Search a database to update the results - php

Question

I am taking a users input and storing it in a database, however I want to be able to update the records if a user adds more information. So I want to search the database find the server with the same name and update the the last downtime and the number of downtimes.
$connect = mysqli_connect("localhost", "Username", "Password","Test_downtime");
if (!$connect)
{
die("Connection failed: " . mysqli_connect_error());
}else
{
echo "Connected successfully\n";
}
$servername = $_GET["server_name"];
$downtime = $_GET["downtime"];
$time_now = time();
$result = mysqli_query($connect, "SELECT COUNT(*) FROM `Test_downtime`.`Downtime` WHERE `Server_Name` = '$servername'");
$row = mysqli_fetch_array($result);
// If no downtime have been reported before
if ($row[0] == 0){
$sql = mysqli_query($connect, "INSERT INTO `Test_downtime`.`Downtime` (ID, Server_name, First_downtime, Last_downtime, Num_of_downtime,Total_downtime) VALUES (NULL, '$servername', '$time_now','$time_now',1,'$downtime'); ");
if ($sql==true) {
echo $servername . " has has its first downtime recorded\n";
}
}
//If users is already in the database
else{
$numdowntime = ($row["Num_of_downtime"] + 1);
$id = ($row["ID"]);
$sqlupdate = "UPDATE `Test_downtime`.`Downtime` SET `Num_of_downtime` = $numdowntime, `Last_downtime` = now() WHERE `Server_Name` = '$servername'";
if ($sqlupdate == TRUE) {
echo "Oh No! " . $servername . " has had ". $numdowntime ." of downtimes" ;
}
}
?>
The program works fine if the server is not already in the database, the problems arise if the server is already in the database. I get the message saying it has been updated yet nothing happens to the database. How do i make it so it search and updates the records for the searched item.

Answer 1

So nothing append since you do not execute the sql statement ^^
Take a look here :
$sqlupdate = "UPDATE `Test_downtime`.`Downtime` SET `Num_of_downtime` = $numdowntime, `Last_downtime` = now() WHERE `Server_Name` = '$servername'";
You need to use :
$sql = mysqli_query($connect, $sqlupdate);
Just after it in order to execute it.
Or at least change it to
$sqlupdate = mysqli_query($connect, "UPDATE `Test_downtime`.`Downtime` SET `Num_of_downtime` = $numdowntime, `Last_downtime` = now() WHERE `Server_Name` = '$servername'" );
Btw there is other problem but here is the main one [check out the other answer in order to found another one ]

Answer 2

you are fetching the result as indexed array
mysqli_fetch_array($result);
and here you are accessing results as associative array
$numdowntime = ($row["Num_of_downtime"] + 1);
change your query to
mysqli_fetch_assoc($result);
use
mysqli_num_rows($result);
to checking if you have any data
change
if ($row[0] == 0){}
to
if(mysqli_num_rows($result) ==0){}

Answer 3

A good approach for increasing a count in a column is using SQL to increase that.
$sqlupdate = mysqli_query($connect, "UPDATE `Test_downtime`.`Downtime` SET `Num_of_downtime` = (`Num_of_downtime` + 1), `Last_downtime` = now() WHERE `Server_Name` = '$servername'" );
This way you can skip your $numdowntime calculation, and the result is more accurate.
In your current setup, two users may fire the event at the same time, they both retrieve the same number from the database (ie. 9), both increasing it with one (ie. 10), and writing the same number in the database.
Making your count one short of the actual count.
SQL takes care of this for you by locking rows, and you are left with a more accurate result using less logic :)

Answer 4

You miss the mysqli_query() function, which actually queries the database.
$sqlupdate = mysqli_query("
UPDATE `Test_downtime`.`Downtime`
SET `Num_of_downtime` = $numdowntime, `Last_downtime` = now()
WHERE `Server_Name` = '$servername'"
);