Re: Why is variable not in scope?

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You probably need to show more code. If you can show how/where auth_array is defined and how nav_list is called then that would help.

The best way to get help is to create a complete small sample that re-creates the problem. That might seem like a lot of work but I have done it. Sometimes I solve the problem myself when I do that.

Saturday, July 27, 2019 6:46 PM
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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