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Re: [PHP] Calling a program





On 30 November 2015 20:40:55 GMT+00:00, Curtis Maurand <curtis@xxxxxxxxxxx> wrote:
>
>
>On 11/30/2015 3:33 PM, Ashley Sheridan wrote:
>>
>> $script_result = shell_exec('p7a.php');
>>
>> There is a lot of useful information here once you get past the snide
>> comments at the top.
>>
>>
>http://stackoverflow.com/questions/8450696/execute-a-php-script-from-another-php-script
>>
>>
>>
>> On 11/29/2015 11:36 PM, Ethan Rosenberg wrote:
>>>> Dear List -
>>>>
>>>> Can you not call a PHP script from a form -- action =
>'program.php'?
>>>> In my hands it does not work...
>>>>
>>>>
>>>> I have tried the following
>>>>
>>>> include program.php;
>>>>
>>>> later in the program ...
>>>>
>>>> form
>>>> .
>>>> .
>>>>
>>>> echo "<input type='hidden' name='next_step' value='step14' />";
>>>>
>>>> .
>>>> .
>>>> .
>>>> .
>>>>
>>>> switch ( $_POST['next_step'] )
>>>> {
>>>>
>>>>      case 'step14':
>>>>      {
>>>>          P7a.php();
>>>>      }                                }
>>>>
>>>> } //end Switch
>>>>
>>>> The result of this is that program.php runs at the start of the
>>>> script, and step14 is never called.
>>>>
>>>> I am totally confused.
>>>>
>>>> If you need more details, please specify.
>>>>
>>>> TIA.
>>>>
>>>> Ethan
>>>>
>> shell_exec is *not* the right method here. As others have pointed
>out, it appears there's a lack of understanding about how the PHP stack
>works with regards to requests and logic.
>Maybe not, but it can be done that way.  I never said it was correct.  
>The right way would be to write a class and use the methods of that 
>class to do what you need based on the choice made.

Generally, if you're in a situation where shell_exec is the solution, something went wrong with the question. I've only used it once, and that was when I didn't know any better 

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