php evaluate code before getting file content - php

I have a file B590.php which is having a lot of html code and some php code (eg logged in username, details of user).
I tried using $html = file_get_content("B590.php");
But then $html will have the content of B90.php as plain text(with php code).
Is there any method where I can get the content of the file after it has been evaluated?
There seems to be many related questions like this one and this one but none seems to have any definite answer.

You can use include() to execute the PHP file and output buffering to capture its output:
ob_start();
include('B590.php');
$content = ob_get_clean();

function get_include_contents($filename){
if(is_file($filename)){
ob_start();
include $filename;
$contents = ob_get_contents();
ob_end_clean();
return $contents;
}
return false;
}
$html = get_include_contents("/playbooks/html_pdf/B580.php");
This answer was originally posted on Stackoverflow

If you use include or require the file contents will behave as though the current executing file contained the code of that B590.php file, too. If what you want is the "result" (ie output) of that file, you could do this:
ob_start();
include('B590.php');
$html = ob_get_clean();
Example:
B590.php
<div><?php echo 'Foobar'; ?></div>
current.php
$stuff = 'do stuff here';
echo $stuff;
include('B590.php');
will output:
do stuff here
<div>Foobar</div>
Whereas, if current.php looks like this:
$stuff = 'do stuff here';
echo $stuff;
ob_start();
include('B590.php');
$html = ob_get_clean();
echo 'Some more';
echo $html;
The output will be:
do stuff here
Some more
<div>Foobar</div>

To store evaluated result into some variable, try this:
ob_start();
include("B590.php");
$html = ob_get_clean();

$filename = 'B590.php';
$content = '';
if (php_check_syntax($filename)) {
ob_start();
include($filename);
$content = ob_get_clean();
ob_end_clean();
}
echo $content;

Related

Include a php file, but return output as a string instead of printing

I want to include a file, but instead of printing output I want to get it as string.
For example, I want include a file:
<?php echo "Hello"; ?> world!
But instead of printing Hello world! while including the file I want to get it as a string.
I want to filter some elements from the file, but not from whole php file, but just from the html output.
Is it possible to do something like this?
You can use php buffers like this:
<?php
ob_start();
include('other.php');
$script = ob_get_contents(); // it will hold the output of other.php
ob_end_clean();
EDIT: You can abstract this into a function:
function inlcude2string($file) {
ob_start();
include($file);
$output = ob_get_contents(); // it will hold the output of other.php
ob_end_clean();
return $output;
}
$str = inlcude2string('other.php');

php - output buffering ob_get_contents not returning anything

I am trying to generate a static HTML from php however, i cannot get anything to output to $page, any ideas? cheers
ob_start();
$pageident = $tempdbid;
include 'newpagegenerator.php';
$page = ob_get_contents();
ob_end_clean();
You need to output it using an echo statement.
ob_start();
$pageident = $tempdbid;
include 'newpagegenerator.php';
$page = ob_get_contents();
echo $page;//<----- echo here
ob_end_clean();

Executing another PHP file and return the output from a method, is that possible?

Suppose if I have a PHP script test.php which have a method
<?php
function execute($filename){
//do something
return $output;
}
?>
and I also have another PHP script executable.php
<?php
echo "I am executed";
?>
then can i run any code to excute the second file and return the output from the first method execute when i call echo execute('executable.php'); ?
I guess you guys can understand what I meant.
You can use output buffering, as long as the file being included doesn't already do that:
ob_start();
require $filename;
$content = ob_get_contents();
ob_end_clean();
return $content;
Use ob_Start and ob_get_contents to capture the output of the script. Something like this should work:
<?php
function execute($filename){
ob_start();
include $filename;
$output = ob_get_contents();
ob_end_clean();
return $output;
}
<?php#test.php
include 'executable.php';
echo $test;
?>
<?php#executable.php
$test = "I am executed";
?>
function execute($filename){
include_once($filename);
}
$filename is the name of your file to be included.. I think this will help you...
This is the function call..
execute('abc.php');

How do I populate a variable using a function?

I have the code below on a page basically what I'm trying to do is fill $content variable using the function pagecontent. Anything inside pagecontent function should be added to the $content variable and then my theme system will take that $content and put it in theme. From the answers below it seems you guys think I want the html and php inside the actual function I don't.
This function below is for pagecontent and is what I'm currently trying to use to populate $content.
function pagecontent()
{
return $pagecontent;
}
<?php
//starts the pagecontent and anything inside should be inside the variable is what I want
$content = pagecontent() {
?>
I want anything is this area whether it be PHP or HTML added to $content using pagecontent() function above.
<?php
}///this ends pagecontent
echo functional($content, 'Home');
?>
I think you're looking for output buffering.
<?
// Start output buffering
ob_start();
?> Do all your text here
<? echo 'Or even PHP output ?>
And some more, including <b>HTML</b>
<?
// Get the buffered content into your variable
$content = ob_get_contents();
// Clear the buffer.
ob_get_clean();
// Feed $content to whatever template engine.
echo functional($content, 'Home');
As you are obviously a beginner here's a much simplified, working version to get you started.
function pageContent()
{
$html = '<h1>Added from pageContent function</h1>';
$html .= '<p>Funky eh?</p>';
return $html;
}
$content = pageContent();
echo $content;
The rest of the code you post is superfluous to your problem. Get the bare minimum working first then move on from there.
Way 1:
function page_content(){
ob_start(); ?>
<h1>Hello World!</h1>
<?php
$buffer = ob_get_contents();
ob_end_clean();
return $buffer;
}
$content .= page_content();
Way 2:
function page_content( & $content ){
ob_start(); ?>
<h1>Hello World!</h1>
<?php
$buffer = ob_get_contents();
ob_end_clean();
$content .= $buffer;
}
$content = '';
page_content( $content );
Way 3:
function echo_page_content( $name = 'John Doe' ){
return <<<END
<h1>Hello $name!</h1>
END;
}
echo_page_content( );

How do you define a PHP include as a string?

I tried:
$test = include 'test.php';
But that just included the file normally
You'll want to look at the output buffering functions.
//get anything that's in the output buffer, and empty the buffer
$oldContent = ob_get_clean();
//start buffering again
ob_start();
//include file, capturing output into the output buffer
include "test.php";
//get current output buffer (output from test.php)
$myContent = ob_get_clean();
//start output buffering again.
ob_start();
//put the old contents of the output buffer back
echo $oldContent;
EDIT:
As Jeremy points out, output buffers stack. So you could theoretically just do something like:
<?PHP
function return_output($file){
ob_start();
include $file;
return ob_get_clean();
}
$content = return_output('some/file.php');
This should be equivalent to my more verbose original solution.
But I haven't bothered to test this one.
Try something like:
ob_start();
include('test.php');
$content = ob_get_clean();
Try file_get_contents().
This function is similar to file(), except that file_get_contents() returns the file in a string.
Solution #1: Make use of include (works like a function): [My best solution]
File index.php:
<?php
$bar = 'BAR';
$php_file = include 'included.php';
print $php_file;
?>
File included.php:
<?php
$foo = 'FOO';
return $foo.' '.$bar;
?>
<p>test HTML</p>
This will output FOO BAR, but
Note: Works like a function, so RETURN passes contents back to variable (<p>test HTML</p> will be lost in the above)
Solution #2: op_buffer():
File index.php:
<?php
$bar = 'BAR';
ob_start();
include 'included.php';
$test_file = ob_get_clean(); //note on ob_get_contents below
print $test_file;
?>
File included.php:
<?php
$foo = 'FOO';
print $foo.' '.$bar;
?>
<p>test HTML</p>
If you use ob_get_contents() it will output FOO BAR<p>test HTML</p> TWICE, make sure you use ob_get_clean()
Solution #3: file_get_contents():
File index.php:
<?php
$bar = 'BAR';
$test_file = eval(file_get_contents('included.php'));
print $test_file;
?>
File included.php:
$foo = 'FOO';
print $foo.' '.$bar;
This will output FOO BAR, but Note: Include.php should not have <?php opening and closing tags as you are running it through eval()
The other answers, for reasons unknown to me, don't quite reach the correct solution.
I suggest using the buffer, but you have to get the contents and then clean the buffer before the end of the page, otherwise it is outputted. Should you wish to use the output from the included file, you should use op_get_contents(), which will return a string of the contents of the buffer.
You also don't need to loop over the includes as each will just add to the buffer (unless you clean it first).
You therefore could use the following;
ob_start();
include_once('test.php');
include_once('test2.php');
$contents = ob_get_contents();
ob_end_clean();
Hope this helps.
You can use the function file_get_contents.

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