are you using __mysqli_ston for?Use this instead $query = mysqli_query($connect, “SELECT * FROM users WHERE username=”$username””); Do not use __mysqli_ston. This should work.If doesn’t work,get back to me. solved Myqli php login [duplicate]
Just change the condition to: if(isset($_REQUEST[‘userid’]) && $_REQUEST[‘userid’] > $user_hack) isset tells is a variable is set, while this statement may be true or false, on which you cannot call isset function. Until you check if(isset($_REQUEST[‘userid’])), you cannot assign it to $userid variable. 2 solved Fatal error in if stament [duplicate]