PHP undefined variable that should clearly be defined

Question

I am wondering how $link can be undefined. It is a global inside class cdb.php and is set when the mysql connect is called. If it were undefined, when mysql connect was called, it would die because I have coded it this way.

<html>
<head>
</head>
<body>

<?php
function justGetCSNumbers($input)
{
$input = preg_replace('/[\D]/',"",$input);
$sp = preg_split("/,/",$input);
$numbs = preg_grep('/^(\d+)(,\d+)*$/',$sp);
$csv = implode(",",$numbs);
#echo $csv;
return $csv;
}
function queryDB($cleaned)
{
$split = preg_split('/,/',$cleaned);
$resAy = array();
for($i=0;$i<count($split);$i++)
{
        if((strlen($split[$i])>5)&&(strlen($split[$i])<10))
        {
        $resAy[$i] = "uid='$split[$i]'";
        }
}
if(count($resAy)>0)
{
$q = 'SELECT * FROM userbase WHERE '.$resAy[0];#.$whereclause;
echo '<br/> query: '.$q.'<br/>';
connectDB();
return mysql_query($q,$link) or die("Couldn't complete query ".mysql_error($link));
}
}
function find(){
$p = $_POST['userToQuery'];
if(isset($p))
{
$csv = justGetCSNumbers($p);
$found= queryDB($csv);
}
}
include('cdb.php');
find();
?>

Sorry for the poorly formatted code, using vi.

My apache2 error log shows that the variable $link is undefined when i use it here, even after calling connectDB(); which is the code that does a mysql_connect and therefore sets up the link. 'mysql_error() expects parameter 1 to be resource, null given' I refactored my code so the link would be defined (this version), but somehow I am having trouble.

[EDIT] Here is the cdb.php class:

<?php
function connectDB()
{
global $link;
$uname = 'site123';
$pass = 'abc123';
$loc = "localhost";
$link = mysql_connect($loc, $uname, $pass) or die("Couldn't connect to the DB");
$dbname = 'jagrail';
$db = mysql_select_db($dbname,$link);
        if(!$db)
        {die("Failed to select db");}
        if (mysqli_connect_errno()) {
                printf("Connect failed: %s\n", mysqli_connect_error());

        }
#return $link;
}
?>

jka6510 · Accepted Answer · 2012-02-15 01:59:30Z

1

EDIT:

Add:

global $link;

to the top of the queryDB() function, so that it knows $link is a global variable, rather than just a local.

edited Feb 15, 2012 at 1:59

answered Feb 15, 2012 at 1:52

jka6510

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5 Comments

YoYoMa Over a year ago

I was using that exact code previously. It always triggered the exact same error. As I said above, "It is a global variable inside class cdb.php and is set when the mysql connect is called. If it were undefined when mysql connect was called inside cdb.php, it would die because I have coded it this way." This method of getting $link works perfectly in another class I have, I don't know why its not working here.

jka6510 Over a year ago

Post your connectDB() method (omitting password, etc).

jka6510 Over a year ago

Ah, global variable, then add: global $link; to the top of the queryDB() function.

YoYoMa Over a year ago

shouldn't calling connectDB() set the variable $link for the same scope that connectDB() was called in since $link is a global? btw, your above comment seems to have fixed it. But do I have to add global $link to the top of queryDB(), couldn't I just add $link to the top?

jka6510 Over a year ago

Nope, it sets it in the global scope (outside of any function), but any variables you immediately use inside a function are assumed to be local, not global - thats what the global keyword is for.

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