Start up with Ubuntu server!

Hi! Here's a short article of how to get going with Ubuntu server (9.10+, maybe works on older).

1. Download Ubuntu server http://www.ubuntu.com/getubuntu/download-server

2. Install, you may burn it on a CD or create an startup-USB with http://unetbootin.sourceforge.net/

3. Install! It's easy, just boot and chose install.

4. Update everything! Write the following in terminal:
sudo apt-get update
sudo apt-get upgrade

Now you're up and running! However it's not that fun to have an empty server! It's easy to install stuff however. For example, install apache, php and mysql: write "tasksel install lamp-server" in terminal. And phpmyadmin, write "sudo apt-get install phpmyadmin". Don't forget ssh-server, so you can login remotely: "sudo apt-get install openssh-server".

Why use Ubuntu (or other linux based server)? Well it's free, it's stable, it's secure and it ain't that hard. All info you need can be found on the internet, and the quality of diffrent tutorials and guides or just being better and better! Give it a try!

PHP 5.3 problems with references

Hi! I have recently updated to Ubuntu Lucid. With this I also got PHP 5.3.2. I thought it would be no problems updating PHP, but now after some hours for debugging I know I was wrong.

First I got a mysql SQLException (My own exception telling my about syntax error). And an warning:
Warning: Parameter 2 to mysqli_stmt::bind_param() expected to be a reference, value given in /var/www/bla/lib/classes/Database/MysqliStmt.class.php on line 68

This was confusing as SQL-syntax shouldn't change. It hadn't. I found that the SQLException was due to another query that returned nothing, and by this causing the other query to fail. The other problem with the warning was trickier. I have built an Database abstraction layer, and I'm using "call_user_func_array(array(&$stmt, 'bind_param'), $params);" in my mysqli-driver, on line 68.

This error was new, I have used my database abstraction layer several times on diffrent servers without any errors. I first tried to solve the problem by changing how $params is created.

From:

foreach($values as $value) {
$params[] = $value;
}

to:

foreach($values as $value) {
$params[] = &$value;
}

Problem solved, if my only aim was to get rid of warnings at least. I didn't get any results from my querys earlier, but now I still didn't get results but no errors or warnings either. When I printed out $params I could see that all values where the same, the last put into the array $params. Then after a while I realized what I was doing. "foreach" copies a value from $values to $value. If PHP is working with references (which 5.3.2 seems to do more), then I'm just placing references into $params to $value. And $value changes on every loop. This explains why all values in $params had the same value, the last entered into $params. The solution was to use this code:

for($i = 0; $i < $max; $i++) {
$params[$i+1] = &$values[$i];
}

Here I go thorugh $values. For each value in $values I put an reference into $params. The reference is pointing into the right position in the array $values.