PHP use variable defined outside of function

Asked Apr 17 '21 at 10:10

Active Apr 17 '21 at 10:14

Viewed 33 times

I have seen a lot of questions already asked about the use of variables in functions or outside functions, but somehow I can't make it work in my case.

I have a PHP-file that declares two variables, and then starts defining a function. I want to use the 2 variables and a couple of arguments that should be passed along to the function.

I saw the "use" command that is available on anonymous functions, but I don't think that is what I want as the function does multiple things, and not just return a value. How can I use the variables in my funtion without resorting to global?

For clarification, my code looks like this:

$var1 = 'variable 1';
$var2 = 'variable 2';

function MyFunction($arg1,$arg2,$arg3){
    // use $var1 and $var2 along with the 3 arguments supplied
}

asked Apr 17 '21 at 10:10

Discoverer

_How can I use the variables in my funtion without resorting to global_? Why not? – nice_dev Apr 17 '21 at 10:13
Pass them as parameters. – Barmar Apr 17 '21 at 10:14
Could you give an example @Barmar – Discoverer Apr 17 '21 at 10:16
`function MyFunction($arg1,$arg2,$arg3, $var1, $var2)` and then add them as additional arguments when calling the function. – Barmar Apr 17 '21 at 10:17
1

If you don't want to access `$var` and `$var2` using `gloabl` keyword you have to pass to the function all elements which are need by the function to perform it task, which is reffered to as **dependency injection** – Yves Kipondo Apr 17 '21 at 10:27

PHP use variable defined outside of function

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