Following is my command line arguments:
php a.php -g UAMS ABC
Now when I get command from command line using getopt then I am getting only first argument i.e UAMS. I want to get both arguments i.e UAMS and ABC.
Following is my piece of code to get argument throug getopt() function.
$options=getopt("g");
echo "Options: ".json_encode($options);
I am only getting UAMS.
When I give argument like
php a.php -g a/b/c
Then I get arguments through getopt then it gives output as a\/b\/c?
So why it is not giving correct answer?
"g" is an option without values "g:" would be one with an optional value. See http://de.php.net/getopt
The backslashes are from json_encode().
Related
I have the following code:
<?php
if(isset($_GET['function'])){
$_GET['function']();
}
?>
So if i entered this url:
http://localhost/?function=phpinfo
I will see the phpinfo function output on the screen.
can i have a way to concatenate 2 function in the url like this example:
http://localhost/?function=shell_exec('ls') AND phpinfo
So i want to see the first function output..
If you may asking why i need this, is because i am pen testing an web application with this situation..
Thanks for the help..
With your given code example it is not possible to do what you want. All your functions, so shellexec('ls') and phpinfo will be interpreted as one string, which is then called as a function in by calling it with added parenthis.
The only way that I can think of is using a variable parameter list, rather than just a single parameter. Get all the GET parameters in the function, and loop through them, executing each one.
I'm using shell_exec to execute a python file with a few variables and then print the real, user, and sys results to the console.
shell_exec("time /Users/$USER/anaconda/bin/python
/Applications/MAMP/cgi-bin/file.py
$var1 $var2 $var3", $result );
print_r($result);
Although this has worked for me before, it's not working now. The error I'm getting is PHP Warning: shell_exec() expects exactly 1 parameter, 2 given
It's the same response whether I just have time or /usr/bin/time.
What's going wrong here?
shell_exec() takes only a single parameter. What you were using before was likely exec().
What is in the command string that you're passing it (e.g. time or /usr/bin/time) is irrelevant to the warning you're getting.
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.
I am trying to use the PHP exec() function.
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.
If the execution was successful, it's 0. However, if there is an error, it can be a multitude of other integers. I can't seem to find anywhere what those integers correspond to. How should I interpret the integer that I get?
Update:
I really should have specified this originally, but I am executing another PHP script. Unlike rsync, which has exit values on its man page, I can't find an equivalent for PHP.
So what I am doing is something like:
$rv = exec('php file.php', $out, $rv);
The return value is dependent on the process/program that you ran with exec. For instance, if you ran grep:
The exit status is 0 if selected lines are found, and 1 if not
found. If an error occurred the exit status is 2. (Note: POSIX
error handling code should check for '2' or greater.)
rsync has about 20 different error exit codes, all painstakingly explained in the man page:
http://linux.die.net/man/1/rsync
so yes, it's program-dependant :)
Even if you're running PHP script, the exit value depends on your program itself. By default php scripts will exit with 0. If you use the exit function you can return different exit codes:
http://php.net/manual/en/function.exit.php
If you want to experimentally determine what your php program exits, call it on the command line:
php file.php
then do
echo $?
this will show you the exit value of your php script.
IMHO, before use exec() function better set output and return_var parameters and read return code execution by return_var.
Don't rely on exec() return value.
Look up the manual page for the command that you are executing. This value has nothing to do with PHP but the actual command.
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.