Here is an example trying to understand exec() function
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
echo exec("id;ls");
?>
When i run this code the result of ls only
Does exec() execute the last command only or it executes both of them and echo the last command ?
You can use shell_exec() instead for this purpose.
On the other hand, exec() returns only last line of the output (by default), but you can provide reference for output array as a second argument.
See the documentation for more info.
exec returns the last line from the result of the command. You have to use output argument. If the output argument is present, then the specified array will be filled with every line of output from the command.
exec("id;ls", $output);
var_dump($output);
you need write a shell script for Linux(excutable with .sh file)
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 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.
I have a program that I can run on my command line but I was wondering if I could actually get it to run in php. Basically my program would have a user insert a couple values to search for, then those values would be passed on into the program for it to run. Then I would want the result of the program to be displayed
I found a function called exec() but I didn't understand it at all so I was wondering if anyone else knows a way or can help me out!
exec() runs a command on the command line, just as you desire. You can capture the output of the command in an array named as the second argument.
For example:
exec("whoami", $output);
var_dump($output);
This runs linux's "whoami" command and captures the result in the array $output. The second line displays the contents of the array. Is that similar to what you want to do?
I am running this test code using exec() in PHP
exec('#!/bin/bash');
exec('abc=10');
echo exec('echo $abc'); // no output
echo exec('whoami'); // this works fine
If I run the first 3 lines in terminal, the output is '10'.
But PHP does not output anything. What am I doing wrong?
http://php.net/manual/en/function.exec.php
Manual says use output_arr to capture output.
<?php
Exec ($cmd, $output_arr);
Print_r ( $output_arr);
?>
#zarathuztra is right. Try combining the various commands in a single exec with ; in between. Also check my answer if output_arr gets what you want.
I want to see the output of the following code in the web browser:
code:
<?php
$var = system('fdisk -l');
echo "$var";
?>
When I open this from a web browser there is no output in the web browser. So how can I do this? Please help!
Thanks
Puspa
you can use passthru, like so:
$somevar = passthru('echo "Testing1"');
// $somevar now == "Testing1"
or
echo passthru('echo "Testing2"');
// outputs "Testing2"
use exec('command', $output);
print_r($output);
First of all, be sure that you (=user under which php runs) are allowed to call the external program (OS access rights, safe_mode setting in php.ini). Then you have quite a few options in PHP to call programs via command line. The most common I use are:
system
This function returns false if the command failed or the last line of the returned output.
$lastLine = system('...');
shell_exec or backtick operators
This function/operator return the whole output of the command as a string.
$output = shell_exec('...'); // or:
$output = `...`;
exec
This function returns the last line of the output of the command. But you can give it a second argument that then contains all lines from the command output.
$lastLine = exec('...'); // or capturing all lines from output:
$lastLine = exec('...', $allLines);
Here is the overview of all functions for these usecases: http://de.php.net/manual/en/ref.exec.php