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Using PHP As A Shell Scripting Language
Darrell Brogdon picture of Darrel Brogdon
As most of us already know, PHP is the best language for developing dynamic web pages available today. Not many people are aware that it can be used as a shell scripting language as well. While PHP as a shell script isn't as robust as Bash or Perl it does have definite advantages, especially if you're like me and are more proficient in PHP than you are in Perl.
The requirements for using PHP as a shell language is that you must compile PHP as a CGI binary instead of as an Apache module. There are certain security issues related to this so please refer to the PHP Manual when doing so.
Where the code is concerned, the only difference between a PHP shell script and a regular PHP web page is the existence of the standard shell call at the top of the script:
#!/usr/local/bin/php -q
We're using the '-q' switch so that the HTTP headers are suppressed. Also, you're still required to begin and end the script with the standard PHP tags:
<?php ?>
So let's delve into the standard code sample we all know and love:
#!/usr/local/bin/php -q
<?php

    print("Hello, world!\n");

?>
This, as most of you know already (about a billion times over) will simply output to the screen "Hello, world!".
Passing arguments to the shell script
Commonly with a shell script you need to pass arguments to the script. This is easily done using the built-in '$argv' array as show in the following example:
#!/usr/local/bin/php -q
<?php
    $first_name = $argv[1];
    $last_name = $argv[2];

    print("Hello, $first_name $last_name!  How are you today?\n");

?>
So in the above script we're printing out the first two arguments to the script which would be called like this:
[dbrogdon@artemis dbrogdon]$ scriptname.ph Darrell Brogdon
Which would print out:
Hello, Darrell Brogdon!  How are you today?
[dbrogdon@artemis dbrogdon]$
The only major difference with the '$argv' array between a shell script and a web page is that in a shell script, '$argv[0]' is the name of your script. In a web page it is the first argument in your query string.
Making a script more interactive
But how do we wait for user input? How do we create a truly interactive script? Well, PHP has no native functions like the 'read' command in shell but we can always emulate it using the following PHP function:
*Note that this function will only work for Unix.

<?php

function read() {
    $fp=fopen("/dev/stdin", "r");
    $input=fgets($fp, 255);
    fclose($fp);

    return $input;
}

?>
This function opens a file pointer to Standard In (/dev/stdin on Linux) and reads anything from this pointer up to 255 bytes, newline, or EOF. In this case a newline is most likely to occur. It then closes the file pointer and returns the data.
So now let's modify our previous script to wait for user input using the newly created 'read()' function:
#!/usr/local/bin/php -q
<?php

    function read() {
        $fp=fopen("/dev/stdin", "r");
        $input=fgets($fp, 255);
        fclose($fp);

        return $input;
    }

    print("What is your first name? ");
    $first_name = read();

    print("What is your last name? ");
    $last_name = read();

    print("\nHello, $first_name $last_name!  Nice to meet you!\n");

?>
You may notice, however, that when you execute this script the last line to be printed is broken into three lines instead of one as it should be. This is because our 'read()' function also takes in the newline character. This is easily fixed by stripping off the newline before we return the data:

<?php

function read() {
    $fp=fopen("/dev/stdin", "r");
    $input=fgets($fp, 255);
    fclose($fp);

    return str_replace("\n", "", $input);
}

?>
Embedding PHP shell scripts within a regular shell script
Sometimes it might be handy to embed a PHP shell script within a script written in Bash or other shell. This is fairly simple but can get a tad tricky.
First, how to embed the PHP code:
#!/bin/bash
echo This is the Bash section of the code.

/usr/local/bin/php -q << EOF
<?php
    print("This is the PHP section of the code\n");
?>
EOF
Pretty simple huh? Until you add a variable that is. This is the tricky part. Try running the following code segment:
#!/bin/bash
echo This is the Bash section of the code.

/usr/local/bin/php -q << EOF
<?php
    $myVar = "PHP";
    print("This is the $myVar section of the code.\n");
?>
EOF
You'll get the following error:
<b>Parse error</b>: parse error in <b>-</b> on line <b>2</b><br>
To fix this you have to escape all of the '$' characters in your PHP code:
#!/bin/bash
echo This is the Bash section of the code.

/usr/local/bin/php -q << EOF
<?php
	\$myVar = "PHP";
	print("This is the \$myVar section of the code.\n");
?>
EOF
So that should get you started on creating your own shell scripts using PHP!
--Darrell