create a php file using exec with variables

[moosey_man1988]

Hi Everyone

 

im currently creating an install script and i was hoping to touch and cat to a lot of files for them to be created, but im having troubles.

 

You will easily see in the code below what im trying to achieve.

//Lets create the mysqli_connection page for all other php pages to use with the new DB user
exec ('touch database/mysqli_connection.php');
exec ('echo "<?php
$mysqli_connection = mysqli_connect('.$servername.','. $dbnewuser.','. $newuserpass.','. $dbname.'); 
if (!$mysqli_connection){
die("Connection Failed: " . mysqli_connect_error());
}
?>" > database/mysqli_connection.php'); 

the issue I have is this is how the file comes out, its having trouble passing $mysqli_connection as a string.

 

see the mysqli_connection.php below

<?php
 = mysqli_connect(localhost,testuser,testPassword,testDB);
if (!){
die(Connection Failed:  . mysqli_connect_error());
}
?>

I did try using single quotes around the $mysqli_connection variable like so

 

'.$mysqli_connection.' but the file then comes out blank.

 

any help given would be greatly appreciated, and i thank everyone for their help in advance

Edited October 6, 2015 by moosey_man1988

[Jacques1]

Well, nesting three languages just isn't a good idea: You have a PHP code generating shell code generating PHP code. No wonder you have syntax conflicts.

 

This can be made a lot simpler. PHP can in fact write files, there's no need to delegate this to the shell. There's also no need to generate an entire PHP script. Just define a bunch of constants and put the application logic into static scripts:

<?php

const DB_USER = <insert string here>;
const DB_PASSWORD = <insert string here>;
...

You should also consider using a simple data format like JSON, YAML or XML instead of a full-blown PHP script.

[moosey_man1988]

those are yet languages on my list to learn this script will ever be improved upon with new langues as I learn more languages etc etc.

 

Thanks for the advice, im still not clear on ALL of a functions that PHP can do, but this is what forums and guides are for

 

I will test out the put_file_contents, Thanks for a swift response

[moosey_man1988]

i created a test script, still no luck

<?php

$servername = "localhost";
$username = "root";
$DBpassword = "testPassword";
$dbname = "testDB";
$invoiceDB = "invoices";

$dbnewuser = "testUser";
$newuserpass = "testPass";

$file = "database/mysqli_connection.php";

$myString = "<?php 
$mysqli_connection = mysqli_connect($servername,$dbnewuser,$newuserpass,$dbname); 
if (!$mysqli_connection){
die('Connection Failed: ' . mysqli_connect_error());
}
?>";

file_put_contents($file, $myString);
$getData = file_get_contents($file);
echo $getData;

?>

Ignore the unused variables, this was just something quick to see if can be accomplished.

 

here's the error

 

PHP Notice:  Undefined variable: mysqli_connection in /var/www/html/test.php on line 14

PHP Notice:  Undefined variable: mysqli_connection in /var/www/html/test.php on line 19

 

I really do hope to make a connection script, as once the installation finished it, all the rest of my script call upon the mysqli_connection.php

Edited October 6, 2015 by moosey_man1988

[Jacques1]

This makes no sense.

 

You want $servername, $dbnewuser etc. to be replaced with the corresponding values, but at the same time you want $mysqli_connection to show up as a literal variable in the generated script. How is PHP supposed to tell the difference? If you want a literal dollar sign within a double-quoted string, you need to escape it with a backslash:

echo "\$some_variable";

Secondly, you can't just insert strings into your code template, because those will end up as raw words. What you want are string literals, and those have quotes:

$string_content = 'yada yada yada';
echo "String literal: '".addslashes($string_content)."'";

You should escape the content with addslashes() so that a literal quote within the string content won't blow up your generated PHP script.

Edited October 6, 2015 by Jacques1

[moosey_man1988]

Hey Jaques, Thank you for your response again your answer solved it for me VERY quickly, I will remember in future that magical backslash

 

Resulting Code below

//Lets create the mysqli_connection page for all other php pages to use with the new DB user

$mysqli_file = "database/mysqli_connection.php";

$mysqli_connection_setup = "<?php
\$mysqli_connection = mysqli_connect($servername, $dbnewuser, $newuserpass, $dbname); 
if (!\$mysqli_connection){
die('Connection Failed: ' . mysqli_connect_error());
}
?>";

file_put_contents($mysqli_file, $mysqli_connection_setup);

Thanks again you must grow tired of PHP newbs

[seandisanti]

 

Thanks again you must grow tired of PHP newbs

 

At one point or another we were all there.

Edited October 7, 2015 by seandisanti