PHP 7.2.7 Released

Executing PHP files

The CLI SAPI has three different ways of getting the PHP code you want to execute:

  1. Telling PHP to execute a certain file.

    $ php my_script.php
    $ php -f my_script.php

    Both ways (whether using the -f switch or not) execute the file my_script.php. You can choose any file to execute, and your PHP scripts do not have to end with the .php extension but can have any name or extension you wish.


    If you need to pass arguments to your scripts you need to pass -- as the first argument when using the -f switch.

  2. Pass the PHP code to execute directly on the command line.

    $ php -r 'print_r(get_defined_constants());'

    Special care has to be taken in regards of shell variable substitution and quoting usage.


    Read the example carefully, there are no beginning or ending tags! The -r switch simply does not need them. Using them will lead to a parser error.

  3. Provide the PHP code to execute via standard input (stdin).

    This gives the powerful ability to dynamically create PHP code and feed it to the binary, as shown in this (fictional) example:

    $ some_application | some_filter | php | sort -u > final_output.txt
You cannot combine any of the three ways to execute code.

Like every shell application, the PHP binary accepts a number of arguments but your PHP script can also receive arguments. The number of arguments which can be passed to your script is not limited by PHP (the shell has a certain size limit in the number of characters which can be passed; usually you won't hit this limit). The arguments passed to your script are available in the global array $argv. The first index (zero) always contains the name of the script as called from the command line. Note that, if the code is executed in-line using the command line switch -r , the value of $argv[0] will be simply a dash (-). The same is true if the code is executed via a pipe from STDIN.

The second registered global variable is $argc which contains the number of elements in the $argv array (not the number of arguments passed to the script).

As long as the arguments you want to pass to your script do not start with the - character, there's nothing special to watch out for. Passing an argument to your script which starts with a - will cause trouble because PHP itself thinks it has to handle it. To prevent this, use the argument list separator --. After this separator has been parsed by PHP, every argument following it is passed untouched to your script.

# This will not execute the given code but will show the PHP usage
$ php -r 'var_dump($argv);' -h
Usage: php [options] [-f] <file> [args...]

# This will pass the '-h' argument to your script and prevent PHP from showing it's usage
$ php -r 'var_dump($argv);' -- -h
array(2) {
  string(1) "-"
  string(2) "-h"

However on Unix systems, there's another way of using PHP for shell scripting. You can write a script where the first line starts with #!/usr/bin/php (substitute with the path to your PHP CLI binary if necessary. Following this you can place normal PHP code included within the PHP starting and end tags. Once you have set the execution attributes of the file appropriately (e.g. chmod +x test) your script can be executed like a normal shell or perl script:

Example #1 Execute PHP script as shell script


Assuming this file is named test in the current directory, we can now do the following:

$ chmod +x test
$ ./test -h -- foo
array(4) {
  string(6) "./test"
  string(2) "-h"
  string(2) "--"
  string(3) "foo"

As you see, in this case no care needs to be taken when passing parameters which start with - to your script.

The PHP executable can be used to run PHP scripts absolutely independent from the web server. If you are on a Unix system, you should add a special first line to your PHP script, and make it executable, so the system will know, what program should run the script. On a Windows platform you can associate php.exe with the double click option of the .php files, or you can make a batch file to run the script through PHP. The first line added to the script to work on Unix won't hurt on Windows, so you can write cross platform programs this way. A simple example of writing a command line PHP program can be found below.

Example #2 Script intended to be run from command line (script.php)


if ($argc != || in_array($argv[1], array('--help''-help''-h''-?'))) {

This is a command line PHP script with one option.

  <?php echo $argv[0]; ?> <option>

  <option> can be some word you would like
  to print out. With the --help, -help, -h,
  or -? options, you can get this help.

} else {

In the script above, we used the special first line to indicate that this file should be run by PHP. We work with a CLI version here, so there will be no HTTP header printouts. There are two variables you can use while writing command line applications with PHP: $argc and $argv. The first is the number of arguments plus one (the name of the script running). The second is an array containing the arguments, starting with the script name as number zero ($argv[0]).

In the program above we checked if there are less or more than one arguments. Also if the argument was --help , -help , -h or -? , we printed out the help message, printing the script name dynamically. If we received some other argument we echoed that out.

If you would like to run the above script on Unix, you need to make it executable, and simply call it as script.php echothis or script.php -h. On Windows, you can make a batch file for this task:

Example #3 Batch file to run a command line PHP script (script.bat)

@echo OFF
"C:\php\php.exe" script.php %*

Assuming you named the above program script.php, and you have your CLI php.exe in C:\php\php.exe this batch file will run it for you with your added options: script.bat echothis or script.bat -h.

See also the Readline extension documentation for more functions you can use to enhance your command line applications in PHP.

If you are on Windows, PHP can be configured to run without the need to supply the C:\php\php.exe or the .php extension, as described in Command Line PHP on Microsoft Windows.

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User Contributed Notes 7 notes

php at richardneill dot org
5 years ago
On Linux, the shebang (#!) line is parsed by the kernel into at most two parts.
For example:

1:  #!/usr/bin/php
2:  #!/usr/bin/env  php
3:  #!/usr/bin/php -n
4:  #!/usr/bin/php -ddisplay_errors=E_ALL
5:  #!/usr/bin/php -n -ddisplay_errors=E_ALL

1. is the standard way to start a script. (compare "#!/bin/bash".)

2. uses "env" to find where PHP is installed: it might be elsewhere in the $PATH, such as /usr/local/bin.

3. if you don't need to use env, you can pass ONE parameter here. For example, to ignore the system's PHP.ini, and go with the defaults, use "-n". (See "man php".)

4.  or, you can set exactly one configuration variable. I recommend this one, because display_errors actually takes effect if it is set here. Otherwise, the only place you can enable it is system-wide in php.ini. If you try to use ini_set() in your script itself, it's too late: if your script has a parse error, it will silently die.

5. This will not (as of 2013) work on Linux. It acts as if the whole string, "-n -ddisplay_errors=E_ALL" were a single argument. But in BSD, the shebang line can take more than 2 arguments, and so it may work as intended.

Summary: use (2) for maximum portability, and (4) for maximum debugging.
gabriel at figdice dot org
1 year ago
Regarding shebang:

In both Linux and Windows, when you execute a script in CLI with:

    php script.php

then PHP will ignore the very first line of your script if it starts with:


So, this line is not only absorbed by the kernel when the script file is executable, but it is also ignored by the PHP engine itself.

However, the engine will NOT ignore the first #! line of any included files withing your "outer" script.php.
Any "shebang" line in an included script, will result in simply outputting the line to STDOUT, just as any other text residing outside a <?php ...  ?> block.
david at frankieandshadow dot com
1 year ago
A gotcha when using #!/usr/bin/php at the start of the file as noted above:

if you originally edited the file on Windows and then attempt to use it on Unix, it won't work because the #! line requires a Unix line ending. Bash gives you the following error message if it has DOS line endings:
"bash: /usr/local/bin/wpreplace.php: /usr/bin/php^M: bad interpreter: No such file or directory"

(In Emacs I used "CTRL-X ENTER f" then type "unix" and ENTER to convert)
petruzanautico at yah00 dot com dot ar
6 years ago
As you can't use -r and -f together, you can circumvent this by doing the following:
php -r '$foo = 678; include("your_script.php");'
spencer at aninternetpresence dot net
6 years ago
If you are running the CLI on Windows and use the "-r" option, be sure to enclose your PHP code in double (not single) quotes. Otherwise, your code will not run.
gar37bic at gmail dot com
2 years ago
For Linux and similar, the actual path to php may differ according to the particular distribution's layout.  Sometimes it is in /usr/local/bin/php, sometimes /usr/bin/php, and even /home/user/bin/php.  So #/usr/bin/php may need to be changed.

But there is an alternative:  Use #/usr/bin/env php instead.  This uses the environmental setup, and will ordinarily find PHP in one of your $PATH locations, searching down the path to the first hit. (At least I think that's how it works.)  If your CLI script will be run in different environments on different machines, this allows it to work without changes.  I believe this also works on BSD (which also means MacOS?) but I don't have one around any more to confirm.
andrew-by at hotmail dot com
6 years ago
If you want your batch file's window stay opened after executing the PHP script just add "pause" at the last line. So it should look like:

@echo OFF
"C:\php\php.exe" script.php %*
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