Code works untill it is in a function

[joesaddigh]

Hi,

 

I have some code which works but when I created a function and call this same code it doesn't. The error I get is as follows:

 

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/fhlinux010/l/languageschoolsuk.com/user/htdocs/admin/email.php on line 42

Error retrieving schools

 

The code

function CreateSchoolCheckboxes()
{
	echo '<div style="height:400px;width:400px;font:16px/26px Georgia, Garamond, Serif;overflow:scroll;">';

	$querySchools = "SELECT * FROM school";
	$result = mysql_query($querySchools, $conn)
		or die ("Error retrieving schools ".mysql_error());

	while($row = mysql_fetch_array($result))
	{
		$schoolname = $row['name'];    
		echo '<input type="checkbox" name="school" value="'.$schoolname.'">';
		echo $schoolname . '<br>';
	}							  		  

	echo '</div>';
}

 

Im sure that this is probably something simple but any suggestions would be much appreciated.

 

Thanks,

Joe

[PFMaBiSmAd]

You need to pass $conn into the function when you call it -

 

CreateSchoolCheckboxes($conn);

[joesaddigh]

Excellent Thanks!

 

I thought that the $conn would still be in scope?

 

Joe

[PFMaBiSmAd]

Functions have their own private variable scope so that they don't have any unintended interaction with the program where you use the function. Writing larger programs would become impossible if you needed to keep track of all the variables you used in and out of functions.

[joesaddigh]

That makes sense I can see the good reason for this.

Cheers