PHP Form update without logout - php

Question

As i am newbie to PHP kindly pardon me if i looks silly ,
I created a form in php , while i do the update part of the form the update reflects in db whereas in the form it still shows the same old value . i tried refresh and force refresh but nothing changes .
Whereas if i logout and login again , the form shows the updated value .
I tried using die(); after mysql_close($link); but it logs out the session and needs to re-login .
Kindly help me on viewing the changes while i am still inside the login .
My code is as follows :
<?php
if(isset($_POST['update'])) {
$name_a = $_POST['name'];
$email_a = $_POST['email'];
$pass_a = $_POST['password'];
$sql = "UPDATE admin SET a_name = '$name_a', a_email = '$email_a', password = '$pass_a' where aid='$update_id' ";
$retval = mysql_query($sql,$link);
if(! $retval ) {
die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully\n";
mysql_close($link);
}else {
?>
<!-- Widget: user widget style 1 -->
<div class="box box-widget widget-user-2">
<!-- Add the bg color to the header using any of the bg-* classes -->
<div class="widget-user-header bg-yellow">
<div class="widget-user-image">
<?php echo '<img src="' . $img . '" class="img-circle" alt="User Image">'; ?>
</div>
<!-- /.widget-user-image -->
<h3 class="widget-user-username"><?php echo "$name"; ?></h3>
<h5 class="widget-user-desc"><?php echo "$role"; ?></h5>
</div>
<div class="box-footer no-padding">
<form role="form" method = "post" action = "<?php $_PHP_SELF ?>">
<div class="box-body">
<div class="form-group">
<label for="exampleInputName1">Name</label>
<input type="text" class="form-control" id="exampleInputName1" name="name" value="<?php echo "$name"; ?>">
</div>
<div class="form-group">
<label for="exampleInputEmail1">Email address</label>
<input type="email" class="form-control" id="exampleInputEmail1" name="email" value="<?php echo "$email"; ?>">
</div>
<div class="form-group">
<label for="exampleInputPassword1">Password</label>
<input type="password" class="form-control" id="exampleInputPassword1" name="password" value="<?php echo "$password"; ?>">
</div>
</div>
<!-- /.box-body -->
<div class="box-footer">
<button type="submit" name="update" id="update" class="btn btn-primary">Submit</button>
</div>
</form>
</div>
</div>
<!-- /.widget-user -->
<?php
}
?>

Answer 1

SOLUTION
1) use the updated value like $name_a instead of $name because $name_a contain updated value and $name contain old value
2) reload page after update and get new value from database on page load and store that value in $name , $email etc variable (if new data update successfully in database then only you get new value )
3) if You store your data in session or cookie then update session and cookie value also when you update in database

Answer 2

Try this:
<?php
$name = '';
$email = '';
$password = '';
$update_id = '';
//$img = '';
//$role = '';
//$link = null;
if(
isset($_POST['update']) &&
isset($_POST['id']) &&
isset($_POST['name']) &&
isset($_POST['email']) &&
isset($_POST['password'])
) {
$update_id = mysql_real_escape_string($_POST['id']);
$name = mysql_real_escape_string($_POST['name']);
$email = mysql_real_escape_string($_POST['email']);
$password = mysql_real_escape_string($_POST['password']);
$sql = 'UPDATE admin SET a_name = \'' . $name . '\', a_email = \'' . $email . '\', password = \'' . $password . '\' WHERE aid = \'' . $update_id . '\'';
$result = #mysql_query($sql, $link);
if(!$result)
die('Could not update data: ' . mysql_error($link));
echo 'Updated data successfully', "\n";
}
elseif(isset($_GET['id'][0])) {
$update_id = mysql_real_escape_string($_GET['id']);
$sql = 'SELECT a_name,a_email,a_password FROM admin WHERE aid = \'' . $update_id . '\'';
$result = #mysql_query($sql, $link);
if($result) {
$result = mysql_fetch_row($result);
$name = $result[0];
$email = $result[1];
$password = $result[2];
}
else {
echo 'Could not find the id.' . "\n";
$update_id = '';
}
}
unset($result);
if(isset($update_id[0])) {
mysql_close($link);
?>
<!-- Widget: user widget style 1 -->
<div class="box box-widget widget-user-2">
<!-- Add the bg color to the header using any of the bg-* classes -->
<div class="widget-user-header bg-yellow">
<div class="widget-user-image">
<img src="<?php echo htmlspecialchars($img); ?>" class="img-circle" alt="User Image">
</div>
<!-- /.widget-user-image -->
<h3 class="widget-user-username"><?php echo htmlspecialchars($name); ?></h3>
<h5 class="widget-user-desc"><?php echo htmlspecialchars($role); ?></h5>
</div>
<div class="box-footer no-padding">
<form action="<?php $_SERVER['PHP_SELF']; ?>" method="POST">
<input type="hidden" name="id" value="<?php echo htmlspecialchars($update_id); ?>">
<div class="box-body">
<div class="form-group">
<label for="exampleInputName1">Name</label>
<input type="text" class="form-control" id="exampleInputName1" name="name" value="<?php echo htmlspecialchars($name); ?>">
</div>
<div class="form-group">
<label for="exampleInputEmail1">Email address</label>
<input type="email" class="form-control" id="exampleInputEmail1" name="email" value="<?php echo htmlspecialchars($email); ?>">
</div>
<div class="form-group">
<label for="exampleInputPassword1">Password</label>
<input type="password" class="form-control" id="exampleInputPassword1" name="password" value="<?php echo htmlspecialchars($password); ?>">
</div>
</div>
<!-- /.box-body -->
<div class="box-footer">
<button type="submit" name="update" id="update" class="btn btn-primary">Submit</button>
</div>
</form>
</div>
</div>
<!-- /.widget-user -->
<?php }
else {
$sql = 'SELECT aid,a_name FROM admin';
$result = #mysql_query($sql, $link);
if($result) {
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo '' . $row['a_name'] . '<br />' . "\n";
}
}
mysql_close($link);
}
?>

Answer 3

As #DivyeshSavaliya mentioned in the comment the issue is ,
I didn't Used Select query after update . Once done that the issue solved
The new working code is
<?php
if(isset($_POST['update'])) {
$name_a = $_POST['name'];
$email_a = $_POST['email'];
$pass_a = $_POST['password'];
$sql = "UPDATE admin SET a_name = '$name_a', a_email = '$email_a', password = '$pass_a' where aid='$update_id' ";
$retval = mysql_query($sql,$link);
if(! $retval ) {
die('Could not update data: ' . mysql_error());
}
}
$result = mysql_query("SELECT * FROM admin where aid='$update_id' ",$link);
while($row = mysql_fetch_array($result)){
$name = $row['a_name'];
$email = $row['a_email'];
$password = $row['password'];
}
mysql_close($link);
?>
Thanks to #DivyeshSavaliya