If I execute an R program through macOS Terminal it works fine, but if I execute that program through PHP exec() I get an error which says "Rscript not found".
submit.php:
<?php
// print "hello";
$a = $_POST['a'];
$b = $_POST['b'];
// echo shell_exec("calc");
$output = exec("Rscript /xampp/htdocs/demo/1.R ".$a." ".$b);
print $output
?>
1.R:
setwd("C:/xampp/htdocs/demo")
print("hello")
args=commandArgs(trailingOnly = TRUE)
a = args[1]
b = args[2]
c = as.numeric(a)+as.numeric(b)
cat(c)
You need to tell the command interpreter where Rscript is located. In a normal login, you'd have a long PATH variable to search through, but when running from PHP you have a much more limited path. To figure out where the executable is located, type which Rscript from Terminal. Use the resulting path in your PHP script.
Also note that dumping raw user input into shell commands is a very bad idea. You should always use escapeshellarg() to make sure the input is sanitized. I would also suggest capturing the full output in the event that R outputs more than one line.
<?php
$a = escapeshellarg($_POST["a"]);
$b = escapeshellarg($_POST["b"]);
$r = "/xampp/htdocs/demo/1.R";
$lastline = exec("/usr/local/bin/Rscript $r $a $b", $output, $return);
// you could check the value of $return for non-zero values as well
// full output is returned as an array
echo implode("\n", $output);
Finally, you'll want to check your setwd() command in the R script. Might have some trouble with C: on a Mac! ;)
Related
I have following JAVA program which has been converted to JAR file and placed in the same directory as my PHP file.
So basically it takes an argument passed by PHP and displays it
public class Test {
public static void main(String[] args) throws Exception {
//Takes the value passed from the PHP
String Name = (new String(args[0])).toString();
//This will be treated as Output Parameter which will be returned to PHP
System.out.println("Return to PHP");
}
}
Below is my PHP code which will execute the JAR file and pass the required 1 parameter to the JAR.
<?php
$arg1 = "My_INPUT_PARAMETER";
shell_exec("java -jar TEST.jar $arg1");
echo "Done";
?>
I read somewhere that what ever the placed in Sysout (System.out.println) will be treated as output parameter or Return value to PHP.
So in my case it will be String "Return to PHP".
But I am not able to get the value to PHP and display it.
I tried placing a output value in the exec statement but its not working.
I tried below code but its throwing me error.
<?php
$arg1 = "My_INPUT_PARAMETER";
$output = '';
shell_exec("java -jar TEST.jar $arg1", $output);
echo "Done";
echo $output;
?>
Can anyone help me out here, How can I get a return value from PHP or output parameter from PHP and display it or use it in my PHP and continue with other execution part.
Thanks #Roland Starke.
So basically we can use 2 statements to run the JAR file from PHP:
EXEC and SHELL_EXEC.
EXEC will hold all the return values from JAR file and we can use it as Array and Display the required output parameter.
SHELL_EXEC will hold all the output parameters and it will display all at once.
<?php
$arg1 = "Multi Return";
exec("java -jar TEST.jar $arg1",$output);
echo $output[0]."<br/>";
echo $output[1];
echo "-------------------------------";
$shell_out = shell_exec("java -jar TEST.jar $arg1");
echo $shell_out;
?>
I am trying to execute a C program using the shell_exec command, which needs arguments to be passed. It is working for one input, but not working for others. I tried to run the C program through terminal, it is working for all the inputs.
This is my execprog.php file. I have to give 2 inputs as command line arguments to file. /var/www/project is the path.
$query = "/var/www/project/./a.out /var/www/project/constraints.txt /var/www/project/constraints_keyword.txt /var/www/project/FIB.txt /var/www/project/ANS.txt";
echo $query;
$var = shell_exec($query);
echo $var;
<?php
$query = "/var/www/project/./a.out";
$arguments = array
(
'/var/www/project/constraints.txt',
'/var/www/project/constraints_keyword.txt',
'/var/www/project/FIB.txt',
'/var/www/project/ANS.txt'
);
$string = '';
for($i=0;$i<count($arguments);$i++)
$string.= ' %s';
$command = vsprintf("{$query}{$string}", $arguments);
$var = shell_exec($command);
echo $var;
As you it works on the terminal and not on apache then apache's php.ini file may be disabling the use of shell_exec().
See http://www.php.net/manual/en/ini.core.php#ini.disable-functions
Your apache's php.ini file may look something like
disable_functions=exec,passthru,shell_exec,system,proc_open,popen
Remove shell_exec from this list and restart the web server, although this is a security risk and I don't recommend it.
In general functions such as exec,shell_exec and system are always used to execute the external programs. Even a shell command can also be executed. If these two functions are enabled then a user can enter any command as input and execute into your server. So usually people disable in apache config as disable_functions to secure their site.
It works for me - Here is test run
Sample test c code
[akshay#gold tmp]$ cat test.c
#include<stdio.h>
int main(int args, char *argv[]) {
int i = 0;
for (i = 0; i < args; i++)
printf("Arg[%d] = %s\n",i, argv[i]);
return 0;
}
Compile
[akshay#gold tmp]$ gcc test.c
Sample php script
[akshay#gold tmp]$ cat test.php
<?php
$query = "/tmp/./a.out /var/www/project/constraints.txt /var/www/project/constraints_keyword.txt /var/www/project/FIB.txt /var/www/project/ANS.txt";
$var = shell_exec($query);
echo $var;
?>
Execution and output
[akshay#gold tmp]$ php test.php
Arg[0] = /tmp/./a.out
Arg[1] = /var/www/project/constraints.txt
Arg[2] = /var/www/project/constraints_keyword.txt
Arg[3] = /var/www/project/FIB.txt
Arg[4] = /var/www/project/ANS.txt
Is it possible to pass BASH associative arrays as argv to PHP scripts?
I have a bash script, that collects some variables to a bash associative array like this. After that, I need to send it to PHP script:
typeset -A DATA
DATA[foo]=$(some_bash_function "param1" "param2")
DATA[bar]=$(some_other_bash_function)
php script.php --data ${DATA[#]}
From PHP script, i need to access the array in following manner:
<?php
$vars = getopt("",array(
"data:"
));
$data = $vars['data'];
foreach ($data as $k=>$v) {
echo "$k is $v";
}
?>
What I've tried
Weird syntax around the --data parameter follows advice from a great post about bash arrays from Norbert Kéri how to force passed parameter as an array:
You have no way of signaling to the function that you are passing an array. You get N positional parameters, with no information about the datatypes of each.
However this sollution still does not work for associative arrays - only values are passed to the function. Norbert Kéri made a follow up article about that, however its eval based solution does not work for me, as I need to pass the actual array as a parameter.
Is the thing I'm trying to achieve impossible or is there some way? Thank you!
Update: What I am trying to accomplish
I have a few PHP configuration files of following structure:
<?php
return array(
'option1' => 'foo',
'option2' => 'bar'
)
My bash script collects data from user input (through bash read function) and stores them into bash associative array. This array should be later passed as an argument to PHP script.
php script.php --file "config/config.php" --data $BASH_ASSOC_ARRAY
So instead of complicated seds functions etc. I can do simple:
<?php
$bash_input = getopt('',array('file:,data:'));
$data = $bash_input['data'];
$config = require($config_file);
$config['option1'] = $data['option1'];
$config['option2'] = $data['option2'];
// or
foreach ($data as $k=>$v) {
$config[$k] = $v;
}
// print to config file
file_put_contents($file, "<?php \n \n return ".var_export($config,true).";");
?>
This is used for configuring Laravel config files
Different Approach to #will's
Your bash script:
typeset -A DATA
foo=$(some_bash_function "param1" "param2")
bar=$(some_other_bash_function)
php script.php "{'data': '$foo', 'data2': '$bar'}"
PHP Script
<?php
$vars = json_decode($argv[1]);
$data = $vars['data'];
foreach ($data as $k=>$v) {
echo "$k is $v";
}
?>
EDIT (better approach) Credit to #will
typeset -A DATA
DATA[foo]=$(some_bash_function "param1" "param2")
DATA[bar]=$(some_other_bash_function)
php script.php echo -n "{"; for key in ${!DATA[#]}; do echo - "'$key'":"'${DATA[$key]}'", | sed 's/ /,/g' ; done; echo -n "}"
this does what you want (i think) all in one bash script. You can obviously move the php file out though.
declare -A assoc_array=([key1]=value1 [key2]=value2 [key3]=value3 [key4]=value4)
#These don't come out necesarily ordered
echo ${assoc_array[#]} #echos values
echo ${!assoc_array[#]} #echos keys
echo "" > tmp
for key in ${!assoc_array[#]}
do
echo $key:${assoc_array[$key]} >> tmp # Use some delimeter here to split the keys from the values
done
cat > file.php << EOF
<?php
\$fileArray = explode("\n", file_get_contents("tmp"));
\$data = array();
foreach(\$fileArray as \$line){
\$entry = explode(":", \$line);
\$data[\$entry[0]] = \$entry[1];
}
var_dump(\$data);
?>
EOF
php file.php
the escaping is necessary in the cat block annoyingly.
I'm trying to run a C program of adding two numbers with PHP in a web browser. But when I run the command
exec"gcc name.c -o a & a" it
returns some garbage result like sum is : 8000542.00. It doesn't ask for any input.
I want to give inputs to scanf from the browser. Please suggest to me how can I resolve my problem.
I have tried this but couldn't handle it successfully.
$desc = array(0=> array ('pipe','w'), 1=> array ('pipe','r'));
$cmd = "C:\xampp\htdocs\add.exe";
$pipes=array();
$p = proc_open($cmd,$desc,$pipes);
if(is_resource($p))
{
echo stream_get_contents($pipes[0]);
fclose($pipes[0]);
$return_value=proc_close($p);
echo $return_value;
I'm trying to execute a py file from php.
here is my code:
//usage python my.py var1 var2
$libre = 'python ../../../../root/py/my.py '.$var1.' '.$var2.'';
$cleanlibre = escapeshellarg($libre);
echo exec($cleanlibre);
What is wrong?
Why is it returning nothing?
I also need to know how to secure exec well. Thanks.
--edit--
Used passthru
$libre = 'python ../../../../root/py/mech.py '.$var1.' '.$var2.'';
$cleanlibre = escapeshellarg($libre);
passthru($cleanlibre, $result);
echo $result;
//returned 127 <- i don't know where thats from.
escapeshellarg shall only be used to escape the arguments, not the whole command.
//usage python my.py var1 var2
$libre = 'python ../../../../root/py/my.py '.escapeshellarg($var1).' '.escapeshellarg($var2).'';
echo exec($libre );
exec return result in second argument of function, see http://php.net/manual/en/function.exec.php