Execute a string of PHP code on the command line

I'd like to be able to run a line of PHP code on the command line similar to how the following options work:

perl -e "print 'hi';"
python -c "print 'hi'"
ruby -e "puts 'hi'"

I'd like to be able to do:

php "echo 'hi';"

I've read that there is a -r option that can do what I need for php, however it doesn't appear to be available when I try to use it. I've tried using PHP 5.2.13 and PHP 4.4.9 and neither have an -r option available.

I wrote this script (that I called run_php.php) - which works, but I'm not a huge fan of it just because I feel like there should be a more "correct" way to do it.

#!/usr/bin/php5 -q
<?php echo eval($argv[1]); ?>

Is there a -r option? If so, why is it not available when I run --help? If there is no -r option, what is the best way to do this (without writing an intermediary script if possible)?

Because I don't think it was very clear above, the -r option is not available to me. Here is the php -h output for both versions of PHP that I'm running.

PHP 4.4.9

Usage: php [-q] [-h] [-s] [-v] [-i] [-f <file>]
       php <file> [args...]
  -a               Run interactively
  -C               Do not chdir to the script's directory
  -c <path>|<file> Look for php.ini file in this directory
  -n               No php.ini file will be used
  -d foo[=bar]     Define INI entry foo with value 'bar'
  -e               Generate extended information for debugger/profiler
  -f <file>        Parse <file>.  Implies `-q'
  -h               This help
  -i               PHP information
  -l               Syntax check only (lint)
  -m               Show compiled in modules
  -q               Quiet-mode.  Suppress HTTP Header output.
  -s               Display colour syntax highlighted source.
  -v               Version number
  -w               Display source with stripped comments and whitespace.
  -z <file>        Load Zend extension <file>.

PHP 5.2.13

Usage: php [-q] [-h] [-s] [-v] [-i] [-f <file>]
       php <file> [args...]
  -a               Run interactively
  -C               Do not chdir开发者_开发知识库 to the script's directory
  -c <path>|<file> Look for php.ini file in this directory
  -n               No php.ini file will be used
  -d foo[=bar]     Define INI entry foo with value 'bar'
  -e               Generate extended information for debugger/profiler
  -f <file>        Parse <file>.  Implies `-q'
  -h               This help
  -i               PHP information
  -l               Syntax check only (lint)
  -m               Show compiled in modules
  -q               Quiet-mode.  Suppress HTTP Header output.
  -s               Display colour syntax highlighted source.
  -v               Version number
  -w               Display source with stripped comments and whitespace.
  -z <file>        Load Zend extension <file>.

There is no -r option. When I try to use the -r option I get:

Error in argument 1, char 2: option not found r

Sorry for the confusion.

Yep, it's there in PHP 5.2's cli SAPI.

If you cannot upgrade and that's the case that there's no such option in PHP 5.2 (I don't have it at hand to test), you can do this:

echo "<?php echo \"hi\\n\";" | php

hi

Original:

There is indeed a -r option (though I'm not sure about PHP 5.2):

php -r "echo 'hi';";

hi

Just make sure you are using the command line version of PHP. php --version should give you something like this (notice "cli"):

php --version

PHP 5.3.0 (cli) (built: May 20 2010 19:05:12) (DEBUG)
Copyright (c) 1997-2009 The PHP Group
Zend Engine v2.3.0, Copyright (c) 1998-2009 Zend Technologies

In new versions of PHP you just type "php -a" and you hop into an interactive mode, where you can experiment with PHP.

An extra semicolon is not required at the end.

You can mention php -r "echo 'hi';" instead of php -r "echo 'hi';";

Another example (to get the current timestamp at the command line):

php -r 'print time()."\n";'

The easiest way to do it is to use the -r flag. However, I've found that it does not allow multiline inputs. To work around that, you can do this:

php -r "passthru(file_get_contents('php://stdin'));"

Which lets you pipe from standard input, like so:

echo -n "echo 'test';" | php -r "passthru(file_get_contents('php://stdin'));"

However, to answer the original question, if you do not have the -r flag available, it is also possible to use the -f flag - just pass standard input as the file to open: php -f /dev/stdin

If you do this, be aware that a) you need to include a blank space at the start of the input, and b) you have to open with <?php. Example:

echo -ne " <?php\necho 'test';" | php -f /dev/stdin

Take a look at this page on PHP command line features, if you haven't already. There are some posts on issues based on OS and double or single quotes.

I would also check the PHP information

php -i

to see if PHP was compiled with CLI support disabled (--disable-cli).

I recently found psysh, which is an interactive debugger also available from the command line.