I have the following piece of code
copy($source, $target);
I also use
move_uploaded_file($source, $target);
To prevent code reuse, I want to pass copy and move_uploaded_file in via a variable.
If my variable is $var = "copy";, simply putting $var($source, $target);, doesn't seem to work.
Are there any special characters that must surround $var?
Thanks.
The correct syntax is $var (variable functions), so your code should work.
But please don't do that, just write the code in a straightforward and readable manner. There are legitimate use cases for this technique, but this is not one of them.
You want to look at Variable Functions which goes on to explain how to do that.
function foo() {
echo "In foo()<br />\n";
}
$bar = 'foo';
$bar(); //this calls foo()
This can also be done on both object methods and static methods.
Object Methods
class Foo
{
function MyFunction()
{
//code here
}
}
$foo = new Foo();
$funcName = "MyFunction";
$foo->$funcName();
Static Methods
class Bar
{
static function MyStaticFunction()
{
//code here
}
}
$funcName = "MyStaticFunction";
Bar::$funcName();
While maybe not the case in your situation, when dealing with functions dynamically like this, it is important to check whether the function actually exists and/or is callable.
Alternatively to using Variable Functions, you can use call_user_func which will call the function based on the string name and with provided parameters.
You can use the PHP function call_user_func().
You can use call_user_func to do this.
$result = call_user_func($functionToCall, $source, $target)
Documentation: PHP: call_user_func
