Need help with a query and subsequent output

Databases
21

Did you try my variation?
Can you post your query with the JOIN?
And can you post the error it’s giving you?

22

here is the query:

SELECT certnumbers.setdescription , COUNT() AS qty , AVG(certnumbers.grade) AS avgcond, , (COUNT() / sets.setqty) * 100 AS pcomp FROM certnumbers INNER JOIN sets ON certnumbers.setdescription = sets.item AND certnumbers.sport = sets.sport GROUP BY certnumbers.setdescription

This is the error:

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in /usr/local/psa/home/vhosts/verisleeve.com/httpdocs/admin/mysql_send-4.php on line 28

23

I got it working…

SELECT
certnumbers.setdescription
, COUNT() AS qty
, AVG(certnumbers.grade) AS avgcond
, (COUNT() / sets.setqty) * 100 AS pcomp
FROM certnumbers
INNER JOIN sets
ON certnumbers.setdescription = sets.setdescription
AND certnumbers.sport = sets.sport
GROUP BY
certnumbers.setdescription

Now to add a where clause and add to the php output!

24

In the certnumbers table I have a column called owner. I’m having issues getting a where clause to work as follows:

SELECT
certnumbers.setdescription
, COUNT() AS qty
, AVG(certnumbers.grade) AS avgcond
, (COUNT() / sets.setqty) * 100 AS pcomp
FROM certnumbers
WHERE owner=ekramer
INNER JOIN sets
ON certnumbers.setdescription = sets.setdescription
AND certnumbers.sport = sets.sport
GROUP BY
certnumbers.setdescription

25 
SELECT
certnumbers.setdescription
, COUNT(*) AS qty
, AVG(certnumbers.grade) AS avgcond
, (COUNT(*) / sets.setqty) * 100 AS pcomp
FROM certnumbers
INNER JOIN sets
ON certnumbers.setdescription = sets.setdescription
AND certnumbers.sport = sets.sport
WHERE owner='ekramer'
GROUP BY
certnumbers.setdescription

You really should study the syntax a bit more :slight_smile:

26

By the way, this error

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in /usr/local/psa/home/vhosts/verisleeve.com/httpdocs/admin/mysql_send-4.php on line 28

only indicates that your query has gone wrong.
To know what went wrong, use this construction:


$result = mysql_query($query) or die(mysql_error());
27

That works! I had the quotes in my query… i just had it in the wrong place.

I’m learning and this site and users has been a tremendous help!