I have this function to return the filename inside useful.php.
function filename(){
return basename(__FILE__, PATHINFO_EXTENSION);
}
I want to add useful.php using include_once "useful.php"; to another file such as index.php and there I write something like this:
Log Out
It will show the filename of useful.php not this file:
Log Out
How do I fix this. Thank you in advance. :)
PS: Sorry for the long title
Seems there is no simple solution, but you could use debug_backtrace() for that...
1.php
<?php
include('2.php');
echo filename();
2.php
<?php
function filename(){
$bt = debug_backtrace();
return basename($bt[0]['file'], PATHINFO_EXTENSION);
}
Output
1.php
OR change your function to this:
function filename($file){
return basename($file, PATHINFO_EXTENSION);
}
And use it in other files like this:
filename(__FILE__);
I have a file called test1.php with lots of variable and some function definitions. I am trying to include this file to one another file called test2.php and use the variables and the functions.
test1.php
$i = "a";
$ii = "b";
$iii = "c";
function test1a(){ return "lol"; }
test2.php
function test2a(){ include 'test1.php'; return $i; }
function test2b() { include 'test1.php'; return test2c(); }
function test2c(){ include 'test1.php'; return $iii; }
function test2d() { include 'test1.php'; return test1a(); }
index.php
include 'test2.php'
echo test2a();
echo test2b();
echo test2c();
echo test2d();
Reason:
I have the same code base in two different servers. Only the test2.php file will be different.Each server will have different values inside the test2.php but with same variable name. test2.php will act somewhat like a localization file.
My problem is some of the variables or not showing up. Is there a better way to do this. I don't want to include the file in every function.
Thanks.
Just do it the other way round:
put all the different variables into one file in to an array, without any functions:
//config.php
$config['setting1'] = "val1";
$config['setting2'] = "val2";
$config['setting3'] = 42;
...
and further:
//index.php
include_once("config.php");
echo $config['setting1'];
....
now you may have different configs on different servers w/o need to change any functions.
You are trying to include file with one function several times. Functions are always global. So second include gives you an fatal error Fatal error: Cannot redeclare test1a() (previously declared in [..]).
You should put this function to separate file and use include_once.
I want to get the name of the file that includes another file from inside the included file.
I know there is the __FILE__ magic constant, but that doesn't help, since it returns the name of the included file, not the including one.
Is there any way to do this? Or is it impossible due to the way PHP is interpreted?
So this question is pretty old, but I was looking for the answer and after leaving here unsatisfied, I came across $_SERVER['SCRIPT_FILENAME']; Of course this works if the php file doing the including is a web page.
This gives you the full path of the "including file" on the server. eg /var/www/index.php. so if you want just the filename, eg index.php, you will need to use basename() eg
basename($_SERVER['SCRIPT_FILENAME']);
So, if in your index.php you have the following line:
<?php include("./somephp.php"); ?>
and in somephp.php you have
echo "this is the file that included me: " . basename($_SERVER['SCRIPT_FILENAME']);
you will get
this is the file that included me: index.php
output to the browser. This also works if the user is accessing your file without explicitly including the filename in the url, eg www.example.com instead of www.example.com/index.php.
Solution
Knowing that the functions used to include files are include, include_once, require and require_once, also knowing that they are reserved words in PHP, meaning that it will not be possible to declare user functions with the same name, and based on wedgwood's idea of using debug_backtrace you can actually work out from what file the include was called.
What we are going to do is iterate over the backtrace untill we find the most recent call to any of the four include functions, and the the file where it was called. The following code demostrate the technique:
function GetIncludingFile()
{
$file = false;
$backtrace = debug_backtrace();
$include_functions = array('include', 'include_once', 'require', 'require_once');
for ($index = 0; $index < count($backtrace); $index++)
{
$function = $backtrace[$index]['function'];
if (in_array($function, $include_functions))
{
$file = $backtrace[$index]['file'];
break;
}
}
return $file;
}
The above code will return the absolute path of the file where the last include happened, if there hasn't been any include it will return false. Note that the file may has been include from a file that was included from another file, the above function only works for the deepest include.
With a simple modification, you can also get the last included file:
function GetIncludedFile()
{
$file = false;
$backtrace = debug_backtrace();
$include_functions = array('include', 'include_once', 'require', 'require_once');
for ($index = 0; $index < count($backtrace); $index++)
{
$function = $backtrace[$index]['function'];
if (in_array($function, $include_functions))
{
$file = $backtrace[$index - 1]['file'];
break;
}
}
return $file;
}
Observations
Note that __FILE__ is not the included file, but the current file. For example:
file A.php:
<?php
function callme()
{
echo __FILE__;
}
?>
file B.php:
<?php
include('A.php');
callme();
?>
file index.php:
<?php
include('B.php');
?>
In the above example, in the context of the file B.php the included file is B.php (and the including file is index.php) but the output of the function callme is the path of A.php because __FILE__ is in A.php.
Also note that $_SERVER['SCRIPT_FILENAME'] will give the the absolute path to the script requested by the client. If $_SERVER['SCRIPT_FILENAME'] == __FILE__ it means that the current file is the requested one, and therefore there probably hasn't been any includes...
The above method checks if the current file is the requested one, but not if it hasn't been included (below is an example of how the requested file can be included). An actual solution to check if there has not been any inclusions could be to check count(get_included_files()) == 1.
The requested file can be an included file in the following way:
file x.php
<?php
$var = 'something';
include('index.php');
?>
file index.php
<?php
if (!isset($var))
{
include('x.php');
exit;
}
echo 'something';
?>
In the above example, the file index.php will include x.php, then x.php will include index.php back, afterwards index.php outputs "something" and returns control to x.php, x.php returns control to index.php and it reaches exit;
This shows that even if index.php was the requested script, it was also included from another script.
I can't find the easy way to cover this.
But If the including one is really important to you, you could hack it with some global variable and your own include function.
e.g.
<?php
$g_including_files = array();
function my_include($file) {
$bt = debug_backtrace();
global $g_including_files;
$g_including_files[basename($file)] = $bt[0]['file'];
return include($file);
}
May that be helpful for you :)
This is actually just a special case of what PHP templating engines do. Consider having this function:
function ScopedInclude($file, $params = array())
{
extract($params);
include $file;
}
Then A.php can include C.php like this:
<?php
// A.php
ScopedInclude('C.php', array('includerFile' => __FILE__));
Additionally, B.php can include C.php the same way without trouble.
<?php
// B.php
ScopedInclude('C.php', array('includerFile' => __FILE__));
C.php can know its includer by looking in the $params array.
<?php
// C.php
echo $includerFile;
abspath()
function abspath()
{
echo $_SERVER['DOCUMENT_ROOT'];
}
directory()
function directory()
{
echo '/folder/';
}
Code Line:-
header('Location:'.abspath().directory());
Gives me the following as output:
C:/xampp/htdocs/folder/
When I use
header('Location:'.$_SERVER['DOCUMENT_ROOT'];.directory());
It sends me to my index.php in C:/xampp/htdocs/folder/index.php
Why its not working with functions?
I want to go to C:/xampp/htdocs/folder/index.php using this
header('Location:'.abspath().directory());
- Is there any problem?
The problem is that your functions are echoing your output and not returning it. You'll want to change your functions to:
function abspath()
{
return $_SERVER['DOCUMENT_ROOT'];
}
function directory()
{
return '/folder/';
}
So you can use the returned value (namely $_SERVER['DOCUMENT_ROOT'] or '/folder/', in this case) in your string concatentiation.
Your path should URI when using with header .
Soething like that
header('Location:http://yourpath.com/folder');
In the answer you are using a physical location where file residing
try $_SERVER["REQUEST_URI"]
I understand that the question is rather hard to understand, I didn't know how to ask it better, so I'll use this code example to make things more clear:
If I have the following files:
test.php:
<?php
include('include.php');
echo myClass::myStaticFunction();
?>
include.php
<?php
__autoload($classname){
include_once("class/".$classname.".php"); //normally checking of included file would happen
}
?>
class/myClass.php
<?php
class myClass{
public static function myStaticFunction(){
//I want this to return test.php, or whatever the filename is of the file that is using this class
return SOMETHING;
}
?>
the magic FILE constant is not the correct one, it returns path/to/myClass.php
in case you need to get "test.php" see $_SERVER['SCRIPT_NAME']
I ended up using:
$file = basename(strtolower($_SERVER['SCRIPT_NAME']));
I am using
$arr = #debug_backtrace(false);
if (isset($arr))
foreach ($arr as $data)
{
if (isset($data['file']))
echo $data['file'];
// change it to needed depth
}
This way you don't need to modify the file from which your file is included. debug_backtrace might have some speed consenquencies.