On 12/29/06, Jess Robinson <[EMAIL PROTECTED]> wrote:
On Tue, 26 Dec 2006, Matt LeBlanc wrote:
> On 12/26/06, Jess Robinson <[EMAIL PROTECTED]> wrote:
>> $schema->resultset('boards')->search({
>> 'hiddenboards.userId' => 3,
>> 'me.active' => 1,
>> }, {
>> 'join' => 'hiddenboards',
>> '+select' => [ 'hiddenboards.userId' ],
>> '+as' => [ 'hidden' ],
>> 'order_by' => [ 'pos' ],
>> })
>>
>>
>> .. and set up the relationship between board and hiddenboards to use an
>> inner join, not a left join.
>>
>> I've not seen the syntax "SELECT .. hiddenboards.userId IS NOT NULL as
>> hidden" .. it's odd, why not use the proper join type?
>>
>> Jess
>
> ... this isn't the same thing. He does not want all of the hidden
> boards, as your change to an inner join would return. He wants all
> boards with an extra field that tells him whether or not the board is
> hidden (if the left join doesn't work, hiddenboards.boardId is null
> and as such hidden is false).
>
Oh! That makes a lot more sense.. OTOH the JOIN .. ON ..
hiddenboards.userId = 3 surely negates that effect?
Jess
Sorry, I just realized I used the wrong field. Of course, this doesn't
matter as if the join condition fails, anything from hiddenboards is
null. Here is a full example of how this would work:
CREATE TABLE boards (
id int,
active int,
pos int
);
INSERT INTO boards VALUES(1,1,1);
INSERT INTO boards VALUES(2,1,2);
INSERT INTO boards VALUES(3,1,3);
CREATE TABLE hiddenboards (
userid int,
boardId int
);
INSERT INTO hiddenboards VALUES(1,1); -- not userid 3, so no join
INSERT INTO hiddenboards VALUES(3,2); -- userid 3, so valid join
-- no entry for id 3 so no join
SELECT boards.id,CASE WHEN hiddenboards.userid is not null THEN 1 ELSE
0 END AS hidden
FROM boards LEFT JOIN hiddenboards
ON hiddenboards.userid = 3
AND hiddenboards.boardid = boards.id
WHERE boards.active = 1
ORDER BY pos;
In this particular case, the rows returned are as follows:
id,hidden
1,0
2,1
3,0
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