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mysql - 连接两个 Substring_index

转载 作者:行者123 更新时间:2023-11-29 21:20:02 27 4
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我需要连接下面选择子字符串的 2 个结果。有什么想法吗?

BEGIN
select SUBSTRING_INDEX(SUBSTRING_INDEX(oldVal, '-', n.n), '-', -1) AS oldVal5
FROM audits oldVal
CROSS JOIN ( SELECT a.N + b.N * 10 + 1 n
FROM (SELECT 0 AS N
UNION ALL SELECT 1
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
UNION ALL SELECT 8
UNION ALL SELECT 9) a ,
(SELECT 0 AS N
UNION ALL SELECT 1
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
UNION ALL SELECT 8
UNION ALL SELECT 9) b ORDER BY n ) n
WHERE n.n <= 1 + (LENGTH(oldVal) - LENGTH(REPLACE(oldVal, '-', ''))) and
dbUser=34 and tableName='persona' and task='u';

select SUBSTRING_INDEX(SUBSTRING_INDEX(newVal, '-', n.n), '-', -1) AS newval5
FROM audits newVal
CROSS JOIN ( SELECT a.N + b.N * 10 + 1 n
FROM (SELECT 0 AS N
UNION ALL SELECT 1
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
UNION ALL SELECT 8
UNION ALL SELECT 9) a ,
(SELECT 0 AS N
UNION ALL SELECT 1
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
UNION ALL SELECT 8
UNION ALL SELECT 9) b ORDER BY n ) n
WHERE n.n <= 1 + (LENGTH(newVal) - LENGTH(REPLACE(newVal, '-', ''))) and
dbUser=34 and tableName='persona' and task='u';

END

最佳答案

只需在同一查询中选择两者即可。

select SUBSTRING_INDEX(SUBSTRING_INDEX(oldVal, '-', n.n), '-', -1) AS oldVal5,
SUBSTRING_INDEX(SUBSTRING_INDEX(newVal, '-', n.n), '-', -1) AS newVal5
FROM audits oldVal
CROSS JOIN ( SELECT a.N + b.N * 10 + 1 n
FROM (SELECT 0 AS N
UNION ALL SELECT 1
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
UNION ALL SELECT 8
UNION ALL SELECT 9) a ,
(SELECT 0 AS N
UNION ALL SELECT 1
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
UNION ALL SELECT 8
UNION ALL SELECT 9) b ORDER BY n ) n
WHERE (n.n <= 1 + (LENGTH(oldVal) - LENGTH(REPLACE(oldVal, '-', '')))
or
n.n <= 1 + (LENGTH(newVal) - LENGTH(REPLACE(newVal, '-', ''))))
and dbUser=34 and tableName='persona' and task='u';

关于mysql - 连接两个 Substring_index,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35761739/

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