This question already has an answer here:
My code:
$name=$_POST["name"];
This is the error
Notice: Undefined index: name in /home/u615903880/public_html/reg3.php on line 4
code :
<?php
$con = new mysqli("xxxxx", "xxxx", "xxxx");
$name = $_POST["name"];
$username = $_POST["username"];
$emailaddress = $_POST["emailaddress"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO users (name, username, emailaddress, password)VALUES (?,?,?,?)");
mysqli_stmt_bind_param($statement, "ssss", $name, $username, $emailaddress, $password);
mysqli_stmt_execute($statement);
$response = array();
$response["success"] = true;
echo json_encode($response);
mysqli_stmt_close($statement);
mysqli_close($con);
?>
Error lines 4,5,6,9,10,14
</div>
Try
$name = isset($_POST['name']) ? $_POST['name'] : 'nothing provided';
You can put anything instead of 'nothing provided' like NULL or something
Its Notice not an error:
you can use
error_reporting(1);
on top of page this will hide warnings and notices.
or you can use code this way
if(isset($_POST["name"])){
$name=$_POST["name"];
}
You didn't send a POST variable called name. If it was from an HTML form, make sure the input is not disabled or it won't post. Also, make sure your form field has a name and not just an id, as it is the name that is used in a post request:
<input type="text" id="notused" name="used" />
In which case $_POST['used'] would work upon submission.
I suspect the key 'name' is not defined in the $_POST array. to check for missing key you could use:
$name = $_POST["name"]??null; // PHP /7+
or
$name = isset($_POST["name"])?$_POST["name"]:null; // older