I'm having trouble executing the lessc compiler from PHP. I'm using Symfony, and have tried using the sfLESSPlugin, but have been unsuccessful. I've put my code in a filter that executes before the page renders, so that every time the page is refreshed, my LESS files are compiled into one CSS file (don't want to have recompile manually every time I make a change, at least while I'm developing). Here are the different variations that I've attempted:
$fs = new sfFilesystem();
$command = '/Users/jordanb/node/node_modules/less/bin/lessc less/bootstrap.less css/bootstrap.css';
try
{
$fs->execute($command, null, array($this, 'throwCompilerError'));
}
catch (RuntimeException $e)
{
return false;
}
This returns an error: "Problem executing command", with an error code of 127. Digging deeper into Symfony's execute(), it calls proc_open() and then proc_close(). Some research online told me that an error code of 127 means that the command was not found.
Running the exact same command on the command line works just fine.
To be extra sure, I executed chmod 777 on /Users/jordanb/node/node_modules/less/bin/lessc, just to make sure it wasn't a permissions issue. Still didn't work.
I also tried just "lessc" instead of the full path, which didn't work. I've added lessc to my classpath, so typing "which lessc" gives me "/Users/jordanb/node/node_modules/less/bin/lessc".
I also tried a simple:
shell_exec('lessc less/bootstrap.less css/bootstrap.css');
which didn't seem to do anything. I printed the output to the PHP error log as well as to a text file, and the output was empty in both cases. I also tried using the full path in this case.
You can make a shell script that compiles things the way you want:
compile_css.sh
/Users/jordanb/node/node_modules/less/bin/lessc less/bootstrap.less css/bootstrap.css
Then
chmod +x compile_css.sh
And finally call it from your script
$command = '/path/to/compile_css.sh';
This will give you more control over what you can execute, but if you still want to do it from php, try this:
$command = '/Users/jordanb/node/node_modules/less/bin/lessc "less/bootstrap.less css/bootstrap.css"';
// note the quotes around the arguments.
Related
I'm having trouble running the "exec" function in my PHP website. I am able to run it several times with an executable that just takes in a variable argument and returns some test message. However, when I use an executable that does some image processing, where I want to pass an image from the website that a user uploads as an argument, it does not seem to be executing the executable at all. I even have some cout commands in the executable to ensure its running, but these are not being displayed on the website. So I think that for some reason the php cannot run the executable? I am able to run it fine from my desktop...
Here's an example of the code that isn't working on my PHP website:
$imgtest1="/uploaded_files/me.jpg";
$imgtest2="/uploaded_files/clusteroutput.jpg";
$nosuppix = 400;
$noweight = 100;
$executabletest = exec("ImgProc $imgtest1 $nosuppix $noweight $imgtest2");
echo $executabletest;
Is there a way to debug or get an error output from the exec function? Is there something I'm missing when passing an image to the executable? The executable uses several DLL files which are in the same folder as the executable. Do they need to be packaged together for some reason? I apologize, but I really don't know what's left to test...
***Edit: I'm now able to get it to run if I write out all the code in the escapeshellcmd itself... how come I'm not able to just pass the variables?
$cmdinput = escapeshellcmd('SuperpixelsFinal "D:/WebPages/TALIA ART/TALIA ART/uploaded_files/me.jpg" 400 100 "D:/WebPages/TALIA ART/TALIA ART/uploaded_files/clusteroutput.jpg"');
For anyone else having this issue, it is because I was using escapeshellcmd when I should have used escapeshellarg for each individual string and then run the exec with double quotes using only variables like so:
$executabletest = exec("SuperpixelsFinal $imgtest1 $nosuppix $noweight $imgtest2");
I am trying to make an online judge for c programming. When user enters the c code and submits it, my form redirects to judge.php which is the action file for the form.
Here is what I have written in judge.php
<?php
$text=$_POST['code'];
//echo $text;
$var_str = var_export($text, true);
file_put_contents('code.c', $text);
$ans=exec('pwd');
$ans= exec('gcc code.c');
echo $ans;
?>
I have captured user input in $text and wrote it to a c file(code.c). Till now, it is fine.
But exec(gcc code.c) is not working and not giving any output. I tried other linux commnads like pwd, date, etc. They are working fine. What may be the reason for this and how to fix it?
It is not a directory issue i tried exec(pwd) and it gave the output as the same directory in which code is present.
I tried to run same code.c file from terminal and it is running fine. So, it is also not a 'permission' problem.
One more thing, how to echo the error message generated if any exec() command is not working properly?
After getting suggestion from the answer below, i tried
$cmd="gcc -std=c99 code.c -g -Wall mysql_config --libs --cflags -o db_obj.o --pedantic";
exec($cmd,$out,$status);
But it is also not working. The status returned is 1
Most probably it is permission issue. "whoami" says nobody. Please tell how to change the owner from nobody to root or how to assign the permission to execute gcc from nobody
Three main aspects to my answer
improper use of the exec function.
Look at the man pages. First, the exec function's signature is:
string exec ( string $command [, array &$output [, int &$return_var ]] )
So exec can take up to 3 arguments. It returns the last line of the command's output, like the docs state quite clearly:
The last line from the result of the command. If you need to execute a command and have all the data from the command passed directly back without any interference, use the passthru() function.
To get the output of the executed command, be sure to set and use the output parameter.
So in your case:
$lastLine = exec($command, $fullOutput, $status);
Is what you're looking for. If $status is anything else than 0, your command was unsuccessful. That's what you should check to react accordingly.
The full output of any command can be found in $fullOutput as a line-per-line array.
Output like:
all went well
except for this
Will look like this in the $fullOutput array:
array('all went well', 'except for this');
permissions can be an issue, still.
You say permissions aren't likely to be the cause of the problem, because you can run gcc from the command-line. All fine and dandy, but what user is running the PHP script on the server?
In the case of web-servers, that user is often called nobody, apache or something, and that user is very likely not permitted to run gcc. It's PHP that runs a new instance of whatever default shell it has set up (bash, probably), and it's PHP's user that logs in to that shell, and it's that user that is calling gcc...
Know who you are, and what groups you belong to. Try adding this to your script:
echo 'Script is running under user: ', exec('whoami'), '<br>', PHP_EOL;
echo 'member of the following groups: ', exec('groups'), '<br>', PHP_EOL;
And before you ask: yes, those are comma's... no need to concatenate, you can pass multiple variables/values to echo, separated by a comma. It's actually faster (think of it as C++'s std::cout << some_var << another_var;)
general issues + security
This all said and done: compiling C code from a php script isn't as simple as you seem to think it is. Suppose I were to write this:
#include <stdio.h>
#include <time.h>
int main ( void )
{
time_t t = time(NULL);
if (t%2)
{
float val = (float) t/2.0;
//do stuff with float...
}
else
{
unsigned long long val = t/2;
//do stuff with unsigned long long...
}
}
Your gcc test.c command would fail, because you failed to pass the argument -std=c99, for example.
If I wanted a script to compile a given file, I'd also expect that script to allow me to choose which arguments I compiled my code with, too -g, -Wall and, not to mention: cflags and libs (the output of pkg-config or mysql_config --cflags --libs, to name a specific example I recently used).
Basically, your script simply cannot deal with my wanting to compile something with a commind like
gcc -std=c99 code.c -g -Wall `mysql_config --libs --cflags` -o db_obj.o --pedantic
Which still is a simplified version of what many compilation commands look like, especially when debugging code under development. For stable releases, you'd probably drop -g and --pedantic, but you get my point...
Just think of what it means, allowing the user to pass a set of cli arguments, along with the code. They might pass an argument like -DSOME_MACRO or -O0, which means they might also pass -O0 && rm -Rf *. That means you'll have to call escapeshellcmd or escapeshellarg. Both of which will prohibit me from passing a valid argument, being:
`mysql_config --libs --cflags`
Which contains back-ticks, and thus will be escaped.
To be frank, I struggle to see the point of this exercise... and I'm leaving a lot out, still: the dangers of compiling (let alone running) user-provided code on your machine, for example, are not to be overlooked. You can't just compile code, and run it on your server: memory leaks, segfaults... heck, pure evil code is all getting compiled on your server unchecked if this is the code you have. Really, save yourself a lot of tears, and include an iframe that loads codepad or some similar service...
Recap:
always check the man for a function, see if you're getting all information it returns
check the permissions and runtime for the user that is actually executing the commands
Never trust the network, don't blindly assume people will submit valid, harmless code for you to compile.
Don't reinvent the wheel: compilation services exist, just forward those (but ask for permission first)
Try this code to execute c program from PHP file
<?php
// used to compile the c file using exec() in php
exec('gcc helloworld.c -o helloworld', $out, $status);
if (0 === $status) {
var_dump($out);
// used to execute the c file using exec() in php
exec('./helloworld', $out, $status);
if (0 === $status) {
var_dump($out);
} else {
echo "Command failed with status: $status";
}
} else {
echo "Command failed with status: $status";
}
?>
From inside my index.php file, say, I'd like to check if another PHP file executes without error (and include it, if so), and if it does in fact fail and returns a fatal error, I'd obviously like to not include it. Any suggestions? Thanks...
You may use -l parameter of php CLI:
php -l filename.php
and parse the output.
$o = `php -l filename.php`;
if (strpos($o, 'No syntax errors detected') !== false) {
echo 'No errors';
} else {
echo 'There are errors';
}
You probably don't want to run the second file separately. That said, you can do one of two things...
1.) if you really want to use it as an include that executes separately you could call it with something such as CURL and have it output either the expected result or a failure message that would then be read and acted on accordingly.
2.) Include your function/class/etc execution in a try/catch statement to properly handle any errors encountered. http://php.net/manual/en/language.exceptions.php might help you a little more with this method.
This runs the script as a separate process, with no shared variables/functions/state/scope etc...
$cmd = 'php file.php';
exec($cmd, $ar, $exit_status);
$wasFatal = $exit_status == 255;
There's a real good chance it gets run with a php.ini that's different than whatever your webserver php.ini is, so expect differences in config and maybe even php version.
I am attempting to write a PHP script that will allow for me to select a few files to download from a predetermined location. I'd like my script to pass an array to a Powershell script that id written earlier and have my Powershell script handle the downloading (basically the php file just needs to tell the powershell file what needs to be downloaded).
I've looked at a few options, and it seems that exec is the command I should use for this (as I do not care about command line output I shouldnt need shell_exec).
So far I've turned OFF safe mode to allow me to use this command. I should also note that the php file will be run from a server, however the powershell files are located on a local machine.
A snippet of the code so far to handle the param passing looks like this:
if(isset($_POST['formSubmit']))
{
$choosePlugin = $_POST['wpPlugin'];
$chooseTheme = $_POST['wpTheme'];
if(isset($_POST['wpTheme']))
{
echo("<p>You selected: $chooseTheme</p>\n");
exec('powershell.exe C:\Wordpress Setup\setupThemes.ps1 $chooseTheme');
}
else
{
echo("<p>You did not select a theme</p>\n");
}
I am a bit confused as to what I should put inside the exec. When I run the above code there are no errors however nothing happens. I am a bit new to this so I apologize if more information is required. Any help is appreciated thank you.
Try to do:
echo exec('powershell.exe C:\\Wordpress Setup\\setupThemes.ps1 $chooseTheme');
to see the results of powershell.exe (remember the double \), also make sure to put the absolute path to the exe file:
echo exec('c:\\PATH_TO_POWERSHELL.EXE\\powershell.exe C:\\Wordpress Setup\\setupThemes.ps1 $chooseTheme');
If you want to pass the contents of the variable you should use double quotes to actually expand it, I guess. Furthermore you should quote the script name because the path contains spaces:
exec("powershell.exe \"C:\Wordpress Setup\setupThemes.ps1\" $chooseTheme");
I have a PHP script that creates other PHP files based on user input. Basically, there are files containing language specific constants (define) that can be translated by the user. In order to avoid runtime errors, I want to test newly written files for parse errors (due to "unusual" character sequences). I have read several posts here on SO (like PHP include files with parse errors) and tried a function that uses
$output = exec("php -l $filename");
to determine whether a file parses correctly. This works perfectly on my local machine, but at on the provider's machine, the output of calls to exec("php ...") seems to be always empty. I tried a call to ls and it gives me output, leading me to the assumption that PHP is somehow configured to not react to command line invocations or so. Does anyone know a way around this?
EDIT: I forgot to mention, I had already tried shell_exec and it gives no result, either. In response to sganesh's answer: I had tried that too, sorry I forgot to mention. However, the output (second argument) will always be an empty array, and the return value will always be 127, no matter if the PHP file to test has syntax errors or not.
I had the same problem. The solution that worked for me was found in running-at-from-php-gives-no-output. I needed to add output redirection.
$output = exec("php -l $filename 2>&1");
You can try with exec second and third arguments.
second argument will have the output of the command.
third argument will have the return value.
And exec will return only last line of the command.
$filename = "a.php";
$output = exec("php -l $filename",$op,$ret_val);
print $output."\n";
print $ret_val."\n";
var_dump($op);
By executing shell_exec(), you can see the output as if you executed that file via command line. You can just see if there is an error right here.
<?php
if (strpos(shell_exec('php -l file.php'), 'Syntax Error')) {
die('An error!');
}
There may also be a possibility that shell_exec() or exec() may be disable by your host.
Nice idea to check the file validity :-)!
Now, from the PHP manual for exec():
Note: When safe mode is enabled, you can only execute files within the safe_mode_exec_dir. For practical reasons, it is currently not allowed to have components in the path to the executable.
Can you check if this is not the case for you?
Also, can you check by providing the full path of the PHP interpreter in the exec() instead of only php. Let me know how you fare.
Pinaki
the correct way is to add >2&1 as tested on a windows system using imagemagick!
I worked around my original problem by using a different method. Here is what I do now:
Write a temporary file with contents <?php include "< File to test >"; echo "OK"; ?>
Generate the correct URL for the temporary file
Perform HTTP request with this URL
Check if result equals "OK". If yes, the file to test parses without errors.
Delete temporary file
Maybe this could be done without the temporary file by issuing an HTTP request to the file to test directly. However, if there is a parse error and errors are suppressed, the output will be empty and not discernible from the output in the case of a file that gives no parse errors. This method is risky because the file is actually executed instead of just checked. In my case, there is only a limited number of users who have access to this functionality in the first place. Still, I'm naturally not entirely happy with it.
Why the exec() approach did not work, I still do not know exactly. pinaki might be right by suggesting to provide the full path to the PHP executable, but I cannot find out the full path.
Thank you everyone for answering, I upvoted you all. However, I cannot accept any of your answers as none of your suggestions really solved my problem.