How can I solve this error? I have to try anything, but the result is the same error. Please help me to solve this problem error
[My error][1]
And this is my Code in update.php:
<?php
include_once 'config.php';
$employee_id=$_GET['employee_id'];
$name=$_POST['name'];
$date_of_birth=$_POST['date_of_birth'];
$gender=$_POST['gender'];
$marital_status=$_POST['marital_status'];
$nationality=$_POST['nationality'];
$present_address=$_POST['present_address'];
$city=$_POST['city'];
$country=$_POST['country'];
$phone=$_POST['phone'];
$email=$_POST['email'];
$nip=$_POST['nip'];
$status=$_POST['status'];
$designation=$_POST['designation'];
$joining_date=$_POST['joining_date'];
$leaving_date=$_POST['leaving_date'];
$picture = basename($_FILES['picture']['name']);
if (!empty($_FILES['picture'])) {
$path = "admin/gambar/";
$path = $path . basename($_FILES['picture']['name']);
if (move_uploaded_file($_FILES['picture']['tmp_name'], $path)) {
echo "The file " . basename($_FILES['picture']['name']) .
" has been uploaded";
} else {
echo "There was an error uploading the file, please try again";
}
}
$query = "UPDATE employee_list set name='$name', date_of_birth='$date_of_birth', gender='$gender', marital_status='$marital_status', nationality='$nationality', present_address='$present_address', city='$city', country='$country', phone='$phone', email='$email', nip=$nip, status='$status', designation='$designation', joining_date='$joining_date', leaving_date='$leaving_date', picture='$picture' where employee_id=$employee_id";
?>
Thank you :)
You need to use isset()
to avoid these errors. something like given below.
<?php
include_once 'config.php';
if (isset($_POST['employee_id'])) {
$employee_id=$_POST['employee_id'];
$name=$_POST['name'];
$date_of_birth=$_POST['date_of_birth'];
$gender=$_POST['gender'];
$marital_status=$_POST['marital_status'];
$nationality=$_POST['nationality'];
$present_address=$_POST['present_address'];
$city=$_POST['city'];
$country=$_POST['country'];
$phone=$_POST['phone'];
$email=$_POST['email'];
$nip=$_POST['nip'];
$status=$_POST['status'];
$designation=$_POST['designation'];
$joining_date=$_POST['joining_date'];
$leaving_date=$_POST['leaving_date'];
$picture = basename($_FILES['picture']['name']);
if (!empty($_FILES['picture'])) {
$path = "admin/gambar/";
$path = $path . basename($_FILES['picture']['name']);
if (move_uploaded_file($_FILES['picture']['tmp_name'], $path)) {
echo "The file " . basename($_FILES['picture']['name']) .
" has been uploaded";
} else {
echo "There was an error uploading the file, please try again";
}
}
$query = "UPDATE employee_list set name='$name', date_of_birth='$date_of_birth', gender='$gender', marital_status='$marital_status', nationality='$nationality', present_address='$present_address', city='$city', country='$country', phone='$phone', email='$email', nip=$nip, status='$status', designation='$designation', joining_date='$joining_date', leaving_date='$leaving_date', picture='$picture' where employee_id=$employee_id";
}
?>
Note:Use prepared query to avoid sql injection attack.
put all your code from line #4 to end of the file in following if statement
if (!empty($_GET['employee_id'])) {
// your code
}
also, your code is not secure for SQL injection, XSS, CSRF attacks. You can use http://csrf.htmlpurifier.org/ library to protect your code against CSRF attacks.
Use PDO (http://php.net/manual/en/book.pdo.php) to protect SQL injection.