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;
Related
Is it possible to pass a function by reference? So everytime the reference variable is called the function will be called aswell. Take a look at my code.
<?php
class Documents {
private static $docs = array(
'She went to the toilet and on her way back, opened the wrong door',
'She had found something that would mean she\'d never be poor again - but there was a catch',
'It was just for one night',
'He watched, helpless, as the door closed behind her'
);
public static function get() {
return self::$docs[array_rand(self::$docs)];
}
}
class Printer {
public static $content;
}
Printer::$content = &Documents::get();
echo Printer::$content;
echo "\n" . Printer::$content;
echo "\n" . Printer::$content;
Right now it'll print 3 similar lines but i would like it to call Documents::get() everytime Printer::$content is printed because Printer::$content = **&**Documents::get(); it is by reference.
No, you cannot have a variable which you treat as a variable which nonetheless runs code behind the scenes. If you write $foo, that's using the value of a variable. Only of you write $foo() are you explicitly executing a function.
Having said that, there are some situations in which object methods will be called implicitly. For instance, if $foo is an object and you use it in a string context:
echo $foo;
This will (try to) implicitly call $foo->__toString().
Please do not get the idea to somehow abuse this implied method call to do anything fancy. If you want to call functions, call functions. You can even return functions from other functions, so there's no lack of possibility to pass around functions. You will have to call them explicitly with () however.
There are variable functions:
php > function foo($bar) { echo $bar; }
php > $baz = 'foo';
php > $baz('qux');
qux
But you cannot have PHP automatically execute that "referenced" function when the variable is simply accessed, e.g:
php > $baz;
php >
As you can see, the foo function was not called, and no output was performed. Without the () to signify a function call, that variable is like any other - it's just a string whose contents happen to be the same as a particular functions. It's the () that makes the string "executable".
Note that variable functions, while useful in some limited circumstances, should be avoided as they can lead to spaghetti code and difficult-to-debug bugs.
You can use the magic method __get().
class Printer {
public function __get($name)
{
switch($name) {
case "content":
return Documents::get();
break;
default:
return $this->$name;
break;
}
}
}
But this cannot be done in static context. So you would have to have an instance of Printer. (Perhaps use a singleton?)
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 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 $this-> then PHP will look for that variable within the function's local scope
Anyone has an idea if this is at all possible with PHP?
function foo($var) {
// the code here should output the value of the variable
// and the name the variable has when calling this function
}
$hello = "World";
foo($hello);
Would give me this output
varName = $hello
varValue = World
EDIT
Since most people here 'accuse' me of bad practices and global variables stuff i'm going to elaborate a little further on why we are looking for this behaviour.
the reason we are looking at this kind of behaviour is that we want to make assigning variables to our Views easier.
Most of the time we are doing this to assign variables to our view
$this->view->assign('products', $products);
$this->view->assign('members', $members);
While it would be easier and more readable to just be able to do the following and let the view be responsible to determining the variable name the assigned data gets in our views.
$this->view->assign($products);
$this->view->assign($members);
Short answer: impossible.
Long answer: you could dig through apd, bytekit, runkit, the Reflection API and debug_backtrace to see if any obscure combination would allow you to achieve this behavior.
However, the easiest way is to simply pass the variable name along with the actual variable, like you already do. It's short, it's easy to grasp, it's flexible when you need the variable to have a different name and it is way faster than any possible code that might be able to achieve the other desired behavior.
Keep it simple
removed irrelevant parts after OP edited the question
Regardless of my doubt that this is even possible, I think that forcing a programmer on how to name his variables is generally a bad idea. You will have to answer questions like
Why can't I name my variable $arrProducts instead of $products ?
You would also get into serious trouble if you want to put the return value of a function into the view. Imagine the following code in which (for whatever reason) the category needs to be lowercase:
$this->view->assign(strtolower($category));
This would not work with what you're planning.
My answer therefore: Stick to the 'verbose' way you're working, it is a lot easier to read and maintain.
If you can't live with that, you could still add a magic function to the view:
public function __set($name, $value) {
$this->assign($name, $value);
}
Then you can write
$this->view->product = $product;
I don't think there is any language where this is possible. That's simply not how variables work. There is a difference between a variable and the value it holds. Inside the function foo, you have the value, but the variable that held the value is not available. Instead, you have a new variable $var to hold that value.
Look at it like this: a variable is like a bucket with a name on it. The content (value) of the variable is what's inside the bucket. When you call a function, it comes with its own buckets (parameter names), and you pour the content of your bucket into those (well, the metaphor breaks down here because the value is copied and still available outside). Inside the function, there is no way to know about the bucket that used to hold the content.
What you're asking isn't possible. Even if it was, it would likely be considered bad practice as its the sort of thing that could easily get exploited.
If you're determined to achieve something like this, the closest you can get would be to pass the variable name as a string and reference it in the function from the $GLOBALS array.
eg
function this_aint_a_good_idea_really($var) {
print "Variable name: {$var}\n";
print "Variable contents: {$GLOBALS[$var]}\n";
}
$hello="World";
this_aint_a_good_idea_really('hello');
But as I say, that isn't really a good idea, nor is it very useful. (Frankly, almost any time you resort to using global variables, you're probably doing something wrong)
Its not impossible, you can find where a function was invoked from debug_backtrace() then tokenize a copy of the running script to extract the parameter expressions (what if the calling line is foo("hello $user, " . $indirect($user,5))?),
however whatever reason you have for trying to achieve this - its the wrong reason.
C.
Okay, time for some ugly hacks, but this is what I've got so far, I'll try to work on it a little later
<?php
class foo
{
//Public so we can test it later
public $bar;
function foo()
{
//Init the array
$this->bar = array();
}
function assign($__baz)
{
//Try to figure out the context
$context = debug_backtrace();
//assign the local array with the name and the value
//Alternately you can initialize the variable localy
//using $$__baz = $context[1]['object']->$__baz;
$this->bar[$__baz] = $context[1]['object']->$__baz;
}
}
//We need to have a calling context of a class in order for this to work
class a
{
function a()
{
}
function foobar()
{
$s = "testing";
$w = new foo();
//Reassign local variables to the class
foreach(get_defined_vars() as $name => $val)
{
$this->$name = $val;
}
//Assign the variable
$w->assign('s');
//test it
echo $w->bar['s'];
}
}
//Testrun
$a = new a();
$a->foobar();
impossible - the max. ammount of information you can get is what you see when dumping
debug_backtrace();
Maybe what you want to do is the other way around, a hackish solution like this works fine:
<?php
function assign($val)
{
global $$val;
echo $$val;
}
$hello = "Some value";
assign('hello');
Ouputs: Some value
What you wish to do, PHP does not intend for. There is no conventional way to accomplish this. In fact, only quite extravagant solutions are available. One that remains as close to PHP as I can think of is creating a new class.
You could call it NamedVariable, or something, and as its constructor it takes the variable name and the value. You'd initiate it as $products = new NamedVariable('products', $productData); then use it as $this->view->assign($products);. Of course, your declaration line is now quite long, you're involving yet another - and quite obscure - class into your code base, and now the assign method has to know about NamedVariable to extract both the variable name and value.
As most other members have answered, you are better off suffering through this slight lack of syntactic sugar. Mind you, another approach would be to create a script that recognizes instances of assign()'s and rewrites the source code. This would now involve some extra step before you ran your code, though, and for PHP that's silly. You might even configure your IDE to automatically populate the assign()'s. Whatever you choose, PHP natively intends no solution.
This solution uses the GLOBALS variable. To solve scope issues, the variable is passed by reference, and the value modified to be unique.
function get_var_name(&$var, $scope=FALSE) {
if($scope) $vals = $scope;
else $vals = $GLOBALS;
$old = $var;
$var = $new = 'unique'.rand().'value';
$vname = FALSE;
foreach ($vals as $key => $val) {
if($val === $new) $vname = $key;
}
$var = $old;
return $vname;
}
$testvar = "name";
echo get_var_name($testvar); // "testvar"
function testfunction() {
$var_in_function = "variable value";
return get_var_name($var_in_function, get_defined_vars());
}
echo testfunction(); // "var_in_function"
class testclass {
public $testproperty;
public function __constructor() {
$this->testproperty = "property value";
}
}
$testobj = new testclass();
echo get_var_name($testobj->testproperty, $testobj); // "testproperty"