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Re: [PHP] How Do I make Global Scope Variables Available to Functions





On May 27, 2018, at 5:46 PM, John wrote:

I am writing a PHP script that has a number of variables that (I think) are in Global scope; that is, they are defined inline before the first command of the script. Note that these are not necessarily constants as shown in the example I
included; most of them can be changed by the script.

I have functions defined following these variables and before the commands in the script appear, but when they are called, the functions report the variables as undefined. This means I have to pass all the Globals to each function as a
call argument.

A trivial script illustrating the problem is below.

This seems odd but I don't see what I am doing wrong. Any suggestions?

PHP 5.6.30 called through php-fpm on Apache.

Thanks in advance.

John
------------------------------------
test.php - working version

This variant passes a variable in global scope explicitly as a parameter. It
displays correctly on the browser screen.


<?php

$bad_line = "This line should display on screen.";

// sample function that works
function sample1($bad_line)
{
return "\n" . $bad_line . "\n";
}

This function could also be written
function sample1($a)
  {
 return "\n" . $a . "\n";
}
echo sample1($bad_line);

But it looks like you've answered your own question.
If you wanted to have the function, or another function
alter $bad_line, it would need to be passed by reference

echo sample1(&$bad_line) ....  as I understand it.

// this call does work
echo sample1($bad_line);

exit;
?>



test.php - doesn't work

This variant does not work; it throws error:

[Sun May 27 20:17:46.779348 2018] [proxy_fcgi:error] [pid 893:tid
139838497978112] [client 192.168.1.104:37732] AH01071: Got error 'PHP message: PHP Notice: Undefined variable: bad_line in /httpd/iliffe/yrarc/ test.php on
line 8\n'

<?php

$bad_line = "This line should display on screen.";

// sample function that doesn't work
function sample1()
{
return "\n" . $bad_line . "\n";
}

//  this call doesn't work
echo sample1();

exit;

?>


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