This is odd... I want to include a file inside a variable to print it later but it's already including the file without printing the variable...
$var = include('file.php');
// stuff
echo $var;
file.php:
echo $stuff;
Output:
Notice: Undefined variable: stuff in file.php
I commented the echo $var to make sure but it's included anyways...
Is there any way to just load the content inside the variable instead of including the file?
try file_get_contents
<?php
$var = file_get_contents('file.php');
echo $var;
?>
You can do it like you did
$var = include 'file.php';
The included file needs to return a value, though.
Last in file.php:
// Do some stuff (but don't echo anything)
return 'This will be put in the var-variable to be echoed or used when you see fit';
Related
At first, I know that using eval() is a bad choice,
but my team already built an application based on templates stored in a database.
I am appointed to edit a template that should use a variable in the script outside the eval() function.
Here is an code example:
template.php
function template($arg)
{
if($arg == "show")
{
$template = str_replace("\\'", "'", addslashes($row['template']));
}
return $template;
}
$row['template'] = "this is template from database that shows {$row[data]}";
$row[data] = {$variable};
show.php
<?php
include 'template.php'
$variable = "thisShouldAppear";
eval("\$var = \"".template("show")."\";");
echo $var;
?>
The output in show.php is:
{$variable}
And it's supposed to show:
thisShouldAppear
NOTE:
If I assign the variable from outside the database it works but I'm limited to including it from the template inside the database.
Example:
<?php
include 'template.php'
$variable = "thisShouldAppear";
$row[data] = "{$variable}";
eval("\$var = \"".template("show")."\";");
echo $var;
?>
In this case it works , but the aim is to have the value from the template not from the script.
Any help would be extremely appreciated, for I'm running out of schedule.
Thank you.
I want to use my index.php page as my template for all my other pages. So I'm printing it out with the code below.
echo file_get_contents("index.php");
I've added this piece of code into the template (index.php) where i want to display the contents. of whichever page im on.
<?php
echo $index_content;
?>
So when I use
echo file_get_contents("index.php");
to get my page template, on for example users.php. In the users.php file I want to use the code below
$index_content = echo "string";
to then print out my page contents where I added this variable
<?php
echo $index_content;
?>
My problem is when I say $index_contents = echo ("string");
it's not printing anything out. onto my template. or it prints the stuff out but at the end or the beginning of the template. not where i've inserted my variable. Why wont it echo out my stuff where I've inserted my variable.
file_get_contents() give you the source of your file.
If I get you right you want to use include instead. Also don't echo in a variable but assign the value and echo it in the template.
users.php:
$content = 'what ever';
include 'template.php';
other.php:
$content = 'other page';
include 'template.php';
template.php:
echo $content
If you call users.php output will be "what ever". If you call other.php output will be "other page".
You are storing the return value of "echo" in $index_content, which is empty.
Just omit the echo when assigning the string to the variable.
The other problem is, with file_get_contents you don't evaluate the php expression where you echo out the $index_content.
Instead, you should use include('index.php') in users.php, and set the variable $index_contents before that.
The below code will not display any output as the variable is declared below the echo as PHP gets executed line by line. Is there any way to search for the variable in the whole page and then execute the code?
<?php
include "header.php";
$title = "Test";
?>
header.php
<html>
<head>
<title><? echo $title ?></title>
</head>
You need to learn how compilers/interpreters works. PHP is interpreted language and The binary that lets you interpret PHP is compiled.
PHP run from top to bottom.
so its like
<?php // start from here
echo "$title"; <-- $title is undefined here
$title = "Test"; <-- now you declared $title with value so it goes in memory now
//end
So you need to first check weather $title is set or not than respond according to it
if(isset($title)){
echo $title;
}
According to your logic, I suggest you to use contants like below:
Create a separate file, let's say constant.php and include it on all other pages
<?
define("TITLE", "This is title");
?>
Use it like below:
<?php echo TITLE;?>
Thanks
I am trying to reach the contents of a PHP file without the file actually outputting what it would usually do. Here is my test code:
File1 (test1.php)
<?php
ob_start();
include_once './test2.php';
$test = ob_get_contents();
echo $test;
?>
and here is file2 (test2.php)
<?php
$testVar = 'Name!';
?>
<div class="testClass"><?php echo $testVar?></div>
<p>Spam2</p>
and I want it to only do this because of the
echo $test
line NOT because the file is outputting the content.
<p>Spam2</p><div class="testClass">Name!</div>
<p>Spam2</p></body>
due to the echo, but it returns this
<p>Spam2</p><div class="testClass">Name!</div>
<p>Spam2</p></body>
<p>Spam2</p><div class="testClass">Name!</div>
<p>Spam2</p></body>
So how do I get it to only return the content once?
Don't echo $test;. PHP is executing as it should. Since ./test2.php shows Spam in HTML it appears on the page, then you assign the page contents to a variable and echo it. What do you expect?
If you have 2 files say: app/index.php and app/config.php you can just use the return keyword to return some content from the config.php file. And then, when you include the file whatever you returned from config.php can be saved to a variable.
Example:
First return whatever you want from the config.php file (could be an array, string, etc).
<?php
return ['name' => 'Spam'];
Then in the index.php:
<?php
$contents = include_once('config.php');
echo $contents;
I wrote a php page which has two php tags and one script tag inside it .
<?php
$value = $_GET['hash'];
?>
<script>
function execute(){
<?php
$readfile = file($value);
for ($k=0;$k<=count($readfile)-1;$k++){
$cmd = $readfile[$k];
echo $cmd;}
?>
}
</script>
I want to use $value inside another php tag ( like above it has the file I want to open ), but I am not able to do it.Is the scope of variable limited to one php tag ? if yes how can I solve this problem Please help
Your code works perfectly. The variables in one PHP tag is accessible from all other tags, unless you define them inside a PHP function.
The reason you are not seeing the echo on the screen is because the echo prints to the Javascript function.
If you view the source of the generated page, the file contents will be there.
Try this:
function execute(){
<?php
$readfile = file($value);
for ($k=0;$k<=count($readfile)-1;$k++){
$cmd = $readfile[$k];
?>
alert( <?php echo $cmd; ?> );
<?php
}
?>
}
execute();
if $value is a get then you don't need to access it as a file, it should just be a short string.
just above line 7 (the one with $readfile = file...
type:
echo "alert(The hash value is: ".$value.")";
This will make an alert display (as it is in a script tag)
p.s you should have in your opening tag