This is the error i received when executing my update button
Fatal error: Uncaught mysqli_sql_exception: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ” at line 1 in D:xampphtdocscrudfunctions.php:34 Stack trace: #0 D:xampphtdocscrudfunctions.php(34): mysqli_query(Object(mysqli), ‘UPDATE users SE…’) #1 D:xampphtdocscrudindex.php(5): editUser(Object(mysqli)) #2 {main} thrown in D:xampphtdocscrudfunctions.php on line 34
this is the code which contains the line 34
<?php
function addUser($connection) {
if(isset($_POST['add'])) {
$fname = $_POST['firstname'];
$lname = $_POST['lastname'];
$address = $_POST['address'];
$password = $_POST['password'];
$gender = $_POST['gender'];
$sql = "INSERT INTO users(firstname,lastname,address,password,gender) VALUES ('$fname', '$lname', '$address','$password','$gender')";
$result = mysqli_query($connection,$sql);
if(!$result) {
echo "ERROR";
} else {
echo "User added successfully";
}
}
}
function editUser($connection){
if(isset($_POST['edit'])) {
$fname = $_POST['firstname'];
$lname = $_POST['lastname'];
$address = $_POST['address'];
$gender = $_POST['gender'];
$id = $_POST['id'];
$sql = "UPDATE users SET firstname = '$fname', lastname = '$lname', address = '$address', gender = '$gender' WHERE id = $id";
LINE 34 - $result = mysqli_query($connection,$sql);
if(!$result) {
echo "ERROR";
} else {
echo "Record updated successfully";
}
}
I tried changing this part of the code
$sql = “UPDATE users SET firstname = ‘$fname’, lastname = ‘$lname’, address = ‘$address’, gender = ‘$gender’ WHERE id = $id”;
I changed the WHERE id = 7″;
and it successfully edited my code but when i type it back to WHERE id = $id”;
it errors again
any suggestions please to fix this
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