Question

I’m using the code:

function insertV($deptx, $coursex, $secx, $isbnx,$titlex, $authorx,$usedpx,$newpx)
{
$sql = "INSERT INTO `$deptx` (`course`, `sec`, `isbn`, `title`, `author`, `usedp`, `newp`) 
VALUES ('$coursex','$secx','$isbnx','$titlex','$authorx','$usedpx','$newpx')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }

}

$dept = "ACCOUNTG";
$course = "340";
$sec = "101";
$isbn = "9780324651140";
$title = "FINANCIAL ACCOUNTING";
$author = "STICKNEY";
$usedp = "$129.75";
$newp = "$199.75";
insertV($dept,$course,$sec,$isbn,$title,$author,$usedp,$newp);

But I keep getting an error:

Warning: mysql_query(): supplied
argument is not a valid MySQL-Link
resource in
/home/public_html/mysite.com/myscript.php
on line 14 Error:

I’m guessing it has to do with me trying to access a table using a var?

Any help would be appreciated, I’m still new to PHP if you couldn’t tell 😛

EDIT: Oh yeah, line 14 is

if (!mysql_query($sql,$con))

Answer 1

You aren’t defining your $con within the function. This is part of PHP’s scoping rules. Variables defined outside a function are not visible within a function unless you declare them global, you’d need something like this:

function insertV(...) {
    global $con;

    $query = "...";
    etc...
}

So, the error is due to $con being null within the function, which is “not a valid MySQL-link resource”