Jump to content

Query problem.


Go to solution Solved by Barand,

Recommended Posts

Ok - I have a working query to which I wanted to add one more item from a new table.  Thought it was easy but I must be having a bad day.  I will post the original working query (now commented out) and the new query with the added item 'driver_num':

/*
	$q = "select a.race_winner, w.wins,
			substr(a.race_winner, instr(a.race_winner,' ')+1) as last_name,
			a.race_name, date_format(a.race_date, '%m/%d') as race_date
			from trk_races a
		left outer join
			(select race_winner, count(race_date) as wins from trk_races
				where race_winner > '' and Season='$selyr'
				group by race_winner) w
		on w.race_winner = a.race_winner

		where a.race_winner > '' and a.Season='$selyr'
		order by $orderby";
*/

	$q = "select a.race_winner, w.wins,
			substr(a.race_winner, instr(a.race_winner,' ')+1) as last_name,
			a.race_name,
			date_format(a.race_date, '%m/%d') as race_date,
			d.driver_num
			from trk_races a, drivers d
		left outer join
			(select race_winner, count(race_date) as wins from trk_races
				where race_winner > '' and Season='$selyr'
				group by race_winner) w
		on w.race_winner = a.race_winner

		where a.race_winner > ''
			and a.Season = '$selyr'
			and	d.driver_season = a.Season
			and	a.race_winner = d.driver_name
		order by $orderby";

The only addition is the "drivers d" in the from clause of the first select along with the additions to the ending where clause that connects the "drivers" table to the "trk_races" table in that select.

Here is the error I am getting from this change:

Fatal error: Uncaught PDOException: SQLSTATE[42S22]:
Column not found: 1054 Unknown column 'a.race_winner' in 'on clause' 
in /home/albany/public_html/homejg/nascar/nas_show_drivers.php:129 
Stack trace: #0 /home/albany/public_html/homejg/nascar/nas_show_drivers.php(129): 
PDO->query('select a.race_w...') .....

The error column is already being used elsewhere in the query so I can't figure out why I am getting this error.

Link to comment
https://forums.phpfreaks.com/topic/314699-query-problem/
Share on other sites

  • Solution

Use an explicit join ( trk_races JOIN drivers) instead of trk_races,drivers

SELECT  a.race_winner, 
        w.wins,
        SUBSTRING_INDEX(a.race_winner, ' ', -1) AS last_name,
        a.race_name,
        DATE_FORMAT(a.race_date, '%m/%d') AS race_date,
        d.driver_num
FROM trk_races a
    JOIN
    drivers d ON a.season = d.driver_season
          AND a.race_winner = d.driver_name
    LEFT OUTER JOIN
    (
        SELECT race_winner, 
               COUNT(race_date) AS wins 
        FROM trk_races
        WHERE race_winner > '' AND Season=2021
        GROUP BY race_winner
    ) w ON w.race_winner = a.race_winner
WHERE a.race_winner > ''
AND a.Season = 2021


+-------------+------+-----------+-----------+-----------+------------+
| race_winner | wins | last_name | race_name | race_date | driver_num |
+-------------+------+-----------+-----------+-----------+------------+
| G Harrison  |    2 | Harrison  | Race 1    | 01/01     |         38 |
| G Harrison  |    2 | Harrison  | Race 2    | 01/08     |         38 |
| J Lennon    |    1 | Lennon    | Race 3    | 01/15     |         37 |
| R Starr     |    1 | Starr     | Race 4    | 01/22     |         39 |
| P McCartney |    1 | McCartney | Race 5    | 01/29     |         36 |
+-------------+------+-----------+-----------+-----------+------------+

 

Link to comment
https://forums.phpfreaks.com/topic/314699-query-problem/#findComment-1595332
Share on other sites

But why?  One works and the other doesn't.  Why?  I would never think of this and have not had this issue before.  And - the error message pointed at the join for the outer join that was previously working and was not altered and not at the Where for the newly included table.

Edited by ginerjm
Link to comment
https://forums.phpfreaks.com/topic/314699-query-problem/#findComment-1595339
Share on other sites

Quote

INNER JOIN and , (comma) are semantically equivalent in the absence of a join condition: both produce a Cartesian product between the specified tables (that is, each and every row in the first table is joined to each and every row in the second table).

However, the precedence of the comma operator is less than that of INNER JOIN, CROSS JOIN, LEFT JOIN, and so on. If you mix comma joins with the other join types when there is a join condition, an error of the form Unknown column 'col_name' in 'on clause' may occur. Information about dealing with this problem is given later in this section.

source: Mysql Manual - 13.2.10.2 JOIN Clause

Your query would be interpreted as if it were:

select *
from trk_races a
inner join drivers d
	left outer join	(
		select race_winner, count(race_date) as wins from trk_races
		where race_winner > '' and Season='$selyr'
		group by race_winner
	) w
	on w.race_winner = a.race_winner
on 1=1

Essentially your left join becomes like a sub-join of the drivers table.  When you do a join like that you can only reference the parent join table (d) or some same-level sibling table (none here) in your on clauses.  Table a does not meet either of those conditions (it's a sibling of the parent) so it's invalid in your left join's ON clause.

In case it's unclear, when I say sibling tables I mean something like this.


select *
from a
left join b
	inner join c on c.Id=b.cID
	inner join d on d.Id=c.dID
on a.Id=b.aID

Table c is a sibling to table d, so it can be used in the on condition, as can table b as it's the parent table.  Table a cannot because it's the parent's sibling.

 

  • Like 1
Link to comment
https://forums.phpfreaks.com/topic/314699-query-problem/#findComment-1595343
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.