I have the following scenario (simplified):
function changeFruit($fruit) {
changeAgain($fruit);
}
function changeAgain($fruit) {
$fruit = "Orange";
}
MAIN:
$fruit = "Apple";
changeFruit($fruit);
echo $fruit // Will show up as "Apple", How do I get it to show up as "Orange"??
EDIT: FORGOT TO ADD. THE SCENARIO CANNOT USE RETURN STATEMENTS - JUST &$variable
Thanks! Matt Mueller
When you pass something that is not an object to a function in PHP, php makes a copy of that to use within the function.
To make it not use a copy, you need to tell PHP you are passing a reference.
This is done with the & operator
function changeFruit(&$fruit) {
changeAgain($fruit);
}
function changeAgain(&$fruit) {
$fruit = "Orange";
}
$fruit = "Apple";
changeFruit($fruit);
echo $fruit;
It would be more sensible, and better practice, to use return values of the functions (as this makes things easier to read)
function changeFruit($fruit) {
return changeAgain($fruit);
}
function changeAgain($fruit) {
// do something more interesting with$fruit here
$fruit = "Orange";
return $fruit;
}
$fruit = "Apple";
$fruit = changeFruit($fruit);
echo $fruit
function changeFruit($fruit) {
return changeAgain($fruit);
}
function changeAgain($fruit) {
return $fruit = "Orange";
}
MAIN:
$fruit = "Apple";
$fruit = changeFruit($fruit);
echo $fruit;
Hope that helps!
Note: the return from the changeAgain function and overwriting $fruit = changeFruit($fruit);
You are not returning the values from your functions. Try this:
function changeFruit($fruit) {
return changeAgain($fruit);
}
function changeAgain($fruit) {
$fruit = "Orange";
return $fruit;
}
MAIN:
$fruit = "Apple";
$fruit = changeFruit($fruit);
