Hi. I wants to display logged in user's profile in a page. I wrote all queries for that. The problem is I am unable to fetch data from table "users" from database "videos". Please tell me where I made mistake and what I have to do to overcome those.
Her is my code:
user.php
PHP Code:
<div class="content">
<?php
$sql = "SELECT * FROM users WHERE username='$username'";
$result = mysqli_query($database,$sql) or die(mysqli_error($database));
$rws = mysqli_fetch_array($result);
?>
<?php include 'edit-profile-form.php' ?>
</div>
edit-profile-form.php
In this I have added form like:
PHP Code:
<form action="update-profile.php" method="POST">
<label>Username</label>
<input type="text" name="Username" class="form-control" value="<?php echo $rws['Username']; ?>">
</form>
update-profile.php
PHP Code:
<?php
ini_set("display_errors",1);
session_start();
$id=(isset($_SESSION['Id']) ? $_SESSION['Id'] : '');
if(isset($_POST)){
require '../_database/database.php';
$username = $_POST['Username'];
$email = $_POST['Email'];
$employee = $_POST['EmployeeID'];
$designation = $_POST['Designation'];
$password = $_POST['Password'];
$query = ("UPDATE users SET (Username, Email, EmployeeID, Designation, Password) VALUES ('$username', '$email', '$employee', '$designation', '$password') WHERE Id='$id'");
if(!$query) $con->errno;
if ( !$stmt = $con->prepare("SELECT * FROM users WHERE Id='$id'") )
echo "Prepare Error: ($con->errno) $con->error";
if ( !$stmt->bind_param("sssss", $_POST['Username'], $_POST['Email'], $_POST['EmployeeID'], $_POST['Designation'], $_POST['Password']) )
echo "Binding Parameter Error: ($con->errno) $con->error";
if ( !$stmt->execute() )
echo "Execute Error: ($stmt->errno) $stmt->error";
}
?>
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