i got some trouble to understand scope in OOP. What i want is that $foo->test_item() prints "teststring"...Now it just fails with:
Warning: Missing argument 1 for testing::test_item()
Thanks a lot!
<?php
class testing {
public $vari = "teststring";
function test_item($vari){ //$this->vari doesn't work either
print $vari;
}
}
$foo = new testing();
$foo->test_item();
?>
test_item() should be:
function test_item() {
print $this->vari;
}
There is no need to pass $vari as a parameter.
Well, you've declared a method which expects an argument, which is missing. You should do:
$foo->test_item("Something");
As for the $this->, that goes inside of the class methods.
function test_item(){
print $this->vari;
}
function parameters can not be as "$this->var",
change your class like
class testing {
public $vari = "teststring";
function test_item(){ //$this->vari doesn't work either
print $this->vari;
}
}
$foo = new testing();
$foo->test_item();
And read this Object-Oriented PHP for Beginners
What's happening there is that $foo->test_item() is expecting something passed as an argument, so for example
$foo->test_item("Hello");
Would be correct in this case. This would print Hello
But, you may be wondering why it doesn't print teststring. This is because by calling
print $vari;
you are only printing the variable that has been passed to $foo->test_item()
However, if instead you do
function test_item(){ //notice I've removed the argument passed to test_item here...
print $this->vari;
}
You will instead be printing the value of the class property $vari. Use $this->... to call functions or variables within the scope of the class. If you try it without $this-> then PHP will look for that variable within the function's local scope
Related
How can i reference a class property knowing only a string?
class Foo
{
public $bar;
public function TestFoobar()
{
$this->foobar('bar');
}
public function foobar($string)
{
echo $this->$$string; //doesn't work
}
}
what is the correct way to eval the string?
You only need to use one $ when referencing an object's member variable using a string variable.
echo $this->$string;
If you want to use a property value for obtaining the name of a property, you need to use "{" brackets:
$this->{$this->myvar} = $value;
Even if they're objects, they work:
$this->{$this->myobjname}->somemethod();
As the others have mentioned, $this->$string should do the trick.
However, this
$this->$$string;
will actually evaluate string, and evaluate again the result of that.
$foo = 'bar';
$bar = 'foobar';
echo $$foo; //-> $'bar' -> 'foobar'
you were very close. you just added 1 extra $ sign.
public function foobar($string)
{
echo $this->$string; //will work
}
echo $this->$string; //should work
You only need $$string when accessing a local variable having only its name stored in a string. Since normally in a class you access it like $obj->property, you only need to add one $.
To remember the exact syntax, take in mind that you use a $ more than you normally use. As you use $object->property to access an object property, then the dynamic access is done with $object->$property_name.
If I try this code :
<?php
class ref
{
public $reff = "original ";
public function &get_reff()
{
return $this->reff;
}
public function get_reff2()
{
return $this->reff;
}
}
$thereffc = new ref;
$aa =& $thereffc->get_reff();
echo $aa;
$aa = " the changed value ";
echo $thereffc->get_reff(); // says "the changed value "
echo $thereffc->reff; // same thing
?>
Then returning by reference works and the value of the object property $reff gets changed as the variable $aa that references it changes too.
However, when I try this on a normal function that is not inside a class, it won't work !!
I tried this code :
<?php
function &foo()
{
$param = " the first <br>";
return $param;
}
$a = & foo();
$a = " the second <br>";
echo foo(); // just says "the first" !!!
it looks like the function foo() wont recognize it returns by reference and stubbornly returns what it wants !!!
Does returning by reference work only in OOP context ??
That is because a function's scope collapses when the function call completes and the function local reference to the variable is unset. Any subsequent calls to the function create a new $param variable.
Even if that where not the case in the function you are reassigning the variable to the first <br> with each invocation of the function.
If you want proof that the return by reference works use the static keyword to give the function variable a persistent state.
See this example
function &test(){
static $param = "Hello\n";
return $param;
}
$a = &test();
echo $a;
$a = "Goodbye\n";
echo test();
Echo's
Hello
Goodbye
Does returning by reference work only in OOP context ??
No. PHP makes no difference if that is a function or a class method, returning by reference always works.
That you ask indicates you might have not have understood fully what references in PHP are, which - as we all know - can happen. I suggest you read the whole topic in the PHP manual and at least two more sources by different authors. It's a complicated topic.
In your example, take care which reference you return here btw. You set $param to that value - always - when you call the function, so the function returns a reference to that newly set variable.
So this is more a question of variable scope you ask here:
Variable scope
I have the following piece of code
copy($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().
More info here.
You can use call_user_func to do this.
$result = call_user_func($functionToCall, $source, $target)
Documentation: PHP: call_user_func
as far as i know your code should work
here is the link for your refrence
I have been told that a class cannot be defined within a class in PHP. However, in my own example this seems to work which has me confused:
class_test.php:
require('class_1.php');
new class_1
//Need $missing_variable here.
class_1.php
class class_1{
public function function_1(){
function callback_function(){
echo "A Callback";
$missing_variable = "Where Did I Go?";
}
require('class_2.php');
new class_2('callback_function');
}
public function __construct(){
$this->function_1();
}
}
class_2.php
class class_2{
public function __construct($callback){
echo "Hello World - ";
call_user_func($callback);
}
}
Loading class_test.php prints out
Hello World - A Callback
Question: How do I define $missing_variable such that I can get it where I need it?
In case anyone in the future has a similar problem, however unlikely that may be, I want to link to the codepad from below that shows the $missing_variable echo'd from outside the classes:
http://codepad.org/tRk0XWG7
Thanks again everyone.
Note: This is a follow up.
You can declare a class within a function. That's known as conditional declaration, i.e. only if the function is called will the class be declared. It doesn't make much of a difference then whether you include a file with the class declaration or if you type out the code inside the function.
This does not mean however that the classes share any sort of scope or data. Only the declaration is conditionally nested, it still has the same functionality and scope as explained before.
Your confusion about the callback can be explained by the same thing. When class_1::function_1 is executed the first time, the function callback_function is being defined. This is a regular global function that can be called from anywhere. It's not bound to the class in any way. You will also notice that you cannot execute class_1::function_1 a second time, PHP will complain that callback_function already exists when you're trying to declare it again.
As for the comment in the source code //How do I declare this variable so that it is available where I need it?: You don't. That variable is a local variable inside a function. It's only in scope inside the function. You can return its value from the function like any other return value if you want to. (You could make it global, but for the love of god don't!) If you need that value somewhere else, don't declare it as a variable inside a function, because only the function can access it then.
You would return $missing_variable in a few places. See below. (This isn't the only way to do it, mind you)
http://codepad.org/tf08Vgdx
<?
class class_2{
public function __construct($callback){
echo "Hello World - ";
$missing = $callback();
$this->missing = $missing;
}
}
class class_1{
public function function_1(){
function callback_function(){
echo "A Callback. ";
$missing_variable = "Where Did I Go?";
return $missing_variable;
}
$class2 = new class_2('callback_function');
return $class2->missing;
}
public function __construct(){
$this->missing = $this->function_1();
}
}
$class = new class_1();
echo $class->missing;
How can i reference a class property knowing only a string?
class Foo
{
public $bar;
public function TestFoobar()
{
$this->foobar('bar');
}
public function foobar($string)
{
echo $this->$$string; //doesn't work
}
}
what is the correct way to eval the string?
You only need to use one $ when referencing an object's member variable using a string variable.
echo $this->$string;
If you want to use a property value for obtaining the name of a property, you need to use "{" brackets:
$this->{$this->myvar} = $value;
Even if they're objects, they work:
$this->{$this->myobjname}->somemethod();
As the others have mentioned, $this->$string should do the trick.
However, this
$this->$$string;
will actually evaluate string, and evaluate again the result of that.
$foo = 'bar';
$bar = 'foobar';
echo $$foo; //-> $'bar' -> 'foobar'
you were very close. you just added 1 extra $ sign.
public function foobar($string)
{
echo $this->$string; //will work
}
echo $this->$string; //should work
You only need $$string when accessing a local variable having only its name stored in a string. Since normally in a class you access it like $obj->property, you only need to add one $.
To remember the exact syntax, take in mind that you use a $ more than you normally use. As you use $object->property to access an object property, then the dynamic access is done with $object->$property_name.