Include a php file without displaying its contents - php

Is it possible to include a php file without including its contents? I just want to access the functions and variables in that file without displaying any content. I tried this
<?
ob_start();
include('$file');
ob_end_clean();
?>
But this will hide only contents in php tag. I want to know how to hide others as well.

How to hide? Redesign your solution and separate your concerns! Do not mix logic with UI and so on.
Maybe you should apply the MVC or similar pattern(s).

While I totally agree with Peter's answer
I just tried this because I've never tried it before..
File toinclude.php:
<p>Loads of text</p>
<?php
function my_test()
{
echo 'Hello';
}
?>
Ooh a link
File includer.php:
<?php
ob_start();
include('toinclude.php');
ob_end_clean();
my_test();
?>
And it does work!
Output:
Hello

No you can't.
Include is meant to execute everything inside the file, there are no ways to prevent execution of some part of the file.
The only way is to edit the included file.

If your file is not printing any output You may try eval($fileContent).

Related

PHP - include/require/insert some data from one file to another

I'm not really familiar with php and would like to ask you to help me.
I want to insert css styling data from styles.css file into the <style> tag inside of index.php, and not sure how to do so with secure and optimized way.
Should I use <?php include 'styles.css;?> or <?php require 'styles.css;?> or what exactly should I do in order to insert the data in the optimized and secure way, please?
I use the readfile() for this. I think it does not load the whole file into the memory, so it is safer for outputting a large files. So the code can be:
<?php readfile('styles.css') ?>
require/include is for parsing a PHP file, not copying content.
Just read the file and echo the contents.
<?= file_get_contents('styles.css') ?>
where <?= is a shorthand for <?php echo.
Note that the include path in PHP environment may be different from that in the HTML, as understood by the browser. Better use __DIR__ . '/styles.css' to prevent problems.

Execute the Code in a PHP Page And Echo the HTML Output

I absolutely don't post a question here in SO unless I really can't find a way to solve my problem myself. I did a lot of googling and was not able to find a solution for this one problem I am about to describe.
Here is the problem. I am creating a templated php website. With templated I mean something like below:
<?php include("header.php");?>
<div id="content">
<div id="main">
<h2><?php echo($page_title);?></h2>
<?php
echo ($page_content);
?>
</div>
<?php include("sidebar.php");?>
</div>
<?php include("footer.php");?>
As you can see here page template ECHOES the content of the $page_content variable between header and footer sections to build the page.
To keep the code clean and separated (in my own way) I have been placing the html content in .txt files (let's say page1_content.txt) and assigning the txt content to this variable ($page_content) as below:
$page_content = file_get_contents("page1_content.txt");
My problem starts when I place some php code in page1_content.txt, lets' call this file page2_content.php (yes, I change the file from .txt to .php). Then I assign the content of this file to $page_content variable as below as usual:
$page_content = file_get_contents("page2_content.php");
Now, when the page template ECHOES page2_content.php contents the php code in it is also echoed as string and not executed, but I am trying to query a database and do some stuff in this file with some php code. I mean, I want the php code inside page2_content.php to be executed and the cumulative html code to be echoed by the "echo" line inside the template file.
How can I achieve this?
Please ask me any questions if you need more info/clarification.
Thanks
EDİT:
As many people here suggested the solution was including the file. Actually, I tried including the file before but it didn't look like it was working, it broke my template, so I though I was on the wrong track and quit the "include" way of doing this. Since everybody here is advising to use include I tried that again. I replaced the php code in "page2_content.php" with a basic 1-line code just to see if it gets executed before adding generated html code without breaking the template and it worked. Apparently my php code had a problem at first place and hence broke my template execution.
Now I have changed the template structure slightly and pages using the template, and it seems to work nicely. Thanks a lot everybody. I have up-voted every answer suggesting that I use include :)
As #Ali suggested, you could include the files. The other option which I highly suggest you do not use is the eval() function.
I think what you want to do is to include your content PHP file, not echo it (as you are doing with header.php and footer.php).
echo($page_content);
Would become as below:
include("page2_content.php");
You've already done this in your footer and sidebar, just use include()

PHP includes header or footer location

I was hoping someone could help. I have just started to dabble with PHP includes for time saving in the future. For example I want to change the footer and header on a web page once (using include) instead of copying and pasting the code 30 or 40 times - oh no... a typo start again.
Which brings me to the question(s) where is it best to place this script?
<?php include("includes/headernav.html"); ?>
Can it be placed in a div, or should it be placed at the top of your code under the body?
If I want to make an image/banner include module. Can I
<?php include("includes/image.jpg"); ?>
Or is best to wrap the image in html and apply like this?
<?php include("includes/imagewrapped.html"); ?>
Do not include .jpeg files directly, use a wrapper. Only use include with other PHP files.
As for including the header, do it any way that feels natural as long as it produces valid html. There is no particular reason to declare another div element.
Hope this helps:
<?php include("includes/ui_header.php"); ?>
My page content between header and footer
<?php include("includes/ui_footer.php"); ?>
You can probably save this as a function and call that function wherever you want to display.
It doesn't matter whether you put include in any place. However, it's better to put include in the top or bottom of your code
While including headers/footers/menus on the site, please keep in mind following things:
1) Your header/footer includes(blocks) should be wrapped inside a div.
2) This way then can be differentiated and any new change to them can be done easily.
3) Its always a good practice to include a wrapper div around an element as CSS can use it for styling.
4) Your header/footer includes (blocks) should have a flexibility that even we place them in header,footer or any sidebar, they should not disturb the UI.
1) Because you are including the HTML file, you probably need to include it where you want to display it.
2) Create a function in php where you send only image URL (maybe some other parameters) and function returns the HTML code (String) which you only echo on page where you want to display it. This way you can ensure, that all images will have the same code and styling.
for example
function generateImage($url=null) {
if (isset($url)) return '<img src='.$url.' style="width: 100px; height:100px; border: 1px;" />';
else return false;
}
The better way is to include always a php file.

PHP - How to echo a variable that contains php?

If I have a piece of code that reads a chunk of HTML from a txt file and then echos that html onto the page, how can I accomplish the same task, but when there is PHP inside of the txt file?
ex:
this is the file being read:
<?php
$filecontent = // read some other file
echo($filecontent);
?>
and this is the page that is reading the file:
<?php
$code1 = //reading the above file
?>
<html>
<?php echo($code1); ?>
</html>
When you want to process files containing PHP code you need to use include instead of echo.
<?php include('your_php_file_name'); ?>
If you have the contents of the file in a string you are in a tough spot because the only way to process the code is eval, and in addition you have to properly set up any environment that the code requires. eval itself should be avoided, and the latter is impossible to do in the general case.
Use include instead of echo:
<?php include($file_that_contains_php); ?>
you need to include the first file and echo statement in the first file will get executed.
<html>
<?php require_once("firstfile.php"); ?>
You need to echo htmlentities($code1), because when you echo then browser will not show it contents, because it try to parse it as a html tag, but htmlentities will encode to safe html output this characters.
If you want to evaulate the code, then you need eval($code1) or include it.

php - include_once seems to print the desired page into the html

My problem is that I try to include a class using include_once(); but instead of including the desired file, it prints the code in the file straight into the source html so the code appears on my webpage.
This is the code:
<?
include_once('php/class.loadClients.php');
$client = new loadClient();
$client->connect();
?>
I'm not sure why its doing this as I've used include_once(); elsewhere in this page and it works fine:
include_once('php/class.register.php');
Thanks in advance!
Open php/class.loadClients.php, and add <? to the beginning of it.
To add to Paulpro's answer: if his solution does not work your server might not have short tags enabled ( <? vs <?php ). Use the full <?php tag.

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