exec('wget --progress=bar http://technabob.com/blog/wp-
content/uploads/2011/04/light_up_pi_symbol_1.jpg');
echo "1";
output 1
ie the download is not working.I m sure that the command is correct. I believe it has something to do with my Apache.Any help will be appreciated.
The image is not getting downloaded.
I was wrong about Apache . I think it is about wget because I can run other commands on my server using
$output = `ls -al`;
echo "<pre>$output</pre>";
This has nothing to do with Apache. Also, read the documentation for exec(). Most likely you'll see that that function call replaces the current process with the one specified in the argument, meaning that the rest of your code won't get run.
UPDATE
Perhaps the PATH is incorrect? Try using the full path to the wget command instead. Also, the current working directory will be used as the download destination for your command and the process may not have permissions to write to that directory.
The mistake was that I didnot specify the download location so it was saved at some random location. But still the command was working right.
The correct method could be
$command="wget --output-document=/var/2 --progress=bar http:/"."/technabob.com/blog/wp-content/uploads/2011/04/light_up_pi_symbol_1.jpg";
Related
Ive tried searching for the answer but cant find anything proper in relation to my case.
What i want to execute is very simple...its a call to the SPMF framework jar file.
here is the code -
<?php
exec('java -jar spmf.jar run Apriori /opt/lampp/htdocs/rrugd/ip.txt /opt/lampp/htdocs/rrugd/output.txt %20');
?>
why is this not executing?
it obviously runs perfectly using cli.
the folder having all the files (jar file,ip.txt,output.txt) have permissions set to r/w to all users
I am on Linux - Ubuntu 13.10
I found the solution.
The problem was in the path to the spmf zip file.
Either it needs a full path or should be in the same folder as the script.
I solved it by placing it in the folder same as the script.( under htdocs as i am using lampp)
Also, the folder needs to have permissions for other to be set to "Create and Delete" for php to be able to access and write output to the folder.
Basic debugging: If exec() doesn't work, then you need to check return values:
exec('some command here', $output, $return_var);
var_dump($output, $return_var);
NEVER assume that an external resource succeeded. ALWAYS check return values and output for error messages.
Most likely java isn't in the path of whatever shell PHP is using to handle the exec() call, meaning you'll need exec('/full/path/to/java ...') instead.
I have tried to use exec() with 'whoami' to check if it works and I got the result of
nt authority\system
Now I need to run a .exe file with parameters from php via exec() function.
I tried this in command prompt and it actually runs the program with given parameters. This is the example command.
NOTE the exe file gets 3 inputs (folder, file_name, report_file_nmae)
> ..\..\some_file.exe folder="C:\path_to_folder" param=1.xml report=2.xml
But when I run this command from php file:
exec('..\..\some_file.exe folder="C:\path_to_folder" param=1.xml report=2.xml');
nothing is happening. This is the first time I am using exec() function, so I am not familiar with its details. What is wrong?
I tried using:
\\ instead of \
escapeshellarg() on the directory
added "" around directory folder names
No luck
Addendum:
echo exec($command) // echos < .... why?
or
exec($command, $output);
print_r($output); // Array()
I even changed the permission on the file to full control to all users.
If I call the program from command prompt, I can see the icon appearing next to clock for a second.
But the same call from php will not even call the program.
Edit
Even exec('notepad.exe'); is not working. Something has to be done with php configurations maybe?
I already said that I was new to exec() function. After doing some more digging, I came upon 2>&1 which needs to be added at the end of command in exec().
Thanks #mattosmat for pointing it out in the comments too. I did not try this at once because you said it is a Linux command, I am on Windows.
So, what I have discovered, the command is actually executing in the back-end. That is why I could not see it actually running, which I was expecting to happen.
For all of you, who had similar problem, my advise is to use that command. It will point out all the errors and also tell you info/details about execution.
exec('some_command 2>&1', $output);
print_r($output); // to see the response to your command
Thanks for all the help guys, I appreciate it ;)
You might also try giving the full path to the binary you're trying to run. That solved my problem when trying to use ImageMagick.
I'm currently using the Pygments for PHP plugin that is located here: http://derek.simkowiak.net/pygments-for-php/.
The line that actually calls Pygments from that code is an exec() passed: pygmentize -f html $extra_opts -l $language $temp_name as the command. This all works fine, and I get back the output and it is formatted by the plugin.
What I would like to happen at the same time is for Pygments to create an image of it, so I pass exec() a similar command: pygmentize -f png $extra_opts -l $language -o $full_image_path/$output_file.png $temp_name This is where I run into a problem. The image never shows up in the expected folder.
However, if I var_dump() that command string before I exec() it and take it and run it straight from the command line, it works fine.
I have tried echoing exec('whoami') which tells me that the PHP user is www-data. I've tried giving permissions to www-data and changing ownership to www-data on the folder where I store the images. I've also tried changing permissions to 777 just to see what would happen, and the answer is nothing.
Is there something I'm missing? I'm running out of ideas to try. Thank you!
Edit: Another thing that I've checked is the output from the exec command, and the return value. It outputs an empty array, and it returns 1 as the return value.
Edit 2: After seeing that that directory should be writeable/readable for the PHP user, is it possible that pygments doesn't have permission to write it as a specific user? I'm not sure this makes sense, as when I run it myself it works fine, and in fact, when PHP runs it with the HTML lexer, it is able to run. I'm not very experienced in Python, so I don't know if this is a potential issue.
I guess you cannot do it like this.
$output_file.png
Try
$file = $output_file.".png"
and substitute in the exec
Ended up being an issue with the font that was installed for use by the www-root user. Apparently the one that is used by default for Pygments was installed only for the user that I was running as when I use the command line.
The way I was able to figure this out, was running
exec("$command 2>&1", $out, $code);.
The extra 2>&1 redirects stderr into the output for me to see the issue.
The $out parameter showed the FontNotFound error that pygments was throwing.
I changed the font that Pygments used via the command line using: full,style=manni,cssclass=pygmentize_kbOKBd,font_name='DejaVu Sans Mono' -l php -o /srv/www/path/to/images/uploads/2513732976ad4b7.02729290.png /tmp/pygmentize_kbOKBd after finding which fonts I had available to me.
To see which fonts I had available to me as the user running the script, I just ran fc-list in an exec() command for Ubuntu, and checked the output of that for the list of available fonts.
I am using the wkhtmtoimage library on my server, and I have managed to get the output for the following command when I am running in PuTTY:
wkhtmltoimage www.google.com test.jpg
But, when I use the following shell command I don't get the output, and I don't know why.
$filnename = "test.jpg";
$url = "http://www.google.com";
shell_exec("wkhtmltoimage $url $filename");
Even I tried with this variation instead, but still without getting the desired result:
shell_exec("/usr/local/bin/wkhtmltoimage $url $filename");
What am I doing wrong?
Edit:
I downloaded the Linux binary and put it into the folder then I ran it.Whether I need to restart the server the changes affected?
shell_exec command is allowed because we used it for ffmpeg(already installed one)
I cannot give an exact solution but if you can try out what I write I might be of help. I will list more solutions if the one I am providing does not work.
Solution proposal 1:
Use quotes around url and filename
shell_exec("wkhtmltoimage \"$url\" \"$filename\"");
Solution proposal 2:
Try redirecting streams to files and type the contents.
shell_exec("wkhtmltoimage \"$url\" \"$filename\" > output.txt 2> error.txt");
May be user under which PHP (or Apache) runs have no permissions to run this tool. Try printing output of this command or just use passthru() to directly print output of command.
Another solution is to pass output to files as #Cem_Kalyoncu wrote.
I have a multithreaded cli downloader for ubuntu called Aria2c.
I've been trying to get a php script to run aria2c and download a file using shell_exec, but i can't seem to get it to work. Ultimately I plan to have an input box on a page where I can enter a link and aria would download it.
Here's the code I've come up with (for now im inputting the link manually):
<?php $dl = shell_exec('aria2c -d /home/user/ www.downloadlink.com'); ?>
Note that the aria2c command I specified works well in the shell; and the directory I'm attempting to download to is set to '777'.
I'm baffled as to why it's not working, any ideas?
PS: I prefer to use aria rather than the alternatives because it is multithreaded and it supports cookies.
Check if PHP is running in safe_mode. shell_exec won't work if safe_mode is on.
EDIT: aria2c was not referenced with a full path. Referencing it like this: shell_exec('/path/to/aria2c -d /home/user/ www.downloadlink.com') works.
I'll make the assumption that you are running PHP through a web server. In such case, it's very unlikely that the web server has permission to write into your user's home directory: Apache runs as daemon with the credentials of a limited user. Also, the PATH env variable in Apache is not necessarily the same as your user's PATH. Last but not least, you don't check the return value or the script output.
Try something like this:
<?php
exec('/path/to/aria2c -d /tmp www.downloadlink.com', $output, $return_var);
var_dump($output, $return_var);
?>
You can get the full path for aria2c with:
which aria2c