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');
exit();
}
if(!mysql_select_db('databasename', $link_resource)) {
printf("Error in selecting %s database", 'databasename');
printf("ERROR:%d %s", mysql_errno($link_resource),
mysql_error($link_resource));
exit();
}
}
function runquery($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));
exit();
} 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) {
$dbquery="SELECT id, name from companies where enabled = 'yes' order by
name";
connecttodatabase();
runquery($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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