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 } }
