PHP/Mysqli check before update - php

Question

Hey I am completely new to PHP/MySqli, I would like to check before update if Scanstatus field of given ID is already "Scanned". if its already Scanned display a message as "Already Scanned" else Update.
Below code only update and doesn't check if already exists.
<?php
$id = $_POST['id'];
$connection = mysqli_connect("localhost", "username", "passwd","dbname");
if(mysqli_connect_errno())
{
echo "failed to connect " . mysqli_connect_error();
}
if(isset($_POST['Submit']))
{
$query = "UPDATE `sales` SET `ScanStatus` = 'Scanned' WHERE `id` = $id";
$result = mysqli_query($connection,$query);
if (!$result) {
die('Error' . mysqli_error($connection));
}
else
{
echo "Successfully updated";
}
}
?>

Answer 1

Try following php code
$id = $_POST['id'];
$connection = mysqli_connect("localhost", "username", "passwd","dbname");
if(mysqli_connect_errno())
{
echo "failed to connect " . mysqli_connect_error();
}
else{
if(isset($_POST['Submit']))
{
$sql = "UPDATE sales SET ScanStatus = 'Scanned' WHERE id = '$id'";
$result = $connection->query($sql);
if ($result->num_rows > 0) {
echo "Successfully updated";
} else {
die('Error' . mysqli_error($connection));
}
}
}
When click SUBMIT button "isset($_POST['Submit'])" will be true and direct into if statement. then run sql command and is if there is any result that sql query affected go into next if statement after true the condition "$result->num_rows > 0" then echo "Successfully updated".

Answer 2

As I understand, you want to update only if the ScanStatus field is not Scanned
So you can modify you existing query like this without the need to fetch the record:
$query = "UPDATE `sales` SET `ScanStatus` = 'Scanned' WHERE `id` = $id AND `ScanStatus` != 'Scanned'";
Just change the query to above and use:
if(mysqli_affected_rows($result) > 0 ){
Here is a full code:
<?php
$connection = mysqli_connect("localhost", "username", "passwd","dbname");
if(mysqli_connect_errno()) {
echo "failed to connect " . mysqli_connect_error();
}
if(isset($_POST['Submit'])) {
$id = $_POST['id'];
$query = "UPDATE `sales` SET `ScanStatus` = 'Scanned' WHERE `id` = $id AND `ScanStatus` != 'Scanned'";
$result = mysqli_query($connection, $query);
if(mysqli_affected_rows($connection) > 0 ){
echo "Successfully updated";
}
else {
echo 'Already Scanned';
}
}

Answer 3

You can use this code:
Use if(mysqli_affected_rows($mysqli) > 0 ) or no comparison at all.
Replace your code with this:
<?php
$id = $_POST['id'];
$connection = mysqli_connect("localhost", "username", "passwd","dbname");
if(mysqli_connect_errno())
{
echo "failed to connect " . mysqli_connect_error();
}
if(isset($_POST['Submit']))
{
$my_query = mysqli_query($connection, "SELECT * FROM `sales` WHERE `id` = ". $id . " AND `ScanStatus` = 'Scanned'");
if(mysqli_num_rows($my_query) > 0){
echo "Already ScanStatus is Scanned";
}
else{
$query = "UPDATE `sales` SET `ScanStatus` = 'Scanned' WHERE `id` = ".$id;
//echo $query;die;
$result = mysqli_query($connection, $query);
if(mysqli_affected_rows($connection) > 0 ){
echo "Successfully updated";
/* get new updated data */
$new_query = mysqli_query($connection, "SELECT * FROM `sales` WHERE `id` = '$id' LIMIT 1");
$new_info = mysqli_fetch_array($new_query);
echo "<pre>"; print_r($new_info);
}
else
{
echo "Not updated";
}
}
}
?>