I'm trying to pass a variable into an include file. My host changed PHP version and now whatever solution I try doesn't work.
I think I've tried every option I could find. I'm sure it's the simplest thing!
The variable needs to be set and evaluated from the calling first file (it's actually $_SERVER['PHP_SELF'], and needs to return the path of that file, not the included second.php).
OPTION ONE
In the first file:
global $variable;
$variable = "apple";
include('second.php');
In the second file:
echo $variable;
OPTION TWO
In the first file:
function passvariable(){
$variable = "apple";
return $variable;
}
passvariable();
OPTION THREE
$variable = "apple";
include "myfile.php?var=$variable"; // and I tried with http: and full site address too.
$variable = $_GET["var"]
echo $variable
None of these work for me. PHP version is 5.2.16.
What am I missing?
Thanks!
You can use the extract() function
Drupal use it, in its theme() function.
Here it is a render function with a $variables argument.
function includeWithVariables($filePath, $variables = array(), $print = true)
{
$output = NULL;
if(file_exists($filePath)){
// Extract the variables to a local namespace
extract($variables);
// Start output buffering
ob_start();
// Include the template file
include $filePath;
// End buffering and return its contents
$output = ob_get_clean();
}
if ($print) {
print $output;
}
return $output;
}
./index.php :
includeWithVariables('header.php', array('title' => 'Header Title'));
./header.php :
<h1><?php echo $title; ?></h1>
Option 3 is impossible - you'd get the rendered output of the .php file, exactly as you would if you hit that url in your browser. If you got raw PHP code instead (as you'd like), then ALL of your site's source code would be exposed, which is generally not a good thing.
Option 2 doesn't make much sense - you'd be hiding the variable in a function, and be subject to PHP's variable scope. You'ld also have to have $var = passvariable() somewhere to get that 'inside' variable to the 'outside', and you're back to square one.
option 1 is the most practical. include() will basically slurp in the specified file and execute it right there, as if the code in the file was literally part of the parent page. It does look like a global variable, which most people here frown on, but by PHP's parsing semantics, these two are identical:
$x = 'foo';
include('bar.php');
and
$x = 'foo';
// contents of bar.php pasted here
Considering that an include statment in php at the most basic level takes the code from a file and pastes it into where you called it and the fact that the manual on include states the following:
When a file is included, the code it contains inherits the variable scope of the line on which the include occurs. Any variables available at that line in the calling file will be available within the called file, from that point forward.
These things make me think that there is a diffrent problem alltogether. Also Option number 3 will never work because you're not redirecting to second.php you're just including it and option number 2 is just a weird work around. The most basic example of the include statment in php is:
vars.php
<?php
$color = 'green';
$fruit = 'apple';
?>
test.php
<?php
echo "A $color $fruit"; // A
include 'vars.php';
echo "A $color $fruit"; // A green apple
?>
Considering that option number one is the closest to this example (even though more complicated then it should be) and it's not working, its making me think that you made a mistake in the include statement (the wrong path relative to the root or a similar issue).
I have the same problem here, you may use the $GLOBALS array.
$GLOBALS["variable"] = "123";
include ("my.php");
It should also run doing this:
$myvar = "123";
include ("my.php");
....
echo $GLOBALS["myvar"];
Have a nice day.
I've run into this issue where I had a file that sets variables based on the GET parameters. And that file could not updated because it worked correctly on another part of a large content management system. Yet I wanted to run that code via an include file without the parameters actually being in the URL string. The simple solution is you can set the GET variables in first file as you would any other variable.
Instead of:
include "myfile.php?var=apple";
It would be:
$_GET['var'] = 'apple';
include "myfile.php";
OPTION 1 worked for me, in PHP 7, and for sure it does in PHP 5 too. And the global scope declaration is not necessary for the included file for variables access, the included - or "required" - files are part of the script, only be sure you make the "include" AFTER the variable declaration. Maybe you have some misconfiguration with variables global scope in your PHP.ini?
Try in first file:
<?php
$myvariable="from first file";
include ("./mysecondfile.php"); // in same folder as first file LOLL
?>
mysecondfile.php
<?php
echo "this is my variable ". $myvariable;
?>
It should work... if it doesn't just try to reinstall PHP.
In regards to the OP's question, specifically "The variable needs to be set and evaluated from the calling first file (it's actually '$_SERVER['PHP_SELF']', and needs to return the path of that file, not the included second.php)."
This will tell you what file included the file. Place this in the included file.
$includer = debug_backtrace();
echo $includer[0]['file'];
I know this is an old question, but stumbled upon it now and saw nobody mentioned this. so writing it.
The Option one if tweaked like this, it should also work.
The Original
Option One
In the first file:
global $variable;
$variable = "apple";
include('second.php');
In the second file:
echo $variable;
TWEAK
In the first file:
$variable = "apple";
include('second.php');
In the second file:
global $variable;
echo $variable;
According to php docs (see $_SERVER) $_SERVER['PHP_SELF'] is the "filename of the currently executing script".
The INCLUDE statement "includes and evaluates the specified" file and "the code it contains inherits the variable scope of the line on which the include occurs" (see INCLUDE).
I believe $_SERVER['PHP_SELF'] will return the filename of the 1st file, even when used by code in the 'second.php'.
I tested this with the following code and it works as expected ($phpSelf is the name of the first file).
// In the first.php file
// get the value of $_SERVER['PHP_SELF'] for the 1st file
$phpSelf = $_SERVER['PHP_SELF'];
// include the second file
// This slurps in the contents of second.php
include_once('second.php');
// execute $phpSelf = $_SERVER['PHP_SELF']; in the secod.php file
// echo the value of $_SERVER['PHP_SELF'] of fist file
echo $phpSelf; // This echos the name of the First.php file.
An alternative to using $GLOBALS is to store the variable value in $_SESSION before the include, then read it in the included file. Like $GLOBALS, $_SESSION is available from everywhere in the script.
Pass a variable to the include file by setting a $_SESSION variable
e.g.
$_SESSION['status'] = 1;
include 'includefile.php';
// then in the include file read the $_SESSION variable
$status = $_SESSION['status'];
You can execute all in "second.php" adding variable with jQuery
<div id="first"></div>
<script>
$("#first").load("second.php?a=<?=$var?>")
</scrpt>
I found that the include parameter needs to be the entire file path, not a relative path or partial path for this to work.
This worked for me: To wrap the contents of the second file into a function, as follows:
firstFile.php
<?php
include("secondFile.php");
echoFunction("message");
secondFile.php
<?php
function echoFunction($variable)
{
echo $variable;
}
Do this:
$checksum = "my value";
header("Location: recordupdated.php?checksum=$checksum");
I have a php file on my server that takes in two inputs through the URL and then comes back with a result. When a page is loaded, I'd like to have the result of that calculation already loaded. For example:
$var = load("http://mysite.com/myfile.php?&var1=var1&var2=var2");
I know that load isn't a real function for this, but is there something simple that suits what I'm looking for? thanks
Use file_get_contents
$foo = file_get_contents('http://mysite.com/myfile.php?&var1=var1&var2=var2');
Or, a better solution if the file is located on your server:
include('myfile.php');
and either set the $_GET variables in the included script itself, or prior to including it.
If they are running on the same server, consider calling the script directly?
$_GET["var1"] = "var1";
$_GET["var2"] = "var2";
include "myfile.php";
You could use file_get_contents, but it may be a more practical solution to simply include the file and call the function directly in the file, rather than trying to manually load the file.
I'm trying to use PHP variables into echoed file, and couldn't get where is a trouble at first using that script:
$head = new mod_head("head.php");
$id="ASDSSgdfsfsdfS";
echo $head;
mod_head class:
class mod_head
{
private $out="";
function __construct($arg)
{
$this->out=$this->parts($arg);
}
public function __toString()
{
return $this->out;
}
private function parts($file)
{
return fread(#fopen(PATH . "parts/".$file, 'r'), filesize(PATH . "parts/".$file));
}
}
and the file is "head.php"
<h1><center style="background:orange; border-radius:15px;">LOGO</center></h1>
<br><?php print_r($id)?>
<div>BANNER <div>$id <?php echo $id ;?></div></div>
i dont want to create global vars, why it doest echo $id var?
First, you're going to need to parse the file, not just read it. The second problem you'll have is a scope issue. $id is outside of the scope of the parts() function. In order to return the contents of the required file instead of just printing it I've used the output control functions
Try changing your parts function to this:
private function parts($file)
{
ob_start();
require(PATH . "parts/".$file);
$output = ob_get_contents();
ob_end_clean();
return $output;
}
To fix the scope issue try changing $id="ASDSSgdfsfsdfS"; to $head->id = "ASDSSgdfsfsdfS";, then change head.php to be the following:
<h1><center style="background:orange; border-radius:15px;">LOGO</center></h1>
<br><?php print_r($this->id)?>
<div>BANNER <div>$id <?php echo $this->id ;?></div></div>
Simply reading a file with fread will not parse any PHP contained inside. Perhaps you are looking for something like:
http://php.net/manual/en/function.require.php
Using require() is basically like copying and pasting the required file directly where your require() statement is. This means that the required file would only be able to use variables that are within the scope of where the require() statement is.
Because you are reading the contents of head.php and echoing them verbatim; nowhere do you make PHP compile and run that file as code. You could do that using include('head.php'), but that would not work blindly because you also have to make sure that $id is in scope at the point you do the include.
However that's not as easy as it sounds because it is not possible to automatically "pack the whole local scope" for later use because automatically implies code inside a function, and the very act of calling that function causes the scope to change.
My thought is that you are probably using mod_head is passing in a php file so it can be used as a view, right?
if thats the case, usually there is also a way to include a variable so that way it exists in the other file, head.php.. since there isn't one, you'd need to create a method for that. and then make that fread an include or require instead.
If not, and you're simply loading the file, the other comments about fread not parsing is totally correct and you will not be able to simply access $id from the other file.
im working with arrays and replacing certain values with html code so it gets outputed properly, but stays neat and html-free when being stored. following is the dumbed down code.
<?php
$file = 'somefile.php';
$replace = array('<1>','<2>');
$this = array("<div class=\"post\"><p>",
"</p></div>"
);
str_replace($replace,$this,$file);
?>
problem i have is that i need some php inserted as well, so that only a user with admin privileges will see output. i have tried putting the php into the $this array however it doesnt get processed as php. i would need an
if ($userIsAdmin) {
before the
<div>
and
}
after the
</div>
any suggestions? thanks in advance!
You can generate the unprocessed php code, store it to a variable $code and the evaluate it:
$result = eval("?>" . $code . "<?");
However, I don't think it's a good idea. The code will be unclear and you can introduce a lot of security issues.
With my data files I use with sites I usually include some PHP code in them to prevent them being directly accessed, such as below..
<?php
if (defined("VALID")) {
?>
html code here
<?php
} else {
die('Restricted Access.');
}
?>
Now this works fine when I do a simple include..... however I am using one of these files to do some replacements in & hence need to make use of file_get_contents(); however when using this, not only do I get the HTML code, I obviously also get the PHP code returned with it..... this ends up going in the source, which I do NOT want.
Is there any way around this? Perhaps stripping the PHP code? Any better ways/suggestions?
If you want to make replacements on an output of a script try using output buffering.
Instead of file_get_contents('your-php-script.php') do this:
ob_start();
include('your-php-script.php');
$contents = ob_get_clean();
// do your replacements on a $contents
echo preg_replace("~<\?php(.*?)\?>~", "", $contents);
This should work to erase the PHP code in the file.
Why dont you use a hashed string in a session cookie to check it? I think its the best solution. So add to the cookie a hashed value, then check for that value on the file you need to check if its valid and voila!
Hope it helps!