mysql - mysql - 公用表表达式:模拟全连接
问题描述
我有多个(超过 20 个)cte,我需要将它们连接在一起以并排输出行。
WITH
cte_lnz AS (SELECT n.name, ROW_NUMBER() OVER (ORDER BY e.id) AS rn_lnz FROM entries e LEFT JOIN nodes n ON n.id = e.node_id LEFT JOIN attribs a ON a.id = e.attrib_id
WHERE a.name = 'LOCATION' AND e.value = 'Gruberstrasse' AND DATE(ts) = CURRENT_DATE ORDER BY e.id
),
cte_wbg AS (SELECT n.name, ROW_NUMBER() OVER (ORDER BY e.id) AS rn_wbg FROM entries e LEFT JOIN nodes n ON n.id = e.node_id LEFT JOIN attribs a ON a.id = e.attrib_id
WHERE a.name = 'LOCATION' AND e.value = 'Wienerberg' AND DATE(ts) = CURRENT_DATE ORDER BY e.id
),
cte_gbg AS ( SELECT n.name, ROW_NUMBER() OVER (ORDER BY e.id) AS rn_gbg FROM entries e LEFT JOIN nodes n ON n.id = e.node_id LEFT JOIN attribs a ON a.id = e.attrib_id
WHERE a.name = 'LOCATION' AND e.value = 'Geiselberg' AND DATE(ts) = CURRENT_DATE ORDER BY e.id
),
cte_vie AS ( SELECT n.name, ROW_NUMBER() OVER (ORDER BY e.id) AS rn_vie FROM entries e LEFT JOIN nodes n ON n.id = e.node_id LEFT JOIN attribs a ON a.id = e.attrib_id
WHERE a.name = 'LOCATION' AND e.value IN ('Geiselberg', 'Wienerberg') AND DATE(ts) = CURRENT_DATE ORDER BY e.id
)
这适用于前两个 cte ......
-- simulated full join
SELECT lnz.name LNZ, wbg.name WBG
FROM cte_lnz lnz LEFT JOIN cte_wbg wbg
ON wbg.rn_wbg = lnz.rn_lnz
UNION
SELECT lnz.name LNZ, wbg.name WBG
FROM cte_wbg wbg LEFT JOIN cte_lnz lnz
ON wbg.rn_wbg = lnz.rn_lnz
+----------------------+------------------------------+
| LNZ | WBG |
+----------------------+------------------------------+
| STP17T1_SGKKT1 | AIXHVBDBT |
| AIXSAGRU3 | KUG0167_JBOSS-T4 |
| ARR5S1P9_TIC | KUG0169_RMAN_HA |
| AIXBUILDHOST | KUG0121_ZPVTT1 |
这给出了一个错误......
SELECT lnz.name LNZ, wbg.name WBG
FROM cte_lnz lnz LEFT JOIN cte_wbg wbg
ON wbg.rn_wbg = lnz.rn_lnz
UNION
SELECT lnz.name LNZ, wbg.name WBG
FROM cte_wbg wbg LEFT JOIN cte_lnz lnz
ON wbg.rn_wbg = lnz.rn_lnz
-- just a placeholder
SELECT gbg.name GBG, vie.name VIE
FROM cte_gbg gbg LEFT JOIN cte_vie vie
ON vie.rn_vie = gbg.rn_gbg
UNION
SELECT gbg.name GBG, vie.name VIE
FROM cte_vie vie LEFT JOIN cte_gbg gbg
ON vie.rn_vie = gbg.rn_gbg
ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'SELECT gbg.name GBG, vie.name VIE
FROM cte_gbg gbg LEFT JOIN cte_vie vie
ON vie.' at line 53
那么如何将其他 cte 添加到游戏中?
解决方案
union
完全连接的近似值仍然存在吗?这是一种糟糕的方法——在功能上并不完全等同,而且比必要的要贵得多。
对于您的问题,使用union all
和聚合。一种方法是:
SELECT MAX(CASE WHEN which = 'lnz' THEN name END) as lnz,
MAX(CASE WHEN which = 'wbg' THEN name END) as wbg,
. . .
FROM ((SELECT 'lnz' as which, rn, lnz.name
FROM cte_lnz lnz
) UNION ALL
(SELECT 'wbg', rn, wbg.name
FROM cte_wbg wbg
) UNION ALL
. . .
) n
GROUP BY rn;
添加更多 CTE 就像在子查询中添加更多UNION ALL
子句然后MAX(CASE)
在外部查询中添加更多表达式一样简单。
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