How do I load a new php/html file ?

[juli]

Hi!
I need to call a new php file to be upload on my site but can't find a function for that.
For example, there is a function in java Servlet that loads up another file:

RequestDispatcher rd = 
	    getServletContext().getRequestDispatcher("/login.jsp");
if (rd != null)
	    rd.forward(req,res); 

It must be something like function 'forward' in php too.

Thank you.

[bernouli]

do you mean header() function?

[superwormy]

Call a new file to be uploaded...?

Do you mean you want a user to upload a file, or do you mean you want to include a file?

include(); is what I think you need...

www.php.net/include

[joshishreekar]

Hi

I think the original poster wanted to ask about Server.Transfer (ASP) equivalent in PHP. That is, "how to pass the control to another PHP page from the PHP script by submitting all the form variables to the next page"
Header("Location:xxx.php") simply redirects and the submitted info is lost.
So, whats the alternate ?
As to the answer, I have been searching for the same.
Any and all pointers and help is appreciated

Regards
Shreekar Joshi

[superwormy]

So you have form1.html which posts for form2.php. form2.php then includes form3.php, and form3.php does the processing of the form. Is that what you mean?

[joshishreekar]

Taking the same example given by you,
form1.html contains the payment method like Creadit Card, Cash, Cheque, DD etc.
There is a separate PHP page to handle each payment type.
Plus, there are a bunch of hidden vars to be passed as usual.
form1.html-->POST to form2.php
form2.php decides which page to call
Now, if I use conditional include, I could get away with it. But, suppose the logic where I am checking the type of payment resides after some HTML is o/p to the browser, like the standard header etc., then can I clear this HTML and then tell the Apache Server to simulate a form POST to the PHP form decided in the logic of form2.php.
This might happen when the HTML to be shown is different for each type of payment and it is imperative that the files be separated.
This requirement statement might be a bit flawed but I hope u get the idea.

Regards
Shreekar Joshi

[superwormy]

Well, you can't Redirect the user ( with PHP anyway ) after you have already sent output. BUT, you could use Output Buffering:
http://www.php.net/manual/en/ref.outcontrol.php

But in any case, that's not what I meant :-) I don't think you need to redirect the user anywhere. Do this instead:

------ form1.html --------

<form method="post" action="form2.php">

<inpug type="text" name="persons_name" value="" />

<select name="payment_type">
<option value="credit">Credit Card</option>
<option value="check">Check</option>
<option value="dd">DD Number</option>
</select>

<input type="hidden" value="whatever" name="hidden_value" />

<inpug type="submit" value="submit" />

</form>

---

--------- form2.php ( form handler ) -------

<?php

include ("header_file.php");
include ("menu_file.php");

if ($POST['payment_type'] == "credit") {
include ("credit_card_form.php");
} else if ($POST['payment_type'] == "check") {
include ("check_form.php");
} else {
include ("dd_form.php");
}

?>

---

Now, the form that comes up is different depending on which payment method is used, and whichever payment method form (credit_card_form.php, check_form.php, dd_form.php) is included gets full access to all of the posted values, so it can see the hidden value $_POST['hidden_value'] which is equal to "whatever"

Will that work? Otherwise I would say you need to otherwise use Output Buffering or a Javascript onLoad="submit.thisform" kinda deal. Ugly.