Please Help - MySQL transaction works on command but returns warning through app

[jonabomer]

hi,

I was just wondering if anyone had a suggestion I am trying to display the results of a join on webpage. The selection renders a result when typed directly into the command line but only seems to return an invalid mysql result resource when run through the code.

the selection is as follows

SELECT b.request_id, CONCAT(a.lname, " ", a.fname) as name, b.received
FROM new_contacts as a, new_requests as b
WHERE a.new_contact_id = b.new_contact_id
ORDER BY b.received, name ASC;

...I've echoed out the $sql variable and it is exacly as printed above, and I've checked with an "if" statement...

if(mysql_query($sql)){
echo('it works');
}elseif(!mysql_query($sql)){
echo(mysql_error());
}

...and gotten nothing whatsoever printed to the screen.

1. I know it works from using the command line
2. I know its running as printed from echoing onto the screen
3. I cant seem to generate the mysql_error() msg but get the invalid result resource msg..

...Does anyone have answer or a suggestion (short of handling the data management with more application code VS. an SQL Join)

THANKS!

[jjfletch]

First, I'd check to make sure you're establishing a connection to MySQL.

Then, I'd try this:

$query = "SELECT b.request_id, CONCAT(a.lname, " ", a.fname) as name, b.received
FROM new_contacts as a, new_requests as b
WHERE a.new_contact_id = b.new_contact_id
ORDER BY b.received, name ASC";

$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_assoc($result)) {

print_r($row); 

} else {

echo "query not working";

}