Not quite a good enough JOIN? [duplicate] - mysql

Question

I need to retrieve all default settings from the settings table but also grab the character setting if exists for x character.
But this query is only retrieving those settings where character is = 1, not the default settings if the user havent setted anyone.
SELECT `settings`.*, `character_settings`.`value`
FROM (`settings`)
LEFT JOIN `character_settings`
ON `character_settings`.`setting_id` = `settings`.`id`
WHERE `character_settings`.`character_id` = '1'
So i should need something like this:
array(
'0' => array('somekey' => 'keyname', 'value' => 'thevalue'),
'1' => array('somekey2' => 'keyname2'),
'2' => array('somekey3' => 'keyname3')
)
Where key 1 and 2 are the default values when key 0 contains the default value with the character value.

Answer 1

The where clause is filtering away rows where the left join doesn't succeed. Move it to the join:
SELECT `settings`.*, `character_settings`.`value`
FROM `settings`
LEFT JOIN
`character_settings`
ON `character_settings`.`setting_id` = `settings`.`id`
AND `character_settings`.`character_id` = '1'

Answer 2

When making OUTER JOINs (ANSI-89 or ANSI-92), filtration location matters because criteria specified in the ON clause is applied before the JOIN is made. Criteria against an OUTER JOINed table provided in the WHERE clause is applied after the JOIN is made. This can produce very different result sets. In comparison, it doesn't matter for INNER JOINs if the criteria is provided in the ON or WHERE clauses -- the result will be the same.
SELECT s.*,
cs.`value`
FROM SETTINGS s
LEFT JOIN CHARACTER_SETTINGS cs ON cs.setting_id = s.id
AND cs.character_id = 1

Answer 3

If I understand your question correctly you want records from the settings database if they don't have a join accross to the character_settings table or if that joined record has character_id = 1.
You should therefore do
SELECT `settings`.*, `character_settings`.`value`
FROM (`settings`)
LEFT OUTER JOIN `character_settings`
ON `character_settings`.`setting_id` = `settings`.`id`
WHERE `character_settings`.`character_id` = '1' OR
`character_settings`.character_id is NULL

Answer 4

You might find it easier to understand by using a simple subquery
SELECT `settings`.*, (
SELECT `value` FROM `character_settings`
WHERE `character_settings`.`setting_id` = `settings`.`id`
AND `character_settings`.`character_id` = '1') AS cv_value
FROM `settings`
The subquery is allowed to return null, so you don't have to worry about JOIN/WHERE in the main query.
Sometimes, this works faster in MySQL, but compare it against the LEFT JOIN form to see what works best for you.
SELECT s.*, c.value
FROM settings s
LEFT JOIN character_settings c ON c.setting_id = s.id AND c.character_id = '1'

Answer 5

For this problem, as for many others involving non-trivial left joins such as left-joining on inner-joined tables, I find it convenient and somewhat more readable to split the query with a with clause. In your example,
with settings_for_char as (
select setting_id, value from character_settings where character_id = 1
)
select
settings.*,
settings_for_char.value
from
settings
left join settings_for_char on settings_for_char.setting_id = settings.id;

Answer 6

The way I finally understand the top answer is realising (following the Order Of Execution of the SQL query ) that the WHERE clause is applied to the joined table thereby filtering out rows that do not satisfy the WHERE condition from the joined (or output) table. However, moving the WHERE condition to the ON clause applies it to the individual tables prior to joining. This enables the left join to retain rows from the left table even though some column entries of those rows (entries from the right tables) do not satisfy the WHERE condition.

Answer 7

The result is correct based on the SQL statement. Left join returns all values from the right table, and only matching values from the left table.
ID and NAME columns are from the right side table, so are returned.
Score is from the left table, and 30 is returned, as this value relates to Name "Flow". The other Names are NULL as they do not relate to Name "Flow".
The below would return the result you were expecting:
SELECT a.*, b.Score
FROM #Table1 a
LEFT JOIN #Table2 b
ON a.ID = b.T1_ID
WHERE 1=1
AND a.Name = 'Flow'
The SQL applies a filter on the right hand table.