I included an html file like this, so that it is not displayed when the site loads:
<div id="menugrp0" class="menuhide">
<?php include 'menugrp0.html'; ?>
</div>
Now I want it to be shown at a specific spot. I am using this php code, to get some variables which are transported with the $_SESSION. I am using this kind of question for some simple html links, in which case it works perfectly:
if ($_SESSION['gruppe'] == $h['gruppe']) {
printf(' menugrp0.html');
}
I know that this is not working at all at the moment for this included html. I also tried to add the <?php [...] ?> tag inside the printf, which is also not working.
Is it possible to show a hidden included html file with a printf tag?
Try this one.
<?php
if($_SESSION['gruppe'] == $h['gruppe']){
echo 'Foo';
include ('/path/to/menugrp0.html');
echo 'Example: one';
}
?>
readfile('menugrp0.html'); // Reads a file and writes it to the output buffer. It is like read then "echo"
How to echo the whole content of an .html file in php?
Thanks to Daan Meijer, this works.
echo file_get_contents('...');
}
Related
I am writing an anti-ddos php script, and I want to write a custom load page, but the html page of the orginal html page would be rendered. Is there anyway for me to prevent the html showing up while showing up my php load page
<?php require "antiddos.php"?>
#Samuel Wang, you can try check that out by one of this examples.
below assume code of antiddos.php and other file from where you include it.
1. antiddos.php :
E.g. before your php code or you can keep html inside the file that won't be an issue.
<?php
echo"<p>This is a test html content of antiddos.php</p>";
function testcall($TextReceiver ){
return $TextReceiver;
}
function MathMultiplication($Prt1,$Prt2){
return $Prt1*$Prt2;
}
?>
2. testprocess.php :
You can use the ob_clean(); function next to your include file statement that will discards the contents / html output from the antiddos.php.
<?php require_once("antiddos.php");ob_clean();
echo testcall("Hello this is print call from testprocess.php to antiddos.php");
echo "<br>";
echo "MathMultiplication= ". MathMultiplication(4,3);
?>
Conclusion by both the files output: First from the antiddos.php and second one from the testprocess.php
I'm learning Php
And i made a main index.php page
And at some point it contains the line
<?php
include './BaseTemplate.php';
?>
BaseTemplate.php contains a lot of common plain html that is equal between all pages.
It looks like (but these are only a few lines) :
echo '<script src = "../assets/js/intention.js"></script>';
echo '<script src = "../assets/js/context.js"></script>';
echo '<head><body>';
echo '<table>';
But these echo commands dont get executed how should i resolve this ?
When you are including any file into your page the code for that file will come into the page.
So if the included file and the page where you are including it are in not in same path will create a problem.
It seems that you are getting that only. So please make the path in BaseTemplate.php according to the index.php, and it will work.
I solved it. I was thinking that include would put all code in a page from an external php source file. That's not the case, include files are libraries with functions which can be called.
So to resolve it, I rewrote it the BaseTemplate.php like this
<?php
Function WritePageBase() { // now i put it all inside a callable function
echo '<script src = "../assets/js/intention.js"></script>';
echo '<script src = "../assets/js/context.js"></script>';
echo '<head><body>';
echo '<table>';
}
?>
Now my other pages can call it like
<?php
include './BaseTemplate.php'; // include the file with functions
WritePageBase(); // call the function to echo all html code in page
?>
which reduces a lot of html code which now can be edited by a single file (and all other pages using it will change too).
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.
How would one go about showing PHP code on user end. Sort of like w3School does?
Having lets say a grey area div, and then showing the code in there without activating it?
You can use html entities <?php in the html it will be rendered as <?php
You can use htmlspecialchars to encode your code to use html entities.
Use <pre> or <code> tags to wrap your code.
Take a look at http://php.net/manual/en/function.highlight-string.php to further see how you can make the code look pretty.
Since passing a large block of code to highlight_string() can be messy, you may want to look at output buffering in combination with highlight_string to output colorized php code.
Something like:
<?php
ob_start();
?>
phpinfo();
echo "this echo statement isn't executed";
<?php
$code = ob_get_clean();
highlight_string($code);
?>
Simply you can use following code to display php code on webpage.
highlight_string("<?php print('This is php code.'); ?>");
It will give output like
<?php print('This is php code.'); ?>
The first step is to not wrap that code in PHP tags. So instead of this:
<?
var sample = "code";
?>
You would have this:
var sample = "code";
It's not the code itself which triggers the server-side compile from the PHP engine, it's the tags which indicate to that engine what blocks of the file are code and what are not. Anything that's not code is essentially treated as a string and output to the page as-is for the browser to interpret.
Once you're outputting the code, it's then a matter of formatting it. The old standard is to wrap it in pre tags to get rid of HTML-ish formatting:
<pre>
var sample = "code";
</pre>
You can also apply CSS style to the pre tags (or any other tags you want to use for displaying code, such as div) as you see fit.
There are also very useful code syntax highlighting plugins and tools to make the code a lot "prettier". Google-code-prettify often comes highly recommended.
Typically this is done by showing code within <pre> or <code> tags.
You can use this template........
######################################################################
echo "<h2><br>Source Code of ".basename((string)__FILE__) . "</h2><hr>";
show_source(__FILE__);
echo "<hr>";
echo "<h2>Output of ".basename((string)__FILE__) . "<hr></h2>";
#######################################################################
It will show the source code and output following.
use the header function of php, this will rea
<?php
header("content-type: text/plain");
?>
The PHP code will just be a string that you can echo or print onto the page, no different than any other data you want PHP to display for you. If you want to keep the formatting (ex. the indentation), put it inside a <pre><code> block.
Ex:
$php_code = '<?php $foo = bar; ?>';
echo "<pre><code>$php_code</code></pre>";
Well most is in the title. I wonder if it's supposed to be that way or i can do the same without an if(1) condition I'm doing this because my website pages are all as php includes.
Thank you all
Answer retained:
Okay basically the way to do it is simply to include('file.php') as it will be considered out of the current <?php ?> environment.
Putting
<?php if(1): ?>
...
<?php endif; ?>
around your HTML code in a PHP file will have no effect on the result. You will still be able to include the file without it.
You can think of it like the "default mode" for a PHP file is that it contains HTML content. You only need to add <?php ?> tags if you want to add PHP code. If you're just putting HTML code in a PHP file, they're unnecessary.
The beauty of PHP is that you can move "in" and "out" of PHP very easily. You can do the following without issues:
<?PHP
if(whatever) {
?>
your HTML
<?php
include('whatever.php');
?>
more HTML
<?PHP
}
?>
To build on Zak's answer:
You can also use PHP to echo out things that aren't PHP... as long as you quote it appropriately.
<?php
//HTML
while ($x < 5) {
echo "<p> this is html that you can wrap with html tags! </p>";
$x++;
}
//Javascript
echo "<script type='text/javascript'>
some javascript code
</script>"
?>
Although, it's less confusing to just end the php tag to keep things separate.
And you can even use php as you want within html or javascript as long as you put the tags, and as long as the file is saved as a .php file (so PHP can be processed on the server).
Ex:
<script type="text/javascript">
//set a javascript image array to a php value
var imgArray = [<?php echo implode(',', getImages()) ?>];
</script>
But if you want to do this the other way around (IE, assign a browser-compiled value, such as a javascript value to a php value), you'll need to use AJAX.