Mysql results as PHP named variables

Question

I have a table in mysql called site_settings that looks like this

I am trying to store all of my website settings in mysql and want to pull them into PHP as variables.

I want to pull all values from the variable_name column as the $variable names in PHP and have their values set to whats in the variable_type column.

$site_name = Vexed

$registration_enabled = False

here is my code:

$sql = connect($database_address, $database_username, $database_password, $database);

$query = "SELECT * FROM site_settings";
$result = $sql->query($query);
//$row = $result->fetch_all(MYSQLI_ASSOC);

while($row = $result->fetch_assoc())
{
  $$row['variable_name'] = $row["variable_type"];
}

$arr = get_defined_vars();
print_r($arr);

the last two lines i am using to see if the variable have been created but i cant get it to work. the best result i have got so far is

    [Array] => Array
        (
            [variable_name] => Vexed
        )

Can anyone tell me where i am going wrong?

Thanks in advance to anyone who can help.

I'd recommend against this, as you can clobber existing variables pretty easily. You'll have to make sure you never create a setting that's named the same as any in-use variable. You're better off reading them into a $config array or an instance of a stdClass or similar. — Alex Howansky
– Alex Howansky, Commented Mar 4, 2021 at 22:30

Bill Karwin · Accepted Answer · 2021-03-04 22:37:16Z

1

What you're trying to duplicate is PHP's extract() builtin function.

It's generally considered a bad practice, because it will make your code harder for readers to understand or debug.

answered Mar 4, 2021 at 22:37

Bill Karwin

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Jeffrey Van Alstine · Accepted Answer · 2021-03-04 22:46:13Z

0

What I think is happening is that when you call $$arr['variable_name'] it's actually doing $$arr first (which evaluates to $Array after the string conversion), and then trying to assign into assign the ['variable_name'] key into $Array.

I would expect this minor modification to work:

$sql = connect($database_address, $database_username, $database_password, $database);

$query = "SELECT * FROM site_settings";
$result = $sql->query($query);
//$row = $result->fetch_all(MYSQLI_ASSOC);

while($row = $result->fetch_assoc())
{
  $name = $row['variable_name'];
  $$name = $row["variable_type"];
}

$arr = get_defined_vars();
print_r($arr);

Edit: I'll also echo, that it's a little bit weird to dynamically create variables in this way and it will make your code hard to follow.

answered Mar 4, 2021 at 22:46

Jeffrey Van Alstine

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7 Comments

Sutherland Over a year ago

Really appreciate your help, agreed it a little weird, i am creating a site where i can check and monitor different web apps, to monitor everything in one site i need to store a lot of values, my Mysql skills are much sharper than my php, hence going the route i have, is there a better way to store the values? i'm trying to create the site as open as possible so that i can share with others, or use the base of the site on different projects.

Jeffrey Van Alstine Over a year ago

I'd probably shove everything into a look up dictionary/array like this: $settings = array(); while($row = $result->fetch_assoc()) { $settings[$row['variable_name']] = $row['variable_value']; } Then later on in the code you can access everything through $variables array. e.g. $settings['site_name'] = Vexed, $settings['registration_enabled'] = true, etc.

Sutherland Over a year ago

where would you write this data though?

Jeffrey Van Alstine Over a year ago

I'm not sure what you mean by "write" the data? Long-term persistent storage would still be in the database and all the settings in short-term memory would be accessed through the $settings dictionary.

Sutherland Over a year ago

Sorry, yeah i meant long term persistent storage, so storing in a database is fine but calling them into php the way i do is the concern, is there somewhere i could find some examples on the recommended way to store site values?

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