This question already has answers here:
How to use mysqli prepared statements?
(3 answers)
Closed 2 years ago.
I'm Reviewing this code I got from YouTube to help my understanding about connecting PHP to MySQL
<?php
$username = $_POST['username'];
$password = $_POST['password'];
$gender = $_POST['gender'];
$email = $_POST['email'];
$phoneCode = $_POST['phoneCode'];
$phone = $_POST['phone'];
if (!empty($username) || !empty($password) || !empty($gender) || !empty($email) || !empty($phoneCode) || !empty($phone)) {
$host = "localhost";
$dbUsername = "root";
$dbPassword = "";
$dbname = "youtube";
//create connection
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$SELECT = "SELECT email From register Where email = ? Limit 1";
$INSERT = "INSERT Into register (username, password, gender, email, phoneCode, phone) values(?, ?, ?, ?, ?, ?)";
//Prepare statement
$stmt = $conn->prepare($SELECT);
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->bind_result($email);
$stmt->store_result();
$stmt->store_result();
$stmt->fetch();
$rnum = $stmt->num_rows;
if ($rnum==0) {
$stmt->close();
$stmt = $conn->prepare($INSERT);
$stmt->bind_param("ssssii", $username, $password, $gender, $email, $phoneCode, $phone); //line 32
$stmt->execute();
echo "New record inserted sucessfully";
} else {
echo "Someone already register using this email";
}
$stmt->close();
$conn->close();
}
} else {
echo "All field are required";
die();
}
?>
My question is what "ssssii" on line 32 do?
and also what values (?,?,?,?,?,?) On $insert for?
Source: Code and Coins YouTube.
This is prepared statement. S is string i is integer.
The (?,?,?,?) is for the script to know where to bind the params to
Related
This question already has an answer here:
"No data supplied for parameters in prepared statement"
(1 answer)
Closed 1 year ago.
i'm very new to php. So, I tried to make a simple form to order sandwiches, but when I click the submit button i get this error "No data supplied for parameters in prepared statement".
Btw, I copied most of the code from a YouTube video, and I don't know what some parts of the code actually do.
that's my code:
<?php
if (isset($_POST['submit'])) {
if (isset($_POST['nombre']) && isset($_POST['apellido']) &&
isset($_POST['bocadillo']) && isset($_POST['extra']) &&
isset($_POST['comentario']) && isset($_POST['comentario'])) {
$nombre = $_POST['nombre'];
$apellido = $_POST['apellido'];
$bocadillo = $_POST['bocadillo'];
$extra = $_POST['extra'];
$comentario = $_POST['comentario'];
$host = "localhost";
$dbUsername = "------";
$dbpassword = "------";
$dbName = "------";
$conn = new mysqli($host, $dbUsername, $dbpassword, $dbName);
if ($conn->connect_error) {
die('Could not connect to the database.');
}
else {
$Select = "SELECT extra FROM pedidos WHERE extra = ? LIMIT 1";
$Insert = "INSERT INTO pedidos(nombre, apellido, bocadillo, extra, comentario) values(?, ?, ?, ?, ?)";
$stmt = $conn->prepare($Select);
$stmt->execute();
$stmt->bind_result($resultemail);
$stmt->store_result();
$stmt->fetch();
$rnum = $stmt->num_rows;
if ($rnum == 0) {
$stmt->close();
$stmt = $conn->prepare($Insert);
if ($stmt->execute()) {
echo "New record inserted sucessfully.";
}
else {
echo $stmt->error;
}
}
else {
echo "Someone already registers using this email.";
}
$stmt->close();
$conn->close();
}
}
else {
echo "All field are required.";
die();
}
}
else {
echo "Submit button is not set";
}
?>
You're missing the bind_param statements for both queries
$Select = "SELECT extra FROM pedidos WHERE extra = ? LIMIT 1";
$stmt = $conn->prepare($Select);
$stmt->bind_param("s", $extra);
$stmt->execute();
and then in the insert
$Insert = "INSERT INTO pedidos(nombre, apellido, bocadillo, extra, comentario) values(?, ?, ?, ?, ?)";
$stmt = $conn->prepare($Select);
$stmt->bind_param("sssss", $nombre, $apellido, $bocadillo, $extra, $comentario);
$stmt->execute();
https://www.w3schools.com/php/php_mysql_prepared_statements.asp
There is an html form which I want to store data from in a database. After filling the form, it returns the php script instead of success message. And no data is getting stored in the database. Can you please help?
<?php
if (isset(['submit'])) {
$username = $_POST['username'];
$password = $_POST['password'];
$gender = $_POST['gender'];
$email = $_POST['email'];
}
if (!empty($username) || !empty($password) || !empty($gender) || !empty($email) {
$host = "localhost";
$dbUsername = "root";
$dbPassword = "";
$dbname = "cakeshop";
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_error().)')'. mysqli_connect_error());
}else {
$SELECT = "SELECT email From register Where email = ? Limit 1"
$INSERT = "INSERT INTO register (username, password, gender, email) values(?, ?, ?, ?)";
$stmt = $conn->prepare($SELECT);
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->bind_result($email);
$stmt->store_result();
$rnum = $stmt->num_rows;
if ($rnum==0){
$stmt->close();
$stmt = $conn->prepare($INSERT);
$stmt->bind_param("ssss", $username, $password, $gender, $email);
$stmt->execute();
echo "New record inserted successfully"
}else {
echo "Someone already registered";
}
$stmt->close();
$conn->close();
}
}
?>
I have a form that I want to submit it without reloading the page with jquery. I am trying to make a form that will send the data to MySQL database with PHP.
This is the code in submit button onclick:
e.preventDefault();
var data = JSON.stringify($("form").serialize());
$.post("insert.php?" + data);
and this is the PHP file :
<?php
$username = $_POST['username'];
$password = $_POST['password'];
$gender = $_POST['gender'];
$email = $_POST['email'];
$phoneCode = $_POST['phoneCode'];
$phone = $_POST['phone'];
if (!empty($username) || !empty($password) || !empty($gender) || !empty($email) || !empty($phoneCode) || !empty($phone)) {
$host = "localhost";
$dbUsername = "use";
$dbPassword = "user_pass";
$dbname = "test";
//create connection
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$SELECT = "SELECT email From register Where email = ? Limit 1";
$INSERT = "INSERT Into register (username, password, gender, email, phoneCode, phone) values(?, ?, ?, ?, ?, ?)";
//Prepare statement
$stmt = $conn->prepare($SELECT);
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->bind_result($email);
$stmt->store_result();
$rnum = $stmt->num_rows;
if ($rnum==0) {
$stmt->close();
$stmt = $conn->prepare($INSERT);
$stmt->bind_param("ssssii", $username, $password, $gender, $email, $phoneCode, $phone);
$stmt->execute();
echo "New record inserted sucessfully";
} else {
echo "Someone already register using this email";
}
$stmt->close();
$conn->close();
}
} else {
echo "All field are required";
die();
}
?>
But it's not adding data to the database. I tried also writing the URL then the data like this: insert.php?data=value.But it didn't work. I am new to server-side and PHP, so I appreciate any help
I have been trying to connect to my database (I used Xampp) and made a PHP trying to link the file to the "users" database that I made. There is an issue as whenever I click the "Register" button it has an error saying "Connect Error(2002)Connection refused." Any help would be appreciated!
PHP code
<?php
$name = $_POST['name'];
$username = $_POST['username'];
$password = $_POST['password'];
$email = $_POST['email'];
$phone = $_POST['phone'];
if (!empty($name) || !empty($username) || !empty($password) || !empty($email) || !empty($phone)) {
$host = "localhost:80";
$dbUsername = "root";
$dbPassword = "";
$dbname = "pracdata";
//create connection
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$SELECT = "SELECT email From users Where email = ? Limit 1";
$INSERT = "INSERT Into users (name, username, password, email, phone) values(?, ?, ?, ?)";
//Prepare statement
$stmt = $conn->prepare($SELECT);
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->bind_result($email);
$stmt->store_result();
$rnum = $stmt->num_rows;
if ($rnum==0) {
$stmt->close();
$stmt = $conn->prepare($INSERT);
$stmt->bind_param("ssssii", $username, $username, $password, $email, $phone);
$stmt->execute();
echo "Your account has been registered!";
} else {
echo "This email is already linked to Preak account";
}
$stmt->close();
$conn->close();
}
} else {
echo "All fields are required";
die();
}
?>
This question already has answers here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
How can I prevent SQL injection in PHP?
(27 answers)
Reference - What does this error mean in PHP?
(38 answers)
Closed 4 years ago.
I have created this for a user registration. I am testing this to be input into the SQL database prior to going further. For some reason, this will not input into my database. Any ideas?
I appreciate your assistance.
<?php
$uid = $_POST['uid'];
$pwd = $_POST['pwd'];
$email = $_POST['email'];
if (!empty($uid) || !empty($pwd) || !empty($email)) {
$host = "localhost";
$dbUsername = "connectg";
$dbPassword = "";
$dbname = "my_connectg";
//create connection
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$SELECT = "SELECT email From users Where email = ? Limit 1";
$INSERT = "INSERT Into users (uid, pwd, email) values(?, ?, ?)";
//Prepare statement
$stmt = $conn->prepare($SELECT);
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->bind_result($email);
$stmt->store_result();
$rnum = $stmt->num_rows;
if ($rnum==0) {
$stmt->close();
$stmt = $conn->prepare($INSERT);
$stmt->bind_param("ssi", $uid, $pwd, $email);
$stmt->execute();
echo "New record inserted sucessfully";
} else {
echo "Someone already register using this email";
}
$stmt->close();
$conn->close();
}
} else {
echo "All field are required";
die();
}
?>