Left Join With Where Clause

问题

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.

回答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'

回答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

回答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

回答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'

回答5:

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.

来源：https://stackoverflow.com/questions/4752455/left-join-with-where-clause

标签

sql

left-join

where-clause