What's wrong with this include statement? - php

This function is returning the content of the file rather the result of fetch_link_settings_overide() within it.
The issue is not with the overide function as after the initial error I commented out my modification just to be sure it wasn't something I had done there.
function fetch_link_settings(){
include( plugins_url()."/plugin-child/plugin_overrides.php");
return fetch_link_settings_override();
}
Adding the content of the derived function plugin-child/plugin_overrides.php as we are not getting anywhere currently.
function fetch_link_settings_override(){
global $post;
// If the destination url is set by the user, use that. Otherwise, use the permalink
$destination_url = get_post_meta($post->ID, '_promo_slider_url', true);
// ASAdd additional place to look in the case of the post being via the PODS advert track
if( ! $destination_url )
$destination_url = get_post_meta($post->ID, 'okd_advert_link', true);
if( ! $destination_url )
$destination_url = get_permalink($post->ID);
// If the target attribute is set by the user, use that. Otherwise, set it to _self
$target = get_post_meta($post->ID, '_promo_slider_target', true);
if( ! $target ) $target = '_self';
// Setup the disable links variable
$disable_links = get_post_meta($post->ID, '_promo_slider_disable_links', true);
return compact('destination_url', 'target', 'disable_links');
}

You write this:
include( plugins_url()."/plugin-child/plugin_overides.php");
Why is plugins_url() there? The include function is strictly based on the file system:
The `include` statement includes and evaluates the specified file.
As explained in the WordPress docs, the plugins_url() would give you the full web URL which is 100% different than the file system WordPress is installed on:
Retrieves the absolute URL to the plugins directory (without the
trailing slash) or, when using the $path argument, to a specific file
under that directory.
So perhaps it should be like this:
include("/plugin-child/plugin_overides.php");
Or perhaps you need the plugin_dir_path()?
include(plugin_dir_path( __FILE__ ) . "/plugin-child/plugin_overides.php");
But that seems wrong. Where would /plugin-child/plugin_overides.php? Try doing this:
include("/full/path/to/wordpress/and/this/plugin-child/plugin_overides.php");
Just replace /full/path/to/wordpress/and/this/ with the actual file system path to /plugin-child/plugin_overides.php.
EDIT: Since the original poster is persistent in using plugins_url() despite all of the suggestions otherwise, here is my detailed response:
…you said “you cannot load raw functions via a URL with include” well
this is not relevant because even if I add $some_var = 'smith'; as the
first statement in the included file, it is not visible in the
function using the include.
Apologies. Functions, classes, strings, constants… Just about anything that you want to be raw, unprocessed PHP will simply not be passed via an http:// or https:// URL because Apache will parse the PHP instructions & simply return the output of that file and not the raw, unprocessed contents of the PHP in that file.
Additionally the original poster contents the following:
You can’t help me because what you are saying does not make sense or
you are not explaining yourself adequately. Look at these examples:
include realpath(dirname(FILE) . "/" . "relative_path");
include("data://text/plain;base64,".base64_encode($content));
include("data://text/plain,".urlencode($content));
All taken from the official PHP documentation. They all use
functions returning components that are concatenated with the rest of
the url. I also tried this typing the filepath explicitly and the
result is the same.
The examples cited are as follows:
include realpath(dirname(FILE) . "/" . "relative_path");
This is a filesystem level include which is the most common way PHP files are included into other files.
include("data://text/plain;base64,".base64_encode($content));
include("data://text/plain,".urlencode($content));
These are both data URLs. Not http or https. So again when you use plugins_url() what you are getting is a full http:// or https:// URL in which Apache parses the PHP instructions & simply return the output of that file and not the raw, unprocessed contents of the PHP in that file. Or as very clearly explained in the PHP documentation you are linking to; emphasis mine:
If "URL include wrappers" are enabled in PHP, you can specify the file
to be included using a URL (via HTTP or other supported wrapper - see
Supported Protocols and Wrappers for a list of protocols) instead of a
local pathname. If the target server interprets the target file as PHP
code, variables may be passed to the included file using a URL request
string as used with HTTP GET. This is not strictly speaking the same
thing as including the file and having it inherit the parent file's
variable scope; the script is actually being run on the remote server
and the result is then being included into the local script.
Going back to your example, you say now the contents of plugin_overides.php is $some_var = 'smith';. How exactly? If it is a PHP file like this:
<?php
$some_var = 'smith';
?>
When you call that file via a URL generated by the following code:
include(plugins_url() . "/plugin-child/plugin_overrides.php");
Assuming your website is http://some.cool.website/ the you are basically making a call like this:
http://some.cool.website/plugin-child/plugin_overides.php
So the output of plugin_overides.php would be 100% blank. If you wanted to get output of that file, you could do the following:
<?php
$some_var = 'smith';
echo $some_var;
?>
And that would return smith. Meaning the absolute ONLY output you would get from that call is pure text. Nothing else.
Now I see you actually have posted the contents of plugin_overides.php. My example explanation above is still apt, but still a basic question. This is your function; just the interface & return for example:
function fetch_link_settings_override(){
// Other code removed. Just a structural illustration for now.
return compact('destination_url', 'target', 'disable_links');
}
Do you actually call fetch_link_settings_override() in plugin_overides.php when it runs? Well, if that function does not run, then there is 100% no way you will ever get any output. But assuming good faith, look at your return statement here:
return compact('destination_url', 'target', 'disable_links');
If you are returning compact, then you are returning an array. You cannot simply return a bare array as a URL call like this http://some.cool.website/plugin-child/plugin_overides.php. The output at most would be simply the word Array.
If the goal is to take that array & do something, then you should use json_encode in fetch_link_settings_override and then use json_decode on the receiving side of that. So the return statement would be something like this:
return json_encode(compact('destination_url', 'target', 'disable_links'));

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();

Read parent URL from included file

OK, so I have searched around for long enough to finally post this one here. Sure enough, it has been asked before a zillion time...
What I have, is one file, which includes another. No magic here. The trouble is, the included file then includes another file, which... includes yet another... Yep, a pain. Actually it's all working quite nicely, except that I now wanted to read the URL of the original file in the last of the included files.
So I thought in the original file, file_1.php I just say
$var_foo = dirname(__FILE__);
or
$var_foo = $_SERVER['SCRIPT_NAME'];
and then read that value in the first include, file_2.php, passing it on like
$var_foo_2 = $var_foo;
then in file_3.php
$var_foo_3 = $var_foo_2
etc, until I arrive at the final file, file_4.php, where I'd like to know the exact value of the original file's URL. Passing it on the first level works OK, but then it gets lost somewhere along the way. Tried going GLOBAL in file_1 -- to no avail.
Since file_3 and file_4 must both execute to produce data, setting a breakpoint a la echo / exit to spoof the current value (if any) is no option. I can live without that particular value, but I just would like to have it -- for the fun of it... Any ideas how to accomplish this?
Your examples use filesystem paths, not "URLs";I am assuming the filepath of the parent file is what you actually want.
You don't need to "pass" the variable on each included page. It will automatically be available to the code on the new page.
(If it is not, you may not be in the right scope: e.g., if you're inside a class or function, you'll need to pass it deliberately or use some other method - global, maybe, or even define the filename as a constant instead of a variable.)
main script
$parent_filename = __FILE__;
// alternatively
// define( 'PARENT_FILENAME',__FILE__ );
include "other-file.php";
other-file.php
include "other-dir/somefile.php";
other-dir/somefile.php
print $parent_filename;
// alternatively
// print PARENT_FILENAME;
/* prints something like:
/path/to/main.php
*/
As mentioned before, the issue has been solved like so:
set variable before the first include
add variable to query string
Thanks all for the input, appreciated.

How to Capture PHP Output into a Variable

I've been wanting to do this because my site does about 3 HTTP requests per page load, because each PHP's output is retrieved with cURL. This is too slow, and I want to avoid using cURL. I found this question on Stack Overflow, and it basically does what I want. The accepted answer's suggestion is to use ob_start(); to start getting output then use ob_get_clean(); to put the output into a variable. My issue now is that the PHP scripts I'm trying to capture output from need variables passed to them using HTTP Get. The access those variables like this:
$term = $_GET['term'];
And I don't want to change that, because although I'm going to access these PHP scripts' outputs from another PHP script, I'm also planning on accessing them from elsewhere. So, is there a way to fool these PHP scripts into accepting some arguments through Get, then capturing their outputs with the method suggested above?
You can $_GET variables from any php script if its set (use isset to check that). Then just cURL to such url's will work.
If you have changed the method to POST earlier, you can use CURLOPT_HTTPGET. See the curl_setopt functions page (http://www.php.net/manual/en/function.curl-setopt.php) for more details.
For a non-cURL method, use jQuery ajax. It is quite simple to use, just read the documentation here.
EDIT: This is what you wanted (haven't checked the code though)
<?php
function get_include_contents($filename, $get) {
if (is_file($filename)) {
ob_start();
$_GET = array();
while (list($key, $val) = each($get)) {
$_GET[$key]=$val;
}
include $filename;
return ob_get_clean();
}
return false;
}
$string = get_include_contents('somefile.php', array('param1'=>'x', 'param2'=>'y'));
?>
And I don't want to change that, because although I'm going to access these PHP scripts' outputs from another PHP script, I'm also planning on accessing them from elsewhere. So, is there a way to fool these PHP scripts into accepting some arguments through Get, then capturing their outputs with the method suggested above?
Your question is a bit unclear as to why you're using cURL in the first place. If your scripts are on the same server, you can simply set the correct $_GET variables and use:
<?php
ob_start( );
// include the file.
$_GET['foo'] = 'bar';
include 'the_file.php';
$output = ob_get_clean( );
If your scripts are located on another server, where include is not viable, you will always have to do a HTTP request to get their contents, regardless of whether your this with cURL or Ajax, or sockets for all I care.
well you can access a $_GET from any script loaded as long as its in the URI, the variable $term can be used in any script. You can also include the script.
When you include a script you can access some of its content after the include.

passing URL variables to exec() with php

I have a dedicated server that I use to crunch lots of data. The way I have it now, I can open a script with a process ID like example.php?ex_pid=123 and just let it go. It downloads a small portion of data, processes it, then uploads it into a database then starts again.
Ideally, I would like to call example.php?ex_pid=123 directly and not by passing a variable to example.php like exec('./example.php'.' '.EscapeShellArg($variable)); to keep it from acting globally.
I don't care about the output, if it could execute in the background, that would be brilliant. The server is an Ubuntu distribution btw.
Is this even possible? If so, any help and examples would be more then appreciated.
You could do something like:
exec("./example.php '".addslashes(serialize($_GET))."');
And then in example.php do something like this:
count($_GET) == 0 && $_GET = unserialize(stripslashes($_SERVER['argv'][1]))
The main issue with that is that ?ex_pid is GET data which is generally associated with either including the file or accessing it through a browser. If you were including the file or accessing it from a web browser this would be trivial, but running it as CLI, your only option would be to pass it as an argument, unfortunately. You can pass it as ex_pid=123 and just parse that data, but it would still need to be passed as an argument but doing that you could use parse_str() to parse it.
Depending on what the script does, you could call lynx to call the actual page with the get data attached and generate a hash for an apikey required to make it run. Not sure if that is an option, but it is another way to do it how you want.
Hope that helps!
I had a real problem with this and couldn't get it to work running something like example.php?variable=1.
I could however get an individual file to run using the exec command, without the ?variable=1 at the end.
What I decided to do was dynamically change the contents of a template file , depending on the variables I wanted to send. This file is called template.php and contains all the code you would normally run as a $_GET. Instead of using $_GET, set the value of the variable right at the top. This line of code is then searched and replaced with any value you choose.
I then saved this new file and ran that instead.
In the following example I needed to change an SQL query - the template file has the line $sql="ENTER SQL CODE HERE";. I also needed to change the value of a a variable at the top.
The line in template.php is $myvar=999999; The code below changes these line in template.php to the new values.
//Get the base file to modify - template.php
$contents=file_get_contents("template.php");
$sql="SELECT * FROM mytable WHERE foo='".$bar."'";
$contents=str_replace("ENTER SQL CODE HERE",$sql,$contents);
//Another search
$contents=str_replace("999999",$bar,$contents);
$filename="run_standalone_code".$bar.".php";
//If the file doesnt't exist, create it
if(!file_exists($filename)){
file_put_contents($filename, $contents);
}
//Now run this file
$cmd="/usr/local/bin/php ".$filename." >/dev/null &";
exec($cmd);
I had completely forgotten about this question until #Andrew Waugh commented on it (and I got an email reminder).
Anyways, this question stemmed from a misunderstanding as to how the $argv array is communicated to the script when using CLI. You can pretty much use as many arguments as you need. The way I accomplish this now is like:
if (isset($argv)) {
switch ($argv[1]) {
case "a_distinguishing_name_goes_here":
$pid = $argv[2];
sample_function($pid);
break;
case "another_name_goes_here":
do_something_else($argv[2]);
break;
}
}

php include problem with urls with options ?view=task&others file not found

I really have read the other articles that cover this subject. But I seem to be in a slightly different position. I'm not using modrewrite (other articles).
I would like to 'include' a webpage its a 'Joomla php' generated page inside a php script. I'd hoped to make additions on the 'fly' without altering the original script. So I was going to 'precomplete' elements of the page by parasing the page once it was included I hadent wanted to hack the original script. To the point I can't include the file and its not because the path is wrong -
so
include ("/home/public_html/index.php"); this would work
include ("/home/public_html/index.php?option=com_k2&view=item&task=add"); this would not!
I've tried a variety of alternates, in phrasing, I can't use the direct route "http:etc..." since its a current php version so must be a reference to the same server. I tried relative, these work without the ?option=com_k2&view=item&task=add
It may be the simple answer that 'options' or variables can be passed.
Or that the include can't be used to 'wait' for a page to be generated - i.e. it will only return the html.
I'm not the biggest of coders but I've done alot more than this and I thought this was so basic.
this would work include ("/home/public_html/index.php?option=com_k2&view=item&task=add"); this would not!
And it never will: You are mixing a filesystem path with GET parameters, which can be passed only through the web server (utilizing a http:// call... But that, in turn, won't run the PHP code the way you want.)
You could set the variables beforehand:
$option = "com_k2";
$view = "item";
$task = "add";
include the file the normal way:
include ("/home/public_html/index.php");
this is assuming that you have access to the file, and can change the script to expect variables instead of GET parameters.

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