Passing file url with variables into Postmark causing file not found issue - php

So I'm using Postmark to send emails and the class I have requires a variable as the message body, as below:
$email->to(Input::post('email'))->subject("Verify Your Email Address")->html_message($html)->send();
This works fine if I set $html as just plain html.
What I am trying to do is send the contents of another php file from my site as this html.
I have tried:
$Vdata = file_get_contents('verification.php');
this works fine but as soon as I try and pass variables in it gives me an error:
file not found error
And sends a blank email, for example:
$Vdata = file_get_contents('verification.php?url=blah');
Essentially I just need $html to be the contents of verification.php?url=blah so that I can pass in variables to that file.
Can anyone help?

You're doing a local file inclusion, which means filenames ONLY. URLs are not permitted (query strings in particular) because you're NOT doing an HTTP request. PHP is going to look for a file whose name literally contains ?, u, r, etc... which of course doesn't exist.
If you want to use query strings, then you have to use a full-blown absolute URL, including the protocol:
include('http://....?url=...');
However, this is incredibly inefficient, and also highly dangerous. Since you're now EXECUTING the file specified in the url. you're going to get its output, not the raw PHP code in the file.
If you want to pass data to an included file, then just variables:
$foo = 'bar';
include('test.php');
and use look for/use those variables in the file.

Related

file_get_contents not working when parameters included in a relative url

I am trying to create a simple web service that will give a result depending on parameters passed.
I would like to use file_get_contents but am having difficulties getting it to work. I have researched many of the other questions relating to the file_get_contents issues but none have been exactly the situation I seem to having.
I have a webpage:
example.com/xdirectory/index.php
I am attempting to get the value of the output of that page using:
file_get_contents(urlencode('https://www.example.com/xdirectory/index.php'));*
That does not work due to some issue with the https. Since the requesting page and the target are both on the same server I try again with a relative path:
file_get_contents(urlencode('../xdirectory/index.php'));
That does work and retrieves the html output of the page as expected.
Now if I try:
file_get_contents(urlencode('../xdirectory/index.php?id=100'));
The html output is (should be): Hello World.
The result retrieved by the command is blank. I check the error log and have an error:
[Fri Dec 04 12:22:54 2015] [error] [client 10.50.0.12] PHP Warning: file_get_contents(../xdirectory/index.php?id=100): failed to open stream: No such file or directory in /var/www/html/inventory/index.php on line 40, referer: https://www.example.com/inventory/index.php
The php.ini has these set:
allow_url_fopen, On local and On master
allow_url_include, On local and On master
Since I can get the content properly using only the url and NOT when using it with parameters I'm guessing that there is an issue with parameters and file_get_contents. I cannot find any notice against using parameters in the documentation so am at a loss and asking for your help.
Additional Notes:
I have tried this using urlencode and not using urlencode. Also, I am not trying to retrieve a file but dynamically created html output depending on parameters passed (just as much of the html output at index.php is dynamically created).
** There are several folks giving me all kind of good suggestions and it has been suggested that I must use the full blown absolute path. I just completed an experiment using file_get_contents to get http://www.duckduckgo.com, that worked, and then with a urlencoded parameter (http://www.duckduckgo.com/?q=php+is+cool)... that worked too.
It was when I tried the secure side of things, https://www.duckduckgo.com that it failed, and, with the same error message in the log as I have been receiving with my other queries.
So, now I have a refined question and I may need to update the question title to reflect it.
Does anyone know how to get a parameterized relative url to work with file_get_contents? (i.e. 'file_get_contents(urlencode('../xdirectory/index.php?id=' . urlencode('100'))); )
Unless you provide a full-blown absolute protocol://host/path-type url to file_get_contents, it WILL assume you're dealing with a local filesystem path.
That means your urlencode() version is wrongly doing
file_get_contents('..%2Fxdirectory%2Findex.php');
and you are HIGHLY unlikely to have a hidden file named ..%2Fetc....
call url with domain, try this
file_get_contents('https://www.example.com/inventory/index.php?id=100');
From reading your comments and additional notes, I think you don't want file_get_contents but you want include.
see How to execute and get content of a .php file in a variable?
Several of these answers give you useful pointers on what it looks like you're trying to achieve.
file_get_contents will return the contents of a file rather than the output of a file, unless it's a URL, but as you seem to have other issues with passing the URI absolutely....
So; you can construct something like:
$_GET['id'] = 100;
//this will pass the variable into the index.php file to use as if it was
// a GET value passed in the URI.
$output = include $_SERVER['DOCUMENT_ROOT']."/file/address/index.php";
unset($_GET['id']);
//$output holds the HTML code as a string,
The above feels hacky trying to incorporate $_GET values into the index.php page, but if you can edit the index.php page you can use plain PHP passed values and also get the output returned with a specific return $output; statement at the end of the included file.
It has been two years since I used PHP so I am just speculating about what I might try in your situation.
Instead of trying fetching the parsed file contents with arguments as a query string, I might try to set the variables directly within the php script and then include it (that is if the framework you use allows this).
To achive this I would use pattern:
ob_start -> set the variable, include the file that uses the variable -> ob_get_contents -> ob_end_clean
It is like opening your terminal and running the php file with arguments.
Anyway, I would not be surprised if there are better ways to achieve the same results. Happy hacking :o)
EDIT:
I like to emphasize that I am just speculating. I don't know if there are any security issues with this approach. You could of course ask and see if anyone knows here on stackoverflow.
EDIT2:
Hmm, scrap what I said last. I would check if you can use argv instead.
'argv' Array of arguments passed to the script. When the script is run on the command line, this gives C-style access to the command line parameters. When called via the GET method, this will contain the query string. http://php.net/manual/en/reserved.variables.server.php
Then you just call your php script locally but without the query mark indicator "?". This way you can use the php interpreter without the server.
This is likely to be the most general solution because you can also use argv for get requests if I am understanding the manual correctly.

How to get the name of a file sent to a php://input via wget post-file

I wrote a script that receives an image file through file_get_contents('php://input') and does some other magic. The file is sent from a client using the wget post-file=blahblah.blah command and it's all working fine.
The issue that I have is that I need to use the name of the received image file as a string for processing purposes and file_get_contents() gives me the content of the file but not the name of it.
Would anyone know how I can get the name?
Any answer that can put me in the right direction would be well appreciated.
You can't. The --post-file option to wget sends the contents of the file as a raw POST request; it does not treat it as a file upload, so the name of the file is not transmitted. Per the wget manual page:
--post-data=string
--post-file=file
Use POST as the method for all HTTP requests and send the specified
data in the request body. --post-data sends string as data,
whereas --post-file sends the contents of file. Other than that,
they work in exactly the same way. In particular, they both expect
content of the form "key1=value1&key2=value2", with percent-
encoding for special characters; the only difference is that one
expects its content as a command-line parameter and the other
accepts its content from a file. In particular, --post-file is not
for transmitting files as form attachments: those must appear as
"key=value" data (with appropriate percent-coding) just like
everything else.
There are numerous ways to go about getting the file name. I believe the classic way would be using url parameters ( POST your file to example.com/your-script.php?name=some-file-name-here ) but an alternative and possibly cleaner way would be using custom http headers: 'X-Filename: your-file-name'.
wget --header "X-Filename: your-file-name" --post-file /your/file
Then in PHP check for the headers ( using apache_request_headers() for example ).
PHP has a built in superglobal variable $_FILES, it works similar to $_POST.
$_FILES['/*the html object name goes here*/']['name']
The ['name'] part of the array returns the actual name of the file uploaded from the html object name.
So to clarify, if your html was <input type="file" name="screenshot" />
your PHP would be $name = $_FILE['screenshot']['name']
$name now stores the string that holds the name of the file.

Save binary request body as file

I wrote a program that reads a binary file into the RAM and then sends it using an HTTP request to my server. It uses the PUT method and the binary file is (in) the body.
Now how can I tell my server to receive and safe the file in a folder?
If possible without any additional libraries that I would need to download (unless it's more efficient).
I know, there are some similar threads to this one, but they either they where about receiving text or they were about doing it with libraries or there simply was no sufficient answer.
I'd also like to know, if it would be more efficient or smarter to use the POST method or any other instead of PUT.
You can get at the data by opening a stream to php://input, like so:
$datastr = fopen('php://input',rb);
if ($fp = fopen('outputfile.bin', "wb")){
while(!feof($datastr)){
fwrite($fp,fread($datastr,4096)) ;
}
}
As to whether to use POST or anything else depends on what is happening with the data, and whether you care about being RESTful or such. See other questions/answers, indempotency.
The advantage I would see with using POST is that it's more commonly used (on most submission forms where you upload a file), and therefore has more support from within PHP and html.

Can't pass a PHP variable to file_get_contents()

I am a newbie coder trying to build a simple web app using PHP. I am trying to send an HTML email that has a variable that will change each time it is sent. The code to initiate the email is 'email.php' and contains:
$body = file_get_contents('welcome/green2.html.php');
Within the 'green2.html.php' file, I have a variable called $highlight that needs to be populated. The $highlight variable is defined within the 'email.php' file. I had tried to simply add within the 'green2.html.php' file, however it is not being parsed. I get a blank space where the variable should be when it is output.
Also, I have done an include 'welcome/green2.html.php' within the 'email.php' file. When I echo it, the $highlight var is shown on the resulting page, but not if I echo $body.
Any help would be much appreciated!
Have you tried the str_replace function? http://php.net/manual/en/function.str-replace.php.
Add a placeholder in HTML (for instance #name# for name, #email# for email), and then use the string replace function once you've loaded the content of the file.
$bodytag = str_replace("#name#", $name, $myfile);
Loading a file via file_get_contents() will not cause it to be parsed by PHP. It will simply be loaded as a static file, regardless of whether it contains PHP code or not.
If you want it to be parsed by PHP, you would need to include or require it.
But it sounds like you're trying to write a templating system for your emails. If this is what you're doing, you'd be better off not having it as PHP code to be parsed, but rather having placeholder markers in it, and then using str_replace() or similar functions to inject variables from your main program into the string.
Hope that helps.
Use http://php.net/manual/en/function.sprintf.php put a %s in your code instead of the variable read the content and put the string into the sprintf with the variable you want to put that's it. Hope this will help.

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;
}
}

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