Having php code:
function getChildren($parent_id = 0)
{
....
return $SomeChildrenArray;
}
And Smarty assignation $smarty->assign('children', getChildren($id));
How to write smarty expression to pass some value to the function?
Smarty can be extended through its plugins API. See registerPlugin() for an introduction.
$smarty->registerPlugin('function', 'children', function($params, $template) {
return "some string:" . $params['foobar'];
});
and
{children foobar="hello world"}
would output
some string:hello world
Note that plugin functions have to return a string.
Since Smarty3 you can call and assign arbitrary functions from within a template:
{$children = getChildren(3)}
This way you can call any function and return any value/type you want. See Smarty_Security for details on how to control which functions may be called…
Oh, sorry! I'am was wrong in my post =) Try this please
php file
$smarty->registerPlugin("function","my_supper_closure_function", "my_supper_closure_function");
function my_supper_closure_function($params, $smarty){
$SomeChildrenArray=array($params["parent_id"]);
return print_r($SomeChildrenArray,true);
}
$smarty->display('index.tpl');
tpl file
{my_supper_closure_function parent_id=2}
PHP file:
$smarty->assign('children', function($parent_id = 0) {
....
return $SomeChildrenArray;
});
tpl file for parent_id=2
{children p1=2}
Related
I have a PHP file that can be include'd() in various places inside another page. I want to know whether it has been included inside a function. How can I do this? Thanks.
There's a function called debug_backtrace() that will return the current call stack as an array. It feels like a somewhat ugly solution but it'll probably work for most cases:
$allowedFunctions = array('include', 'include_once', 'require', 'require_once');
foreach (debug_backtrace() as $call) {
// ignore calls to include/require
if (isset($call['function']) && !in_array($call['function'], $allowedFunctions)) {
echo 'File has not been included in the top scope.';
exit;
}
}
You can set a variable in the included file and check for that variable in your functions:
include.php:
$included = true;
anotherfile.php:
function whatever() {
global $included;
if (isset($included)) {
// It has been included.
}
}
whatever();
You can check if the file is in the array returned by get_included_files(). (Note that list elements are full pathnames.) To see if inclusion occurred during a particular function call, check get_included_files before and after the function call.
Is it possible to call only the specific function from another file without including whole file???
There may be another functions in the file and don't need to render other function.
The short answer is: no, you can't.
The long answers is: yes, if you use OOP.
Split your functions into different files. Say you are making a game with a hero:
Walk.php
function walk($distance,speed){
//walk code
}
Die.php
function die(){
//game over
}
Hero.php
include 'Walk.php';
include 'Die.php';
class Hero(){
//hero that can walk & can die
}
You may have other functions like makeWorld() that hero.php doesn't need, so you don't need to include it. This question has been asked a few times before: here & here.
One of the possible methods outlined before is through autoloading, which basically saves you from having to write a long list of includes at the top of each file.
In PHP it's not available to get only a little part of a file.
Maybe this is a ability to use only little parts of a file:
I have a class that calls "utilities". This I am using in my projects.
In my index.php
include("class.utilities.php")
$utilities = new utilities();
The file class.utilities.php
class utilities {
function __construct() {
}
public function thisIsTheFunction($a,$b)
{
$c = $a + $b;
return $c;
}
}
And then i can use the function
echo $utilities->thisIsTheFunction(3,4);
include a page lets say the function is GetPage and the variable is ID
<?php
require('page.php');
$id = ($_GET['id']);
if($id != '') {
getpage($id);
}
?>
now when you make the function
<?php
function getpage($id){
if ($id = ''){
//// Do something
}
else {
}
}
?>
When I execute following code I am getting this error. Why is that? What is the proper use of callbacks?
CODE (simplified)
class NODE {
//...some other stuff
function create($tags, $callback=false) {
$temp = new NODE();
//...code and stuff
if($callback) $callback($temp); //fixed (from !$callback)
return $this;
}
}
$document = new NODE();
$document->create("<p>", function($parent) {
$parent->create("<i>");
});
ERROR
Fatal error: Function name must be a string in P:\htdocs\projects\nif\nif.php on line 36
$document->new NODE();
This is not valid syntax. The accepted format would be:
$document = new NODE();
In addition to this, if you use the unary operator (!) on a false, you get true. If you use it on a Callable, you get false. As such, if (!$callback) $callback() will throw the first error of your script.
As a side note, you are reinventing the wheel. I would strongly recommend you take a look at the DOMDocument family of classes, which are doing exactly what you are currently trying to implement, albeit with fewer callbacks.
if(!$callback) $callback($temp);
If $callback is false, for sure you won't be able to call it as a callback.
if(!$callback) $callback($temp);
should probably be
if($callback) $callback($temp);
And the instanciation:
$document = new NODE();
My 2c here, type hinting may be good to use here as well.
Ex: function create($tags, callable $callback = function())
To do such a thing in php you should use function pointers and tell php which function to execute.
Look at this code.
// This function uses a callback function.
function doIt($callback)
{
$data = acquireData();
$callback($data);
}
// This is a sample callback function for doIt().
function myCallback($data)
{
echo 'Data is: ', $data, "\n";
}
// Call doIt() and pass our sample callback function's name.
doIt('myCallback');
So as you seen you can only pass the name to the function and you should predefine the function..
Similar question: How do I implement a callback in PHP?
I am trying to setup an array that pulls the filename and function name to run, but it not fully working.
The code is
$actionArray = array(
'register' => array('Register.php', 'Register'),
);
if (!isset($_REQUEST['action']) || !isset($actionArray[$_REQUEST['action']])) {
echo '<br><br>index<br><br>';
echo 'test';
exit;
}
require_once($actionArray[$_REQUEST['action']][0]);
return $actionArray[$_REQUEST['action']][1];
Register.php has
function Register()
{
echo 'register';
}
echo '<br>sdfdfsd<br>';
But it does not echo register and just sdfdfsd.
If I change the first lot of code from
return $actionArray[$_REQUEST['action']][1];
to
return Register();
It works, any ideas?
Thanks
Change the last line to:
return call_user_func($actionArray[$_REQUEST['action']][1]);
This uses the call_user_func function for more readable code and better portability. The following also should work (Only tested on PHP 5.4+)
return $actionArray[$_REQUEST['action']][1]();
It's almost the same as your code, but I'm actually invoking the function instead of returning the value of the array. Without the function invocation syntax () you're just asking PHP get to get the value of the variable (in this case, an array) and return it.
You'll find something usefull here:
How to call PHP function from string stored in a Variable
Call a function name stored in a string is what you want...
newbie in PHP here, sorry for troubling you.
I want to ask something, if I want to include a php page, can I use parameter to define the page which I'll be calling?
Let's say I have to include a title part in my template page. Every page has different title which will be represented as an image. So,
is it possible for me to call something <?php #include('title.php',<image title>); ?> inside my template.php?
so the include will return title page with specific image to represent the title.
thank you guys.
An included page will see all the variables for the current scope.
$title = 'image title';
include('title.php');
Then in your title.php file that variable is there.
echo '<h1>'.$title.'</h1>';
It's recommended to check if the variable isset() before using it. Like this.
if(isset($title))
{
echo '<h1>'.$title.'</h1>';
}
else
{
// handle an error
}
EDIT:
Alternatively, if you want to use a function call approach. It's best to make the function specific to activity being performed by the included file.
function do_title($title)
{
include('title.php'); // note: $title will be a local variable
}
Not sure if this is what you're looking for, but you can create a function to include the file and pass a variable.
function includeFile($file, $param) {
echo $param;
include_once($file);
}
includeFile('title.php', "title");
In your included file, you could do this:
<?php
return function($title) {
do_title_things($title);
do_other_things();
};
function do_title_things($title) {
// ...
}
function do_other_things() {
// ...
}
Then, you could pass the parameter as such:
$callback = include('myfile.php');
$callback('new title');
Another more commonly used pattern is to make a new scope for variables to be passed in:
function include_with_vars($file, $params) {
extract($params);
include($file);
}
include_with_vars('myfile.php', array(
'title' => 'my title'
));
The included page will already have access to those variables defined prior to the include. If you require include specific variables, I suggest defining those variables on the page to be included