I'm calling a php script using exec and I'm trying to make a simple log.
Currently I have this :
exec("php script.php $options > temp/log.txt");
If I execute once the result is wrote, but if I execute this multiple times it's always replaced by the last call.
Is there a way to just add the output at the end of the .txt, without replacing all the file ?
Thanks
This has nothing to do with php, you are looking for a shell feature:
exec("php script.php $options >> temp/log.txt");
Note the double >> in there. It appends the redirection instead of overwriting the target.
Related
I have a PHP file that runs a node script using exec() to gather the output, like so:
$test = exec("/usr/local/bin/node /home/user/www/bin/start.js --url=https://www.example.com/");
echo $test;
It outputs a JSON string of data tied to the website in the --url paramater. It works great, but sometimes the output string is cut short.
When I run the command in the exec() script directly, I get the full output, as expected.
Why would this be? I've also tried running shell_exec() instead, but the same things happens with the output being cut short.
Is there a setting in php.ini or somewhere else to increase the size of output strings?
It appears the only way to get this working is by passing exec() to a temp file, like this:
exec("/usr/local/bin/node /home/user/www/bin/start.js --url=https://www.example.com/ > /home/user/www/uploads/json.txt");
$json = file_get_contents('/home/user/www/uploads/json.txt');
echo $json;
I would prefer to have the direct output and tried increasing output_buffering in php.ini with no change (output still gets cut off).
Definitely open to other ideas to avoid the temp file, but could also live with this and just unlink() the file on each run.
exec() only returns the last line of the output of the command you pass to it. Per the section marked Return Value of the following documentation:
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.
https://www.php.net/manual/en/function.exec.php
To do what you are trying to do, you need to pass the function an array to store the output, like so:
exec("/usr/local/bin/node /home/user/www/bin/start.js --url=https://www.example.com/", $output);
echo implode("\n", $output);
I have this line of code to execute a background task (convert several pngs to jpg)
exec("nohup path/to/php path/to/convertToJpg.php >> path/to/convert_to_jpg.log > /dev/null &");
Now I am writing the convertToJpg.php script and I can't figure out how to output information from there so that it will be logged into the convert_to_jpg.log.
When I try to google it all I come up with is how to call a php script from exec or shell_exec, since the words used to describe the two situations is almost the same.
For Clarification
An extra quote got into my code when copying it over to SO. I have fixed it. In my original code the convertToJpg.php is being called as expected, confirmed by error_logs placed in it to check.
A couple answers have pointed to the $output argument in exec(). As I understand it, that totally defeats the purpose of shell redirection using the >> path/to/convert_to_jpg.log.
My question is not how to get the output from the exec() command, but what code do I use to actually output(verb) from the convert_to_jpg.log
More Clarification
If I call
exec("nohup path/to/php path/to/convertToJpg.php >> path/to/convert_to_jpg.log > /dev/null &");
or
$results = shell_exec("path/to/php path/to/convertToJpg.php > /dev/null");
echo $results;
or
$results = "";
exec("path/to/php path/to/convertToJpg.php > /dev/null", $results);
print_r($results);
It doesn't matter which one.
Here is convertToJpg.php
<?php
echo "Will this be in $results?"; // No, this did not end up in results.
error_log("error logs are being logged, so I know that this php file is being called");
//I have tested echo, but that does not work.
//What php function do I use so that a string will be captured in the $output of an exec() call?
?>
$output = "";
$return_var = "";
exec("nohup path/to/php path/to/convertToJpg.php >> path/to/convert_to_jpg.log > /dev/null &", $output, $return_var);
$output
If the output argument is present, then the specified array will be filled with every line of output from the command. Trailing whitespace, such as \n, is not included in this array. Note that if the array already contains some elements, exec() will append to the end of the array. If you do not want the function to append elements, call unset() on the array before passing it to exec().
$return_var
If the return_var argument is present along with the output argument, then the return status of the executed command will be written to this variable.
http://php.net/manual/en/function.exec.php
Ok, I figured it out!
To answer the question plain and simply, use either echo or print.
This was the first thing I tried, but it didn't work, which made me think it was some other function that I was supposed to call. (I obviously haven't worked with this much)
The real problem I was having was the > /dev/null, which was discarding all output. Once I deleted that, it fine. Explained here: What is /dev/null 2>&1?
I had at some point done a copy/paste without really understanding what that did...
And because I was blaming the wrong thing I ended up with this off base question.
I have a C code that I have to execute through PHP,
I have used exec('./sys'), sys is my executable file.
I have also tried system(), passthrough(), shell_exec() and they are not giving output.
When I executed exec('who'); it gives the output.
What can I do to execute sys?
Each of those methods you reference will execute your sys file, but you need to make sure you are executing the correct path. Your working path is determined by what script is actually executing PHP. For example, if you're executing your code through apache or the command line your working directory may be different. Lets assume this file structure:
+ src/
| + script.php
| + sys
I would recommend using PHP's __DIR__ magic variable in your script.php to always reference the current file's directory, and then work from there:
// script.php
exec(__DIR__ . "/sys");
Retrieving output can be done a couple different ways. If you want to store the output of the script in a variable, I would recommend using exec according the the manual:
Parameters ΒΆ
command
The command that will be executed.
output
If the output argument is present, then the specified array will be filled with every line of output from the command. Trailing whitespace, such as \n, is not included in this array. Note that if the array already contains some elements, exec() will append to the end of the array. If you do not want the function to append elements, call unset() on the array before passing it to exec().
return_var
If the return_var argument is present along with the output argument, then the return status of the executed command will be written to this variable.
exec will return the first line of output, but if you want more than that you need to pass a variable by reference:
// script.php
$output = array();
exec(__DIR__ . "/sys", $output);
$output will then contain an array of each line of output from the command. However if you want to run your sys script and directly pass through the output then use passthru(__DIR__ . "/sys"); For example, if you wanted to execute a command that required input on the command line, passthru would be the best option.
How do I properly execute commands in the command line using php? For example I'm using the command below in the command line to convert a docx file into a pdf file:
pdfcreator.exe /PF"D:\Documents\sample.docx
Now using PHP code I want to be able to execute the same command but nothing seems to be happening:
<?php
shell_exec('pdfcreator.exe /PF"D:\Documents\sample.docx"');
?>
Is this possible in PHP?If yes, how do I do it?
system("c:\\path\\to\\pdfcreator.exe /PF\"D:\\Documents\\sample.docx"");
try this.
Don't forget to escape your command with escapeshellcmd(). This will prevent you from having to use ugly backslashes and escape characters.
There are also other alternatives which may work:
`command` // back ticks drop you out of PHP mode into shell
exec('command', $output); // exec will allow you to capture the return of a command as reference
shell_exec('command'); // will return the output to a variable
system(); //as seen above.
Also, make sure your .exe is included within your $PATH variable. If not, include the full path for the command.
I'd like to run something like (in myProgram.sh):
java -cp whatever.jar com.my.program $1
within PHP and read the output.
So far I have something like:
$processOrderCommand = 'bash -c "exec nohup setsid /myProgram.sh ' . $arg1 . ' > /dev/null 2>&1 &"';
exec($processOrderCommand);
But what I'd really like is to be able to get the output of the java program within the PHP script and not just execute it as another thread.
How can this be done?
You can do this :
exec($processOrderCommand, $output);
From the documentation :
If the output argument is present, then the specified array will be filled with every line of output from the command. Trailing whitespace, such as \n, is not included in this array. Note that if the array already contains some elements, exec() will append to the end of the array. If you do not want the function to append elements, call unset() on the array before passing it to exec().
For a better control on your execution you can take a look at proc_open()
Resources :
php.net - exec()
php.net - proc_open()
The key is that the classpaths need to be absolute within the shell_exec
PHP script.
Or at least that's the only way I could get it to correctly work. Basically it's almost impossible to tell from environment to environment what the relative directory is that the php script is running the JVM.
As well, it helped to put the absolute path location for java, such as usr/.../bin/java