[英]Store the results of a sub-query for use in multiple joins
我有以下MySQL查詢,它產生我想要的結果:
SELECT
`l`.`status`,
`l`.`acquired_by`, `a`.`name` AS 'acquired_by_name',
`l`.`researcher`, `r`.`name` AS 'researcher_name',
`l`.`surveyor`, `s`.`name` AS 'surveyor_name'
FROM `leads` `l`
LEFT JOIN (
SELECT '0' AS 'id', 'Unassigned' AS 'name'
UNION ALL
SELECT `id`, `name`
FROM `web_users`
) `r` ON `r`.`id` = `l`.`researcher`
LEFT JOIN (
SELECT '0' AS 'id', 'Unassigned' AS 'name'
UNION ALL
SELECT `id`, `name`
FROM `web_users`
) `s` ON `s`.`id` = `l`.`surveyor`
LEFT JOIN (
SELECT '0' AS 'id', 'Unassigned' AS 'name'
UNION ALL
SELECT `id`, `name`
FROM `web_users`
) `a` ON `a`.`id` = `l`.`acquired_by`
WHERE `l`.`id` = 566
但正如您所看到的,它在其中具有相同的子查詢三次。 有沒有辦法執行此查詢一次並存儲結果,所以我可以使用緩存的結果LEFT JOIN
而不是執行相同的查詢三次?
我已經嘗試將其存儲在變量中:
SET @usercache = (
SELECT '0' AS 'id', 'Unassigned' AS 'name'
UNION ALL
SELECT `id`, `name`
FROM `web_users`
)
......但這給了我一個錯誤:
1241 - 操作數應包含1列
...而一些谷歌搜索這個錯誤讓我沒有更聰明。
有誰知道如何使這個查詢更有效? 或者我只是擔心無關緊要的事情?
如果它有任何區別,我正在使用PHP / MySQLi。
你真的需要子查詢嗎? 這個怎么樣:
SELECT
`l`.`status`,
`l`.`acquired_by`, COALESCE(`a`.`name`, 'Unassigned') AS 'acquired_by_name',
`l`.`researcher`, COALESCE(`r`.`name`, 'Unassigned') AS 'researcher_name',
`l`.`surveyor`, COALESCE(`s`.`name`, 'Unassigned') AS 'surveyor_name'
FROM `leads` `l`
LEFT JOIN `web_users` `r` ON `r`.`id` = `l`.`researcher`
LEFT JOIN `web_users` `s` ON `s`.`id` = `l`.`surveyor`
LEFT JOIN `web_users` `a` ON `a`.`id` = `l`.`acquired_by`
WHERE `l`.`id` = 566
你不能運行一次 - 你實際上使用它三次得到三個不同的結果......
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