Make output of PHP file show in another?

Question

So let's say I have a PHP script which takes the POST data out of a HTML form, like this:

<?php
$data = $_POST['data'];
echo $data;
?>

And I have a HTML form in a PHP file, when the button is pressed it sends the data to the other PHP file, and the data will be echo'd, now I want the data to be echo'd in the file that has the form, not in the single PHP file. I hope I explained correctly. I know I can put the HTML and PHP in one file, but I don't really want to do that.

EDIT: OK, this is a better explanation, I need to take the data that's POSTED to a PHP file and echo it in another PHP file, understand now? I hope so.

You'll be better off using AJAX for this. Ajax will post html forms to php files and then you can determine what is then displayed back to the user within the success function in the ajax call — Andy Holmes, Nov 16 '13 at 17:15
I read this question like 5 times and i still don't understand what you really want to achieve. — Shankar Narayana Damodaran, Nov 16 '13 at 17:16
Have you looked at [`include`](http://php.net/manual/en/function.include.php) and [`require`](http://php.net/manual/en/function.require.php) functions and variations? — ACJ, Nov 16 '13 at 17:17
How about I use $_GET then file_get_contents? Would that work? — user2917204, Nov 16 '13 at 17:18
It's better to use AJAX, it does exactly what you are asking — Andy Holmes, Nov 16 '13 at 17:18
could you please be a bit more be more specific with your question? — , Nov 16 '13 at 17:18
I need to take the data that's POSTED to a PHP file and echo it in another PHP file. — user2917204, Nov 16 '13 at 17:21
Then learn my friend, its really simple. Does exactly what you want, and a bit nicer too — Andy Holmes, Nov 16 '13 at 17:23

jko · Answer 1 · 2013-11-16T17:33:20.530

3

first method: use output buffer

example file1.php

<?php echo $_POST['data']; ?>

example file2.php

<?php 
ob_start();
require "file1.php";
$output1 = ob_get_clean();
?>

read more about output buffers and nesting them: http://us2.php.net/manual/en/ref.outcontrol.php

second method: use return

example file1.php

<?php return $_POST['data']; ?>

example file2.php

<?php 
ob_start();
$output1 = require "file1.php";
?>

edited Nov 16 '13 at 17:33

answered Nov 16 '13 at 17:20

jko

2,068
13
25

Yea I know that, but the data is dynamic as it's a post method. – user2917204 Nov 16 '13 at 17:21

score 0 · Answer 2 · edited May 23 '17 at 11:59

0

If you use AJAX you will be able to easily accomplish what you are asking for. Your front end form will post to a php form and then you can return the values you like inside the AJAX success function. A similar question has been asked here and the answer explains how to use it well - Ajax tutorial for post and get

edited May 23 '17 at 11:59

Community

1
1

answered Nov 16 '13 at 17:21

Andy Holmes

7,817
10
50
83

I'll take a look at that, although I'm not experienced in AJAX, thanks! – user2917204 Nov 16 '13 at 17:24
http://www.cleverweb.nl/php/jquery-ajax-call-tutorial/ - this might help you out a bit. Effectively you set up an ajax call on a form on your page, send that data to your php file which you can read as normal POST data and then set that data to variables, echo it out into the document and the ajax can then show it on success. Theres a bit more to it, but highly suggest learning it :) – Andy Holmes Nov 16 '13 at 17:26
Hmm, nice, thanks, I sure will, is there's a way to do it with pure PHP? – user2917204 Nov 16 '13 at 17:31
There is absolutely no need to do it with just php when you can do it in AJAX easier. In php you will need to post to the same page as the html code is and then go from there. To do reading the contents of a file on post isn't as good as AJAX. AJAX is your best solution for this issue – Andy Holmes Nov 16 '13 at 17:35

Make output of PHP file show in another?

2 Answers2