我有一個mysql
這樣的查詢凡(子查詢)子句警予
(SELECT
`notification`.`id`, `notification`.`user_id` AS `user_id`, `notification`.`activity_type`, `notification`.`source_id`, `event`.`title` as sourceName,concat_ws(" ",v.firstname,v.lastname) as ActorName
FROM
`notification`
INNER JOIN
`event`
ON
event.id = notification.source_id
INNER JOIN
`user` as v
ON
v.id = notification.user_id
AND
notification.activity_type = "checkin"
where
user_id in
(SELECT friend.friend_id from friend WHERE friend.user_id=1 AND friend.is_active=1))
UNION
(SELECT
`notification`.`id`, `notification`.`user_id` AS `user_id`, `notification`.`activity_type`, `notification`.`source_id`, concat_ws(" ",u.firstname,u.lastname) as sourceName,concat_ws(" ",v.firstname,v.lastname) as ActorName
FROM
`notification`
INNER JOIN
`user` as u
ON
u.id = notification.source_id
INNER JOIN
`user` as v ON v.id = notification.user_id
AND
notification.activity_type = "friend"
where user_id in
(SELECT friend.friend_id from friend WHERE friend.user_id=1 AND friend.is_active=1))
,我希望寫yii2
該查詢,我不知道如何寫條子查詢中where
。
到目前爲止,我已經做到了這一點
$query2 ->select(['notification.id', 'notification.user_id AS user_id', 'notification.activity_type', 'notification.source_id', 'concat_ws(" ",u.firstname,u.lastname) as sourceName','concat_ws(" ",v.firstname,v.lastname) as ActorName','user_image.imagepath as image'])
->from('notification')
->innerJoin('user as u', 'u.id = notification.source_id')
->innerJoin('user_image','user_image.user_id = notification.user_id')
->innerJoin('user as v', 'v.id = notification.user_id AND notification.activity_type = "friend"')
->where('user_image.imagetype="profile"')
->andWhere(['user_id'=>('SELECT friend.friend_id from friend WHERE friend.user_id='.$id.' AND friend.is_active=1')]);
$query ->select(['notification.id','notification.user_id AS user_id','notification.activity_type', 'notification.source_id', 'event.title as sourceName','concat_ws(" ",v.firstname,v.lastname) as ActorName','organiser.image as image'])
->from('notification')
->innerJoin('event', 'event.id = notification.source_id')
->innerJoin('organiser','organiser.organiser_id = event.organiser_id')
->innerJoin('user as v', 'v.id = notification.user_id AND notification.activity_type = "checkin"')
->Where(['in', 'user_id', [488,489]])
->union($query2);
生成命令這樣的查詢
(SELECT
`notification`.`id`, `notification`.`user_id` AS `user_id`, `notification`.`activity_type`, `notification`.`source_id`, `event`.`title` AS `sourceName`, concat_ws(" ",v.firstname,v.lastname) as ActorName, `organiser`.`image` AS `image`
FROM
`notification`
INNER JOIN
`event`
ON
event.id = notification.source_id
INNER JOIN
`organiser`
ON
organiser.organiser_id = event.organiser_id
INNER JOIN
`user` `v`
ON
v.id = notification.user_id
AND
notification.activity_type = "checkin"
WHERE
`user_id` IN (:qp0, :qp1))
UNION
(SELECT
`notification`.`id`, `notification`.`user_id` AS `user_id`, `notification`.`activity_type`, `notification`.`source_id`, concat_ws(" ",u.firstname,u.lastname) as sourceName, concat_ws(" ",v.firstname,v.lastname) as ActorName, `user_image`.`imagepath` AS `image`
FROM
`notification`
INNER JOIN
`user` `u`
ON
u.id = notification.source_id
INNER JOIN
`user_image`
ON
user_image.user_id = notification.user_id
INNER JOIN
`user` `v`
ON
v.id = notification.user_id
AND
notification.activity_type = "friend"
WHERE
(user_image.imagetype="profile") AND (`user_id`=:qp2))
但其沒有工作,所以什麼是正確的語法 也隨時給建議如果where
中可以寫inner join
這樣會很容易寫出查詢
謝謝
ü得到什麼誤差R? – Bloodhound