I am new to PHP. Well the text i am referring to say quotes
Functions can be defined anywhere within your program.
the above statement holds fine for code block 1 but not for code block 2. KINDLY EXPLAIN?
CODE Block 1:
<?php
test();
function test()
{
echo "Hello Inside the function";
}
?>
CODE Block 2:
<?php
$no=1;
switch ($no)
{
case "1":
test();
function test()
{
echo "Hello test";
}
}
?>
In theory, yes, functions can be defined "anywhere". In practice, there's a trick to it. The trick is as follows: when PHP reads and compiles the source of your script, it looks for function definitions, and if function definition is in global context (not inside if, switch, etc.) it will be defined immediately. However, if it is inside such construct, or inside another function, etc. it will be defined only when control passes the line on which function() statement resides.
Thus, code block 1 works - because the function is in global context, so PHP will define it before any code is run. But in code block 2, the function is in the context of switch, so it will be defined only when control passes line 7. But since you try to call it on line 6, it is not defined yet! So PHP errors out.
The advice here is never define your functions inside conditionals etc. unless you mean it to be conditional definitions - and then take care not to call them before they are defined.
You can declare function in switch statement, but it's not so good. Your have error because you call function and just then declare it. At first you should declare function and then use it.
<?php
$no=1;
switch ($no)
{
case "1":
function test()
{
echo "Hello test";
}
test();
}
?>
You cannot declare a function in a switch statement.
However what you can do is the following:
<?php
$no=1;
switch ($no)
{
case "1":
test();
break;
}
function test()
{
echo "Hello test";
}
?>
Just remove the function from the switch.
The function only gets executed when called so it doesn't matter.
EDIT
What propably is meant by that quote (Functions can be defined anywhere within your program.) is:
You can declare functions before or even after you call them in your script.
A couple of problems. you need to use
case 1:
for switches, otherwise it will be looking for a string equivalent to "1". "1" != 1 (the first is a string, the second an integer)
While your text did say functions could be defined anywhere, they didn't actually mean anywhere. You cannot define a function inside of a block of code, so you'l have to define the function outside of the switch:
<?php
$no = 1;
switch ($no) {
case 1:
test();
break;
}
function test()
{
echo "I'm inside the test function!";
}
?>
Otherwise things just get crazy.
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?)
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;
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 $this-> then PHP will look for that variable within the function's local scope
I have a function that searches for a string inside a text file. I want to use the same function to assign all lines to an array in case I am going to replace that string. So I will read the input file only once.
I have the search function working but I do not know how to deal with the array thing.
the code is something like that (I made the code sample much simpler,so please ignore the search function that actually isn't below)
function read_ini($config_file_name,$string){
$config_file = file($config_file_name);
foreach($config_file as $line) {
return_string = trim(substr($line,0,15));
$some_global_array = $line'
}
}
echo read_ini('config.ini','database.db')
if ($replaced) {file_put_contents('config.ini', $some_global_array);}
http://php.net/parse_ini_file
I know it doesn't answer the question, but it quite possibly removes the need for even having to ask.
The deal with globals, though, is that they must be defined at the top of the function as globals, or else they're considered part of the function's scope.
function write_global() {
global $foo;
$foo = 'bar';
}
write_global();
echo $foo; // bar
I hava a function that looks something like this:
require("config.php");
function displayGta()
{
(... lots of code...)
$car = $car_park[3];
}
and a config.php that look something like this:
<?php
$car_park = array ("Mercedes 540 K.", "Chevrolet Coupe.", "Chrysler Imperial.", "Ford Model T.", "Hudson Super.", "Packard Sedan.", "Pontiac Landau.", "Duryea.");
(...)
?>
Why do I get Notice: Undefined variable: car_park ?
Try adding
global $car_park;
in your function. When you include the definition of $car_park, it is creating a global variable, and to access that from within a function, you must declare it as global, or access it through the $GLOBALS superglobal.
See the manual page on variable scope for more information.
Even though Paul describes what's going on I'll try to explain again.
When you create a variable it belongs to a particular scope. A scope is an area where a variable can be used.
For instance if I was to do this
$some_var = 1;
function some_fun()
{
echo $some_var;
}
the variable is not allowed within the function because it was not created inside the function. For it to work inside a function you must use the global keyword so the below example would work
$some_var = 1;
function some_fun()
{
global $some_var; //Call the variable into the function scope!
echo $some_var;
}
This is vice versa so you can't do the following
function init()
{
$some_var = true;
}
init();
if($some_var) // this is not defined.
{
}
There are a few ways around this but the simplest one of all is using $GLOBALS array which is allowed anywhere within the script as they're special variables.
So
$GLOBALS['config'] = array(
'Some Car' => 22
);
function do_something()
{
echo $GLOBALS['config']['some Car']; //works
}
Also make sure your server has Register globals turned off in your INI for security.
http://www.php.net/manual/en/security.globals.php
You could try to proxy it into your function, like:
function foo($bar){
(code)
$car = $bar[3];
(code)
}
Then when you call it:
echo foo($bar);
I had the same issue and have been tearing my hair out over it - nothing worked, absolutely nothing - until in desperation I just copied the contents of config.php into a new file and saved it as config2.php (without changing anything in its contents at all), changed the require_once('config.php'); to require_once('config2.php'); and it just started working.