On 29.05.2018 at 02:51, John wrote: > Thanks to all of you who suggested this link. Unfortunately it doesn't actually > work. If I set the variable as: > > global $bad_line = "This line ......"; > > I get error: > > [Mon May 28 20:28:48.491851 2018] [proxy_fcgi:error] [pid 7502:tid > 139838662719232] [client 192.168.1.118:53916] AH01071: Got error 'PHP message: > PHP Parse error: syntax error, unexpected '=', expecting ',' or ';' in > /httpd/iliffe/yrarc/test.php on line 3\n' This is expected, since the global statement does not support initializers. > If I code it this way: > > global $bad_line; > $bad_line = "This line ......"; > > then I get: > > [Mon May 28 20:31:25.553000 2018] [proxy_fcgi:error] [pid 7502:tid > 139838662719232] [client 192.168.1.118:53916] AH01071: Got error 'PHP message: > PHP Notice: Undefined variable bad_line in /httpd/iliffe/yrarc/test.php on line > 9\n' > > Looking at the reference, this is "inside out"; that is, the variable is already > defined in the global scope, that is it is in the scope of <?php ....?> and I > want it to be available inside the function. The reference is for variables > defined within a function to make them available outside the function scope. The global statement is supposed to be used inside a function body, and it creates a local variable which is actually a reference to the global variable with the same name. It does not matter if the global variable is already set or not. -- Christoph M. Becker -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
