Is there a way to achieve the following in PHP, or is it simply not allowed? (See commented line below)
function outside() {
$variable = 'some value';
inside();
}
function inside() {
// is it possible to access $variable here without passing it as an argument?
}
note that using the global keyword is not advisable, as you have no control (you never know where else in your app the variable is used and altered). but if you are using classes, it'll make things a lot easier!
class myClass {
var $myVar = 'some value';
function inside() {
$this->myVar = 'anothervalue';
$this->outside(); // echoes 'anothervalue'
}
function outside() {
echo $this->myVar; // anothervalue
}
}
Its not possible. If $variable is a global variable you could have access it by global keyword. But this is in a function. So you can not access it.
It can be achieved by setting a global variable by$GLOBALS array though. But again, you are utilizing the global context.
function outside() {
$GLOBALS['variable'] = 'some value';
inside();
}
function inside() {
global $variable;
echo $variable;
}
No, you cannot access the local variable of a function from another function, without passing it as an argument.
You can use global variables for this, but then the variable wouldn't remain local.
It's not possible. You can do it by using global. if you just only do not want to define the parameters but could give it inside the function you can use:
function outside() {
$variable = 'some value';
inside(1,2,3);
}
function inside() {
$arg_list = func_get_args();
for ($i = 0; $i < $numargs; $i++) {
echo "Argument $i is: " . $arg_list[$i] . "<br />\n";
}
}
for that see the php manual funct_get_args()
You cannot access the local variable in function. Variable have to set as global
function outside() {
global $variable;
$variable = 'some value';
inside();
}
function inside() {
global $variable;
echo $variable;
}
This works
