Why MySQL database not update when I update something in PHP?

I've recently setup up a MySQL server and an Apache webserver to test my Mysql Database. But there is a problem. PHP won't update the MySql server, or the MySQL server will not update.

I've even gone back and copied and pasted from W3Schools and this seems to do nothing what so ever. What am I doing wrong?

<?php
$servername = "127.0.0.1";
$username = "root";
$password = "password";
$dbname = "form_acceptance";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
  die("Connection failed: " . mysqli_connect_error());
}

$sql = "UPDATE MyGuests SET Player_name='Doe' WHERE id=2";

if (mysqli_query($conn, $sql)) {
  echo "Record updated successfully";
} else {
  echo "Error updating record: " . mysqli_error($conn);
}

mysqli_close($conn);
?>

MySql

CREATE DATABASE form_acceptance;
CREATE TABLE form_acceptance (
    PersonID int,
    Player_Name varchar(255),
    Countries varchar(255),
    Username varchar(255),
    Level_and_rank varchar(255),
    Max_BR varchar(255)
);
INSERT INTO form_acceptance (PersonID, Player_Name, Countries, Username, Level_and_rank, Max_BR)
SELECT 'SayByeBye_exe', 'SayByeBye_exe', 'US', '^GYMP^SayByeBye_exe', '12_Luitenant', '4.7';
select * FROM form_exceptance;

Nothing seems to work. PHP will not update data into MySql. Why not?

Is it maybe because I am using Linux? Or not?

Answer

Solution:

For the query to execute make sure you have at least 2 records in the table

$sql = "UPDATE MyGuests SET Player_name='Doe' WHERE id=2";

if there is no id with value 2 then the query fails so make sure you check that, I don't see anything else wrong except the typing error at last line

Answer

Solution:

There is couple error you are having with your setup. You do not have any MyGuests table so I assume you want to update the form_acceptance table. Then the form_acceptance table doesn't have any id column so either we have to add this or use the PersonID column. Here is the code snippet to fix your issues.

First

Please update your MySQL table creation like this. This will make PersonID as a primary auto incremental column.

CREATE DATABASE form_acceptance;
CREATE TABLE form_acceptance (
    PersonID int NOT NULL AUTO_INCREMENT, //Notice we added NOT NULL AUTO_INCREMENT
    Player_Name varchar(255),
    Countries varchar(255),
    Username varchar(255),
    Level_and_rank varchar(255),
    Max_BR varchar(255),
    PRIMARY KEY (`PersonID `)//define PersonId as primary key
);

Second Insert a few records

INSERT INTO form_acceptance (Player_Name, Countries, Username, Level_and_rank, Max_BR)
Values('Player_Name1', 'US', 'Username`', '12_Luitenant', '4.7'),
('Player_Name2', 'US', 'Username2', '12_Luitenant', '4.7');
select * FROM form_acceptance;//Will show you 2 records having PersonID 1 and 2

Now fix your update query

<?php
$servername = "127.0.0.1";
$username = "root";
$password = "password";
$dbname = "form_acceptance";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
  die("Connection failed: " . mysqli_connect_error());
}

$sql = "UPDATE form_acceptance SET Player_name='Doe' WHERE PersonID=2";

if (mysqli_query($conn, $sql)) {
  echo "Record updated successfully";
} else {
  echo "Error updating record: " . mysqli_error($conn);
}

mysqli_close($conn);
?>

This will successfully update the record of the second row where PersonID is 2.

Source