SQL join table to self to find difference - mysql

Question

Consider the below table
CREATE TABLE `temp` (
`id` int(11) NOT NULL,
`lang` char(2) COLLATE utf8_unicode_ci NOT NULL,
`channel` char(2) COLLATE utf8_unicode_ci NOT NULL,
`name` varchar(20) COLLATE utf8_unicode_ci DEFAULT NULL,
PRIMARY KEY (`id`,`lang`,`channel`)
)
insert into `temp` (`id`, `lang`, `channel`, `name`) values('1','fr','ds','Jacket');
insert into `temp` (`id`, `lang`, `channel`, `name`) values('1','en','ds','Jacket');
insert into `temp` (`id`, `lang`, `channel`, `name`) values('2','en','ds','Jeans');
insert into `temp` (`id`, `lang`, `channel`, `name`) values('3','en','ds','Sweater');
insert into `temp` (`id`, `lang`, `channel`, `name`) values('1','de','ds','Jacket');
The question is how can I find which entries with lang en do not exist for fr?
My head is stuck and I believe this to be a trivial query but I am having one of these days.

Answer 1

There are several ways of achieving this. One way is to use a sub-select with not exists clause:
select id, channel
from temp t1
where t1.lang = 'en'
and not exists (
select 1
from temp t2
where t2.lang = 'fr'
and t1.id = t2.id
)
Alternatively, you can use an outer join:
select t1.id, t1.channel
from temp t1
left outer join temp t2 on t1.id = t2.id
where t1.lang = 'en'
and t2.id is null

Answer 2

based on #AleksG
SELECT t1.id, t1.channel, t1.name
FROM temp t1
LEFT JOIN temp t2 ON t1.id = t2.id AND t2.lang = 'fr'
WHERE t1.lang = 'en' AND t2.lang IS NULL

Answer 3

You can do this with aggregation:
select t.channel, t.name
from temp t
group by t.channel, t.name
having sum(case when t.lang = 'fr' then 1 else 0 end) = 0 and
sum(case when t.lang = 'en' then 1 else 0 end) > 0;
The first condition in the having clause counts up the number of times that French appears. The second counts up the number of times that English appears. When there are none for French and at least one for English, then the channel and name are in the result set.