Jvhcom wrote:
Hello Neil and Chris,

Here are the database-related functions that reside in their own separate file:

function connecttodatabase() {
    global $link_resource;
    if(!($link_resource=mysql_connect('host', 'user_name', 'password'))) {
printf("Error connecting to host %s, by user %s", 'host', 'user_name');

    if(!mysql_select_db('databasename', $link_resource)) {
        printf("Error in selecting %s database", 'databasename');
printf("ERROR:%d %s", mysql_errno($link_resource), mysql_error($link_resource));

function runquery($dbquery, $link_resource) {
    global $result;
    global $numberRows;
    $result = mysql_query($dbquery, $link_resource);
    if (!$result) {
        printf("Error in executing: %s ", $dbquery);
printf("ERROR: %d %s", mysql_errno($link_resource), mysql_error($link_resource));
    } else {
        $numberRows = mysql_num_rows ($result);

Here is the dropdown list function that lives in a separate file with other dropdown functions
in which I use the database functions.

function dd_company($company_id = 0) {
$dbquery="SELECT id, name from companies where enabled = 'yes' order by name";
    runquery($dbquery, $link_resource);

The problem is here.

Inside this function, 'link_resource' doesn't exist.

You can change that easily:

function dd_company($company_id=0)
    global $link_resource;

    $dbquery = "SELECT .....";

Problem solved.

Also note that unless you are using multiple database connections in the one script, you don't need to pass around the $link_resource.

See http://php.net/mysql_query for more info -

If the link identifier is not specified, the last link opened by mysql_connect() is assumed.

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