Re: Why is variable not in scope?

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Bingo! That worked properly.

Does lead to a suggestion: mention that in the Function Arguments section of 
the documentation!

Also, for my curiosity (only) why would this not work when I passed 
$auth_array or &$auth_array as an argument to the call? It should have 
been in scope in the main programme so the value should have been 
available, and certainly the address was for a pass-by-reference.

Thanks to all the people who replied to my question, especially for how quickly
you did so. Made my life MUCH easier!

Regards,

John
==============================

On Sat, 2019-07-27 at 21:47 -0600, LinuxManMikeC wrote:
Functions don't automatically have access to global variables. Use the 'global' keyword to declare what global variables you're going to use in your function.

function nav_list() {
    global $auth_array;
    // Code...
}

https://www.php.net/manual/en/language.variables.scope.php

On Sat, Jul 27, 2019, 19:47 John <john.iliffe@xxxxxxxxx> wrote:
There is probably another way to do this but I have spent a good few hours
trying to resolve it and I think there is something wrong with the way I
understand the scope of variables in PHP, so an answer would be appreciated.

I have a PHP (7.1.3) programme that opens a database during initialization,
gathers an associative array of variables (pg_fetch_array) and then closes the
database.  The array name is $auth_array.  During actual display of the page the
values of the elements of the array are used to control what is (not) displayed
and lookup of the values is done using a separate function.

At this point I can print_r() the array and it contains what I expect.

In the next line, I call a user-defined function nav_list() where this array is
used in the form $auth_array['column name'].  At this point I get a PHP error
"Got error 'PHP message: PHP Notice:  Undefined variable: auth_array in
/httpd/myprogramme/yrarcex.php on line 74\nPHP message: PHP Notice:  Undefined
variable: auth_array .... .  At this point I cannot print_r the array, the same
undefined variable message appears.

I expected that auth_array would be in global scope.

so I tried calling nav_list() with no arguments, then with the name of the array
as the only argument, and with the address of the array (&$auth_array) and also
with the explicit cast nav_list(array $auth_array) and I always get the same
error message.  (not a data type error as suggested in the docs).

So far as I can see I am following the online documentation at

https://www.php.net/manual/en/functions.arguments.php

exactly.

The definition of nav_list() is;

function nav_list()
{
 if ($auth_array['u_....'] == 't')   <---- this is line 74 as shown in the error
   {
      // display something
   }
}



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