PHP - Php Execute Linux Terminal Command (bash Script)

Hi guys,

I'm trying to run a php script from command line in linux.  I installed xampp for linux a.k.a lampp so typing

php /path/to/file.php

Just brings up a message saying I need to install php.

Whats the correct way to do this?

Hi, I have a strange problem using the exec command.
I have the following php code :
<?php
error_reporting(E_ALL);
ini_set('display_errors','On');

$accountBase = "HORAIREMOBILE";
$primaryUser = "DUM";
$secondaryUser = "DUM";
$result = exec(escapeshellcmd("/home/evidian/utils/getAccount ".escapeshellarg($accountBase)." ".escapeshellarg($primaryUser)." ".escapeshellarg($secondaryUser)),$output,$return_val);
echo $result;
?>

When I execute the command from the CLI, with any user, it just works fine, and shows my the result (basically a JSON formated output).
However, when I call the code frome the apache server, it simply returns nothing.

Could anybody help me with this issue ?

This is probably an easy one, but I can't figure it out and it's pretty much not searchable.

on a linux-machine i have installed filezilla

the filezilla runs pretty well and all is ok.

now i need to have the passwd that i have stored years ago. The passprhase is stored in a plain in a file called sitemanager.xmlfile


I want to find that file and open it with a terminal command.

find . -name *.sitemanager

well i thought that this will return the file I'm looking for. Now how do I open it automatically, without typing the name?
 

	find . -name *sitemanager.xm | open

This doesn't work. It says it doesn't found the open command.


question: why it does not work on opensuse?

should i use any other command - eg the following:
 

	find . -name *xyz | xargs open

or
 

find . -name *sitemanager.xml | xargs open

or

	find . -name *.xyz -exec open {} \;

and
 

find . -name *.xyz -exec open {} \; .

any and all help will be greatly appreciate

again: what is wanted and needet is to find out the passphrase in the filezilla-configuration

    

First off I want to thank everyone that is involved here in passing on the knowledge. After all that is what its all about. Okay so I want to do a campaign on password awareness and try to capitalize a little off of it. I came across the Breached_Compilation which is 1.4 billion email and plain text passwords. It has bash scripting in it to search and parse out the info from plain text documents which are in folders then prints the results in terminal. The other thing is that it posts the passwords in clear text. How would I obfuscate a portion of the clear text passwords? What would be the best way to integrate this to PHP? I am new to coding in PHP so any help is very much appreciated. Thanks an advance.

Here is the Bash script.

#!/bin/bash
dir=$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )

if [ "$1" != "" ]; then
	letter1=$(echo ${1,,}|cut -b1)
	if [[ $letter1 == [a-zA-Z0-9] ]]; then
		if [ -f "$dir/data/$letter1" ]; then
			grep -ai "^$1" "$dir/data/$letter1"
		else
			letter2=$(echo ${1,,}|cut -b2)
			if [[ $letter2 == [a-zA-Z0-9] ]]; then
				if [ -f "$dir/data/$letter1/$letter2" ]; then
					grep -ai "^$1" "$dir/data/$letter1/$letter2"
				else
					letter3=$(echo ${1,,}|cut -b3)
					if [[ $letter3 == [a-zA-Z0-9] ]]; then
						if [ -f "$dir/data/$letter1/$letter2/$letter3" ]; then
							grep -ai "^$1" "$dir/data/$letter1/$letter2/$letter3"
						fi
					else
						if [ -f "$dir/data/$letter1/$letter2/symbols" ]; then
							grep -ai "^$1" "$dir/data/$letter1/$letter2/symbols"
						fi
					fi
				fi
			else
				if [ -f "$dir/data/$letter1/symbols" ]; then
					grep -ai "^$1" "$dir/data/$letter1/symbols"
				fi
			fi
		fi
	else
		if [ -f "$dir/data/symbols" ]; then
			grep -ai "^$1" "$dir/data/symbols"
		fi
	fi
else
	echo "[*] Example: ./query name@domain.com"
fi

 

I'm trying to make colourized output of bash script in php. So far I have this:

<?php
$cmd = '/home/thebalk/FiveM/manage.sh restart';
while (@ ob_end_flush()); // end all output buffers if any

$proc = popen($cmd, 'r');
echo '<pre>';
while (!feof($proc))
{
    echo fread($proc, 4096);
    @ flush();
}
echo '</pre>';

//
// Converts Bashoutput to colored HTML
//
function convertBash($cmd) {
    $dictionary = array(
        '[1;30m' => '<span style="color:black">',
        '[1;31m' => '<span style="color:red">', 
        '[1;32m' => '<span style="color:green">',   
        '[1;33m' => '<span style="color:yellow">',
        '[1;34m' => '<span style="color:blue">',
        '[1;35m' => '<span style="color:purple">',
        '[1;36m' => '<span style="color:cyan">',
        '[1;37m' => '<span style="color:white">',
        '[m'   => '</span>'
    );
    $htmlString = str_replace(array_keys($dictionary), $dictionary, $cmd);
    return $htmlString;
}
?>

and my output looks like this:

[1;36m BOT:[0;39m THE [1;32m TheBalkanRP-SRV01-test SERVER [0;39m WAS RUNNING [1;36m BOT:[0;39m STOPPING THE [1;32m TheBalkanRP-SRV01-test SERVER [1;36m BOT:[0;39m GTA V SERVER HAS STOPPED [1;36m BOT:[0;39m REMOVING CACHE FOLDER [1;36m BOT:[1;31m CACHE FOLDER HAS BEEN REMOVED [0;39m [1;36m BOT:[0;39m STARTING THE [1;32m TheBalkanRP-SRV01-test [0;39m SERVER [1;36m BOT:[1;32m SERVER HAS STARTED [0;39m

I know my function is not connected with the code above but what is the right way to do it? Thank you!

Hey y'all.

Probably a dumb question here but I'm at a loss. I've got a PHP service script that has an include, written as so:

$config = new Config('config/config.json');

The config directory is next to the PHP script and does contain the config.json file, so when I run this from the CLI everything works great. However, I've got about 17 PHP services I need to start and really don't feel like typing them all out, so I've written a shell script that does this:

#!bin/bash


cd /path/to/my/script

/usr/local/bin/php ./my-script.php

This does start the service, but it bombs out because it can't find the config.json file.

If I change the PHP to 

$config = new Config('./config/config.json');

it works as expected from the shell script. Updating the files is technically possible, but fraught right now for reasons I can't really get into (sorry). Anybody know what the difference is, or have any ideas on how to get around this?

Edited September 5, 2020 by maxxd

Hi,   I have a cron job that executes this script every 2 minutes:

<?php

// LOAD WP-LOAD.PHP

require('/opt/bitnami/apps/wordpress/htdocs/wp-load.php');

// INCLUDE AND EXECUTE SCHEDULER.PHP

include('/opt/bitnami/apps/wordpress/htdocs/wp-content/themes/yeelloe/scheduler.php');

?>

When I try to include; /opt/bitnami/apps/wordpress/htdocs/wp-content/themes/yeelloe/scheduler.php:

<?php

// EXPLODE AND PARSE WP-CONTENT; FUNCTIONS.PHP

 $parse_uri = explode( '/opt/bitnami/apps/wordpress/htdocs/wp-content', $_SERVER['SCRIPT_FILENAME'] );

// LOAD WP-LOAD.PHP

require_once( $parse_uri[0] . '/opt/bitnami/apps/wordpress/htdocs/wp-load.php' );

// LOAD TEMPLATE FUNCTION

CheckFunction();

?>

<?php

error_reporting(E_ALL);

$psPath = "powershell.exe";
$psDir = "C:\\wamp\\www\\ps\\";
$psScript = "SampleHTML.ps1";
$runScript = $psDir. $psScript;
$prem = "-Action enable";
$runCMD = $psPath. " " .$runScript. " " .$prem;

//var_dump($runCMD);
$output = exec($runCMD);
echo $output;

?>

Hello,   I am working on a small project to get results from powershell script by using PHP. For some reason in PHP logs I get Exec unable to fork. Above is the script I wrote to execute powershell script within php. My webserver is IIS 7, and app pool is using a domain user that has full rights for Powershell to execute and get remote server results.          

Hello,
The following is my situation where I seem to get a 500 error code from the linux server:

i have an 'index' file like this:
Code: [Select]
<?php
require("includes/config.php");

$a = $_REQUEST['a'];

switch ($a)
{
   case "home":
      include("frontpage/main.php");
   case "user-process":
      include("user-process.php");
}
?>
config.php is something like this:
Code: [Select]
<?php

require(includes/classes/session.class.php);
require(includes/classes/user.class.php);
require(includes/classes/db.class.php);
...
?>
Now if we fall into the case "home" it works fine. Instead, if we fall into user-process it writes to the logs file Fatal Error: Class User does not exist bla bla bla. Why doesn't it exist ? every class is included in the config.php file then index.php includes first config.php ( which has all the classes) and then includes the requested page.

I also have a .htaccess file which is as follows:
Code: [Select]
RewriteEngine On
RewriteRule ^([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?$ index.php?a=$1&b=$2&c=$3&d=$4&e=$5&f=$6&g=$7 [NC,L]
which is used to access in a SEO friendly way the pages that users request.

Hi,

 

I am a newbie at php and I recently tried making a small php configuration that runs on the localhost. 

execution.php

<?php
	echo "first script has been executed";
	exec('execution2.php');
?>

execution2.php

<?php
	echo "Second script has been executed";
?>

The script is designed to call another php file whereas on the web page I would expect seeing, "first script has been executed" and "Second script has been executed". I am honestly not sure how the execution method is supposed to work however I am not planning on using "include" or "require" since they do not meet my criteria for another project. I am using xampp localhost server on a windows 10 computer. I tried entering "localhost:8080/dir/execution.php" however it did not work. 

 

Any help would be appreciated, thanks!

Hi all

I have this problem on a server using php5, unix based, safe_mode is On globally, i have turned it off locally through php.ini.

Ok, this is testing example script i used:

$cmd = ( "php -v" );
$out = shell_exec( $cmd );
print $out;

On my own server this returns php version. On this mentioned server i'm using (commercial) this causes complete server breakdown, when logged in with SSH, i can't even issue "ls" command after that, nor find and kill the process.
What could be so wrong with it? I don't think calling php-cli would make any difference.

Seems to be a basic question, but I couldn't find an answer nor figure it out on my own.

Basically I have a script that takes out specific data out of the database, the script works on its own, now I just need a way to make the user execute it with a link or a button.

Example:
Category: [Smileys] - [Category2] - [Category3] - etc.

As soon as the user clicks on [Smileys] all data in the database which contains the word smileys in the category field gets selected and outputted as a list.

Again the script works, I just need to be able to execute it with a button.

If I understood it correctly I have to run the script by adding a
if (isset($_POST['Smileys'])) {
in front of the script.

But how do I build the connection with the text link?

How can I achieve the following scenario with an anchor tag?

Code: [Select]
if(isset($_POST['submit'])) {
     // execute script
}

I do need it for a label system to sort content with labels.

Hi    I've been stuck trying to execute a script for a few weeks now and I really need help.   The script is supposed to schedule social media posts via functions.php and wp-load.php but I get errors.   Can someone do me a favour?   Drop your email and I'll send you the details to connect to my database.   Thanks

Hi,   In reference to my first attached image, I have a form which displays two SELECT/drop-down fields (labeled "Store Name" and "Item Description".....and both of which pull-in values from two separate lookup/master tables, in addition to providing an additional option each for "NEW STORE" and "NEW ITEM").   Now, when first-run, and/or if "NEW STORE" and "NEW ITEM" are not selected from the drop-down's then the two fields in green ("New Store Name" and "New Item Name" are hidden, by means of the following code:  

<div class="new-store-container" id="new-store-container" name="new-store-container" style="display:none;">
    <div class="control-group">
        <div class="other-store" id="new_store_name">
            <?php echo standardInputField('New Store Name', 'new_store_name', '', $errors); ?>
        </div>
    </div>
</div>

  Conversely, if "NEW STORE" and/or "NEW ITEM" are selected from the two drop-down's then one (or both) of the "New Name" fields are unhidden by means of the following two pieces of code, one PHP and the second JS:    

<select class="store-name" name="store_id" id="store_id" onclick="toggle_visibility('store_id','new-store-container')">
    <?php echo $store_options; ?>
                            
    <?php
        if($values['store_id'] == "OTH")
            {
                echo "<option value='OTH' selected> <<<--- NEW STORE --->>> </option>";
            }
        else
            {
                echo '<OPTION VALUE="OTH"> <<<--- NEW STORE --->>> </OPTION>';
            }
    ?>
</select>

 

function toggle_visibility(fieldName, containerName)
{
    var e = document.getElementById(fieldName);
    var g = document.getElementById(containerName);
            
    if (e.value == 'OTH')
        {
            if(g.style.display == 'none')
                g.style.display = 'block';
            else
                g.style.display = 'none';
        }
}

  All  of that is working just fine. The problem I'm having is that when I click the "Create" button, after having left any one of the form fields blank, the two "New Name" fields are hidden again, which I don't want to happen i.e. I want them to remain visible (since the values of "store_id" and/or "item_id" are "OTH"), so that the user can enter values into one or both of them, without havng to click on the drop-down a second time in order to execute the "on-click" code.   The second attached image shows how the fields are hidden, after clicking "Create".   How can I achieve that? It would be greate if someone could cobble-up the required code and provide it to me, since I'm relatively new to this.   Thanks much.    Snap1.png   26.14KB   0 downloads    Snap2.png   149.47KB   0 downloads