PHP undefined variable (possible duplicate?) [duplicate]

Question

I'm making a script for changing e-shop prices. This is my settings panel (key variables)

// --- <SETTINGS PANEL> --- //
$safe_mode = true;
$safer_mode = false;
$safe_cycle = false;
$file_loc = 'cata.txt';
$skuindex = '1';
$nameindex = '2';
$priceindex = '0';
$rowdelimiter = '_';
$celldelimiter = '-';
// --- </SETTINGS PANEL> --- //

Later on I'm using my safe_mode variable here:

function updateit($pricus,$skus,$namef)
{
 if($safer_mode)
 {
 $querytoup = "UPDATE product SET margin = '', vat_id = '1', cost = '', price = '".$pricus."' WHERE sku LIKE '".$skus."' AND name LIKE '".$namef."';";
 }
 else
 {
  $querytoup = "UPDATE product SET margin = '', vat_id = '1', cost = '', price = '".$pricus."' WHERE sku LIKE '".$skus."' AND name LIKE '".$namef."%';";
 };
 echo('Establishing connection...<br>');
 updateexec($querytoup);
};

updateexec(); then executes a bit more complex function I don't want to paste here. The problem is that I'm getting this kind of error/notice:

Notice: Undefined variable: safer_mode in /var/www/doit.php on line 54

Line 54 is exactly the if($safer_mode) line. So where did I go wrong? I know this stuff happens when you declare a variable inside a loop/function that doesn't run under some conditions so it's undeclared, but I'm declaring $safe_mode outside any functions, the settings panel is in the beginning of the file. I'm not running any includes or requires. Thanks for any replies.

Each function has its own local variable scope. Anything declared outside isn't automatically mirrored into functions. — mario
– mario, Commented Jan 31, 2015 at 14:51
PHP scope isn't like JavaScript (I only guess familiarity with JS because of the ; after closing }, which is not used in PHP) — Michael Berkowski
– Michael Berkowski, Commented Jan 31, 2015 at 14:53
You would have to pass that variable as a parameter into the updateit() function. — Michael Berkowski
– Michael Berkowski, Commented Jan 31, 2015 at 14:53

Lorenz Meyer · Accepted Answer · 2015-01-31 15:24:42Z

0

Global variables are not defined inside functions in PHP. In order to use them, you have to declare them using the global keyword.

function updateit($pricus,$skus,$namef)
{
    global $safer_mode;
    if($safer_mode)
    {

In general, using global variables in functions is bad practice. It should be avoided in the majority of cases.

edited Jan 31, 2015 at 15:24

answered Jan 31, 2015 at 14:56

Lorenz Meyer

20.1k23 gold badges83 silver badges130 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Arjan Over a year ago

But it's better to pass the variable as a value into updateit(). The global keyword should be avoided because it causes side effects that are very hard to spot (until something stops working because of a seemingly unrelated change somewhere else).

HajdaCZ Over a year ago

Again, I'm not defining it inside a function, but outside. I'm just using it in the function. Fixed it with passing the variable as a parameter, the weird thing is that it used to work on older versions of PHP.

Lorenz Meyer Over a year ago

You are right, I should have mentionned it.

Collectives™ on Stack Overflow

PHP undefined variable (possible duplicate?) [duplicate]

1 Answer 1

3 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Linked

Related