How to get PHP to interact with bash code ?

[tmyonline]

Hi guys,

 

Is there a way to interact PHP with bash so that the output of a bash program can be used as the input for a PHP program ?  Thanks.

[samshel]

you can call PHP script at the end of your bash code and pass whatever u want as command line parameter. if i understand u correctly..

[corbin]

Do you mean piping?

 

 

You could of course just use exec() inside of PHP, but if you want to pipe data into it, you could read php://stdin.

 

$stdin = fopen('php://stdin', 'r');

echo fread($stdin, 1024);

 

for example.

[samshel]

  #!/bin/bash          
          echo Hello World
/usr/bin/php test.php parameter1 parameter2
[code]

[tmyonline]

Hi guys, thanks for the prompt response.

 

I'm comfortable with PHP but I'm a newbie at bash.  Samshel, in your example

 

#!/bin/bash       

        echo Hello World

/usr/bin/php test.php parameter1 parameter2

 

you are passing two parameters, namely parameter1 and parameter2, to a PHP file called "test.php" located in the php directory under bin.  Is that what you meant ?  If so, if I run test.php, it will automatically output the results which made use of parameter1 and parameter2 as the inputs ?

 

The point I'm trying to make sure here is that I don't have to run each program, bash & PHP, individually.  So, if I run test.php and if it automatically executes the bash program for the output parameter1 and parameter2 which serve as the input for the test.php program, then that's great.  But, is it the case though ?  Many thanks.

 

Hi corbin, yeah, piping could be another method to achieve this.  Do you have the link to this material.  I searched Google on php://stdin but it didn't point me to the right place.  Many thanks.

[corbin]

http://php.net/wrappers.php

 

 

php://stdin is just a stream wrapper.  You would basically treat it like any other stream (like a file stream for example).

 

 

 

 

As for parameters passed on the command line, you would use $_SERVER['argc'] and $_SERVER['argv'] (or you could use $argc and $argv).