php main file not seeing included file variable - php

OK most the answers on S.O. are BACKWARDS from what I am having an issue with. My variable is not seen FROM the included file. I have never ran into this, and it is messing everything up. I was using smarty, but since it cant see the var, I went back to old fashion php templating, and still cannot see the var.. here is the scripts in question (a gaming site ofc).
INCLUDED FILE: prep_feature_game.php
foreach($games->gamexml->game as $game){
$d = parseDate($game['releasedate']);
$game['date_month'] = $d['month'];
$game['date_day'] = $d['day'];
$game['date_year'] = $d['year'];
$game['image_60x40'] = showImage($game['foldername'],'60x40');
$game['genrelist'] = displayGenreList($game['gameid']);
//debugPrint($game);die();
$game_list[] = $game;
}
CONTROL PAGE:
switch ($page) {
default:
include('includes/prep_feature_game.php');
// NOTE $GAME_LIST HAS THE ARRAY ^
display('feature_game');
break;
AND THE DISPLAY FILE: FEATURE_GAME.PHP
<h1><?=SITE_NAME;?>'s Featured Games!</h1>
<table>
<?PHP
// TEST FOR $GAME_LIST
foreach($game_list as $field=>$data){
echo $data['gamename'];
}
/*
this produces the error:
Notice: Undefined variable: game_list in C:\xampp\htdocs\ix2\themes\missingpiece\feature_game.php on line 7
*/
}// end switch page
I have many variations over the years of these templates, and the "included" files have never produced this kind of error. I know that included files act as though the script contained is inserted where the include is at.
why is the receiving script unable to see the variable produced from the included script. These are all included, so it should not be an issue
PS
I noticed this started happening when I upgraded from php v5 - php v7, doubt it has anything to do with that, I havent checked that yet since include files are pretty standard.

Your question is not showing the definition of the display function, but unless you declare $game_list as a global within that function it is out of scope. If you include the template it'll work.
include('includes/prep_feature_game.php');
include('feature_game.php');
Or add parameters to your display function.
display('feature_game', $game_list);
Or if you're using smarty you must assign the variable first.
$smarty->assign('game_list', $game_list);
$smarty->display('feature_game.tpl');

Related

How to know if a script was included inside another script

I am new to PHP and very likely I am using the incorrect approach because I am not used to think like a PHP programmer.
I have some files that include other files as dependencies, these files need to have global code that will be executed if $_POST contains certain values, something like this
if (isset($_POST["SomeValue"]))
{
/* code goes here */
}
All the files will contain this code section, each one it's own code of course.
The problem is that since the files can be included in another one of these files, then the code section I describe is executed in every included file, even when I post trhough AJAX and explicitly use the URL of the script I want to POST to.
I tried using the $_SERVER array to try and guess which script was used for the post request, and even though it worked because it was the right script, it was the same script for every included file.
Question is:
Is there a way to know if the file was included into another file so I can test for that and skip the code that only execute if $_POST contains the required values?
Note: The files are generated using a python script which itself uses a c library that scans a database for it's tables and constraints, the c library is mine as well as the python script, they work very well and if there is a fix for a single file, obviously it only needs to be performed to the python script.
I tell the reader (potential answerer) about this because I think it makes it clear that I don't need a solution that works over the already existant files, because they can be re-generated.
From the sounds of it you could make some improvements on your code structure to completely avoid this problem. However, with the information given a simple flag variable should do the trick:
if (!isset($postCodeExecuted) && isset($_POST["SomeValue"]))
{
/* code goes here */
$postCodeExecuted = true;
}
This variable will be set in the global namespace and therefore it will be available from everywhere.
I solved the problem by doing this
$caller = str_replace($_SERVER["DOCUMENT_ROOT"], "", __FILE__);
if ($_SERVER["REQUEST_METHOD"] === "POST" and $caller === $_SERVER["PHP_SELF"])
performThisAction();

Syntax error when trying to grab and run code from github

I'm trying to get a page from github to be read and executed with PHP upon page load due to it updating often, although while I've managed to format the page to what should be correct, it still doesn't seem to work.
Basically, I've used the file_get_contents to grab all of the information on the page (if I get this working it should potentially work with any page), removed all comments, and now I just need the code to run.
I've heard eval is unsafe, but it's only a personal website and I trust the github page isn't going to use malicious code, but I'm getting a "syntax error, unexpected T_CONSTANT_ENCAPSED_STRING", when the page runs, despite the fact I haven't touched the code and it runs fine if copied and pasted.
Here is the code I've used to remove the comments, I can't see any problem on or around line 391 where it says the error is, http://www.compileonline.com/execute_php_online.php works if you copy and paste it in
#get page data, remove intro comment (unefficient but can be rewritten once working)
$browsercoderaw = explode("*/",file_get_contents("https://raw.githubusercontent.com/cbschuld/Browser.php/master/lib/Browser.php"));
for($i==1;$i<count($browsercoderaw);$i++){
if($i>1){$browsercodejoined.=" */";} # replace */ for other comments so they can be removed later
$browsercodejoined.=$browsercoderaw[$i];}
$browsercode = nl2br(str_replace("?>","",$browsercodejoined));
#remove all /* comments
$commentremove = explode("/*",$browsercode);
$browsercode2 = $commentremove[0];
for($j==1;$j<count($commentremove);$j++){
$commentsplit = explode("*/",$commentremove[$j]);
$browsercode2.=$commentsplit[1];
}
#remove all // comments
$commentremove2 = explode("<br />",$browsercode2);
$browsercode3 = $commentremove2[0];
$linenum = 0;
for($k==1;$k<count($commentremove2);$k++){
$commentsplit = explode("//",$commentremove2[$k]);
if(strlen(trim($commentsplit[0]))>0){
$browsercode3.=$commentsplit[0];
$browsercodereadable.=$linenum.". ".$commentsplit[0]."<br>";
$linenum++;
}
}
echo $browsercodereadable;
eval($browsercode3);
?>
Also, if there is a better way of doing this please say so, I tried include but the webhost doesn't allow fetching urls from other domains. To be fair, I'm not entirely sure if it's the correct use of eval, but it sounds like it should potentially work.

Smarty internal error

Looking through the server PHP error log has revealed that a certain type of PHP Notices comes up as the most frequent, and it has to do with Smarty. I've found a question which seems to describe the same error, but there is actually no answer.
The notice is the following (there are different variables stated as undefined):
PHP Notice: Undefined variable: is_admin in /usr/share/php/Smarty/sysplugins/smarty_internal_data.php on line 291
I wonder how I can possibly debug this one since no data (like template name) is provided.
Below I'm gonna give you some idea of the code (which you can read in full detail here since it's all open-source).
So there is one global smarty object which is created in a file called header.php, and further in the same file some global smarty variables are set, including the one from the notice above:
//init Smarty
require_once('Smarty.class.php');
$smarty = new Smarty();
...
$smarty->assign('is_admin', is_admin() ? 1 : 0);
This header.php is then included in every file that needs to show some HTML by calling $smarty->display(...). So I presume that in any file where the $smarty object is present this object has a variable called is_admin. However, it doesn't seem to be the case.
Additionally, "normal" Smarty warnings about unset variables look differently:
PHP Notice: Undefined index: sent_id in /var/www/smarty_dir/templates_c/a5aab2c66c44442365a39981ba9be18e0a1f11ad.file.history.tpl.cache.php on line 123
Any ideas?
Upd.
I've read some logs and I see that such warnings seemingly arise when a user enters a page and gets 302 HTTP status. This may be due to the following code (which is placed after smarty constructor call but before the variables are assigned:
//cookie check
if (!is_logged() && isset($_COOKIE['auth'])) {
if ($user_id = check_auth_cookie()) {
if (user_login('', '', $user_id, $_COOKIE['auth'])) {
header("Location:".$_SERVER['REQUEST_URI']);
return;
}
}
}
So I guess I should move $smarty initialisation after this block and it is likely to help. Still I'm curious about how it relates to the issue.
I think I got it. The issue is really in the piece of code including
header("Location:".$_SERVER['REQUEST_URI']);
return;
My error is that return does not act like exit in this case, because when you call return from a file which is required or incuded into another one, you just return the control to that another file. I wasn't aware of this feature.
'is_admin', is_admin() you are trying to assign the function is_admin() as a variable {$is_admin} to the Smarty template, which according to me, is impossible.
Assign the return result to a variable i.e.:
$var = is_admin();
$smarty->assign('is_admin', $var);

PHP Call to undefined function

I am trying to call a function from another function. I get an error:
Fatal error: Call to undefined function getInitialInformation()
in controller.php on line 24
controller.php file:
require_once("model/model.php");
function intake() {
$info = getInitialInformation($id); //line 24
}
model/model.php
function getInitialInformation($id) {
return $GLOBALS['em']->find('InitialInformation', $id);
}
Things already tried:
Verified that the require_once works, and the file exists in the specified location.
Verified that the function exists in the file.
I am not able to figure this out. Am I missing something here?
How to reproduce the error, and how to fix it:
Put this code in a file called p.php:
<?php
class yoyo{
function salt(){
}
function pepper(){
salt();
}
}
$y = new yoyo();
$y->pepper();
?>
Run it like this:
php p.php
We get error:
PHP Fatal error: Call to undefined function salt() in
/home/el/foo/p.php on line 6
Solution: use $this->salt(); instead of salt();
So do it like this instead:
<?php
class yoyo{
function salt(){
}
function pepper(){
$this->salt();
}
}
$y = new yoyo();
$y->pepper();
?>
If someone could post a link to why $this has to be used before PHP functions within classes, yeah, that would be great.
This was a developer mistake - a misplaced ending brace, which made the above function a nested function.
I see a lot of questions related to the undefined function error in SO. Let me note down this as an answer, in case someone else have the same issue with function scope.
Things I tried to troubleshoot first:
Searched for the php file with the function definition in it. Verified that the file exists.
Verified that the require (or include) statement for the above file exists in the page. Also, verified the absolute path in the require/include is correct.
Verified that the filename is spelled correctly in the require statement.
Echoed a word in the included file, to see if it has been properly included.
Defined a separate function at the end of file, and called it. It worked too.
It was difficult to trace the braces, since the functions were very long - problem with legacy systems. Further steps to troubleshoot were this:
I already defined a simple print function at the end of included file. I moved it to just above the "undefined function". That made it undefined too.
Identified this as some scope issue.
Used the Netbeans collapse (code fold) feature to check the function just above this one. So, the 1000 lines function above just collapsed along with this one, making this a nested function.
Once the problem identified, cut-pasted the function to the end of file, which solved the issue.
Many times the problem comes because php does not support short open tags in php.ini file, i.e:
<?
phpinfo();
?>
You must use:
<?php
phpinfo();
?>
Your function is probably in a different namespace than the one you're calling it from.
http://php.net/manual/en/language.namespaces.basics.php
I happened that problem on a virtual server, when everything worked correctly on other hosting.
After several modifications I realized that I include or require_one works on all calls except in a file.
The problem of this file was the code < ?php ? > At the beginning and end of the text.
It was a script that was only < ?, and in that version of apache that was running did not work
This is obviously not the case in this Q,
but since I got here following the same error message I though I would add what was wrong with my code and maybe it will help some one else:
I was porting code from JS to PHP and ended up having a class with some public method.
The code that was calling the class (being code that originated from JS) looked something like:
$myObject.method(...)
this is wrong because in PHP it should look like this:
$myObject->method(...)
and it also resulted with "PHP Call to undefined function".
change to use -> and the problem was solved.
Presently I am working on web services where my function is defined and it was throwing an error undefined function.I just added this in autoload.php in codeigniter
$autoload['helper'] = array('common','security','url');
common is the name of my controller.
Please check that you have <?PHP at the top of your code. If you forget it, this error will appear.

Call to undefined function from another php file

Alright this is what my code looks like
index.php
require_once($WebsiteRoot . "/include/testfile.php");
TestFunction();
/include/testfile.php
function TestFunction()
{
echo "It Works";
}
And it gives me the error:
Fatal error:
Call to undefined function TestFunction() in /path/index.php on line 49
Any idea what i'm doing wrong?
Thanks
You haven't included a <?php tag in the included file, so it's just interpreted as plaintext input.
Remember... there's no such thing as a PHP script. There's only files which contain PHP code blocks. Without at least one <?php opening tag, the PHP interpreter will never be invoked and the file's contents will simply be treated as output.
try calling another function from testfile.php, if this is'nt working, its something with the include. Add the code:
error_reporting(E_ALL | E_WARNING | E_NOTICE);
ini_set('display_errors', TRUE);
to the top of index.php and refresh the browser to see your errors, try debugging from there.
The problem that i can forsee is that you are using a URL instead of a path, your $websiteRoot variable should contain a path like:
$websiteRoot = "/var/www/html/websiteName";
OR
$websiteRoot = "C://xampp/htdocs/websiteName";
instead of a URL like:
$websiteRoot = "http://www.somesite.com";
I had a similar issue. I dug into the PHP in the included file and found an invalid PHP tag. I had <? instead of <?php. PHP 7.2 and earlier forgave that, but PHP 7.3 was throwing that same error you faced.
Make sure you're including the file you think you are. If your index.php page looks exactly like you've stated, then it won't return anything.
If you want to link to the same location from anywhere on the site without worrying about relative locations, then at the beginning of the file, put:
$WebsiteRoot=$_SERVER['DOCUMENT_ROOT'];
And it should work fine, provided your file would be located at http://mywebsite.com/include/testfile.php
Try renaming the included file.
I had an included file with the name "system.php". It looked as if the include command was just skipped. Even with the most strict error reporting there was no message and even an echo command in the main body of the included file did not produce output. It had worked ok under PHP 5 but after the upgrade to a 7.2 environment these problems arose. After much effort - I forgot how - I managed to get an error message. It said there was a conflict with a PEAR class with the name "system". Yet my file didn't contain any class, just variables and functions. Anyway, giving the file another name than "system.php" worked for me.
I hope someone else can add a more technical comment on what was going wrong here.

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