In PHP, how do you use an external $var for use within a function in a class? For example, say $some_external_var sets to true and you have something like
class myclass { bla .... bla .... function myfunction() { if (isset($some_external_var)) do something ... } } $some_external_var =true; $obj = new myclass(); $obj->myfunction();
Thanks
You'll need to use the global
keyword inside your function, to make your external variable visible to that function.
For instance :
$my_var_2 = 'glop';
function test_2()
{
global $my_var_2;
var_dump($my_var_2); // string 'glop' (length=4)
}
test_2();
You could also use the $GLOBALS
array, which is always visible, even inside functions.
But it is generally not considered a good practice to use global variables: your classes should not depend on some kind of external stuff that might or might not be there !
A better way would be to pass the variables you need as parameters, either to the methods themselves, or to the constructor of the class...
Global $some_external_var;
function myfunction() { Global $some_external_var; if (!empty($some_external_var)) do something ... } }
But because Global automatically sets it, check if it isn't empty.
that's bad software design. In order for a class to function, it needs to be provided with data. So, pass that external var into your class, otherwise you're creating unnecessary dependencies.
为什么不只在__construct()期间传递此变量,并以该变量的真值为条件使对象在构造期间执行的操作呢?
Use Setters and Getters or maybe a centralized config like:
function config()
{
static $data;
if(!isset($data))
{
$data = new stdClass();
}
return $data;
}
class myClass
{
public function myFunction()
{
echo "config()->myConfigVar: " . config()->myConfigVar;
}
}
and the use it:
config()->myConfigVar = "Hello world";
$myClass = new myClass();
$myClass->myFunction();
http://www.evanbot.com/article/universally-accessible-data-php/24
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