PHP include Variable Fail

Question

PHP include Variable Fail. So i have problem with php variable inclusion, in php files where i need to call for specific variable in another php file, i dont always get the requested variable. And i dont understand why.

settings.php

<?php
$symbol = 'MUSIC';
$key='xxx';
$secret='xxx';
?>

database.php

<?php
include ('settings.php');


function selected($key, $secret){
$nonce=time();
$uri='https://something.com?key='.$key.'&symbol='.$symbol.'&nonce='.$nonce; //Problem is in here


$sign=hash_hmac('sha512',$uri,$secret);
$ch = curl_init($uri);
    curl_setopt($ch, CURLOPT_HTTPHEADER, array('apisign:'.$sign));
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$execResult = curl_exec($ch);
$obj = json_decode($execResult, true);
$info = $obj["result"]["info"];
return $info;

info.php

<?php include "balancedb.php"; 

$url = "https://somthingother.com/";
$fgc = json_decode(file_get_contents($url), true);

$info = selected($key, $secret);
    echo "<br>".$symbol."-----------------".$info."<br>";

?>

I get Notice: "Undefined variable: symbol in ..\database.php on line 7" "MUSIC-----------------"

In one place it takes tha variable but in other it doesnt. Why? How to fix that?

Some sensible indention and formatting would make this code much easier to read, and in turn, easier to troubleshoot. — Qirel, Aug 27 '17 at 12:04
$uri="https://something.com?key='$key'&symbol='$symbol'&nonce='$nonce'"; — Zast, Aug 27 '17 at 12:06
how are you using the GET arrays? Check if all are set/not empty. — Funk Forty Niner, Aug 27 '17 at 12:06
the selected function is being called from inside the balancedb.php file, is the settings.php file being included inn to this file? — Zast, Aug 27 '17 at 12:10
you also seem to be using this with a database, so if you're not showing relevant code for this, we can't help you. How is `info.php` include/used also? and what is `balancedb.php`? — Funk Forty Niner, Aug 27 '17 at 12:10
`function selected($key, $secret, $symbol){` and `$info = selected($key, $secret, $symbol);` — Federkun, Aug 27 '17 at 12:11
@Federkun funny you should post that ^, I deleted the first comment I posted about using 2 arguments in the method, rather than 3. I shouldn't have deleted it. — Funk Forty Niner, Aug 27 '17 at 12:12
function selected($key, $secret) re-declares the $key, $secret variable, which are being passed to it from elsewhere. including the secrets file above the function definition would be irrelevant! — Zast, Aug 27 '17 at 12:14
@user3633383 Right, to which I posted a comment about it most likely related to database work, which we don't know how it's used. The question is unclear for me. I think they're missing an argument in the method. — Funk Forty Niner, Aug 27 '17 at 12:16

score 0 · Answer 1 · answered Aug 27 '17 at 12:20

 <?php 
 include_once( "settings.php" ); //this must be included here, before calling the selected($key, $secret); function below!!!

 include_once( "balancedb.php" ); 

 $url = "https://somthingother.com/";
 $fgc = json_decode(file_get_contents($url), true);

 $info = selected($key, $secret);
 echo "<br>".$symbol."-----------------".$info."<br>";

 ?>

score 0 · Accepted Answer · answered Aug 27 '17 at 12:22

That's because of the Variable scope. You don't have access to settings.php's $symbol (global scope) inside selected.

What to do? Pass it as an argument:

function selected($key, $secret, $symbol) {

Then when you call it

// pass settings.php's $symbol here
$info = selected($key, $secret, $symbol);

For more information, read the documentation about how the scope of a variable works.

PHP include Variable Fail

2 Answers2