这两者在性能方面有什么不同吗?
-- eliminate duplicates using UNION
SELECT col1,col2,col3 FROM Table1
UNION SELECT col1,col2,col3 FROM Table2
UNION SELECT col1,col2,col3 FROM Table3
UNION SELECT col1,col2,col3 FROM Table4
UNION SELECT col1,col2,col3 FROM Table5
UNION SELECT col1,col2,col3 FROM Table6
UNION SELECT col1,col2,col3 FROM Table7
UNION SELECT col1,col2,col3 FROM Table8
-- eliminate duplicates using DISTINCT
SELECT DISTINCT * FROM
(
SELECT col1,col2,col3 FROM Table1
UNION ALL SELECT col1,col2,col3 FROM Table2
UNION ALL SELECT col1,col2,col3 FROM Table3
UNION ALL SELECT col1,col2,col3 FROM Table4
UNION ALL SELECT col1,col2,col3 FROM Table5
UNION ALL SELECT col1,col2,col3 FROM Table6
UNION ALL SELECT col1,col2,col3 FROM Table7
UNION ALL SELECT col1,col2,col3 FROM Table8
) x 发布于 2016-02-25 21:01:28
Union和Union all之间的不同之处在于,UNION ALL不会消除重复的行,而只是从符合您的查询要求的所有表中提取所有行,并将它们合并到一个表中。
UNION语句有效地对结果集执行SELECT DISTINCT。
如果选择Distinct from Union All result set,则输出将等于 Union结果集。
编辑:
CPU上的性能成本:
让我举个例子来解释:
我有两个问题。一个是联合,另一个是联合
SET STATISTICS TIME ON
GO
select distinct * from (select * from dbo.user_LogTime
union all
select * from dbo.user_LogTime) X
GO
SET STATISTICS TIME OFF
SET STATISTICS TIME ON
GO
select * from dbo.user_LogTime
union
select * from dbo.user_LogTime
GO
SET STATISTICS TIME OFF我在SMSS的同一查询窗口中运行了这两个命令。让我们看看SMSS中的执行计划:

实际情况是,使用Union All和Distinct的查询将比使用Union的查询占用更多的CPU开销。
按时性能:
UNION ALL
(1172 row(s) affected)
SQL Server Execution Times:
CPU time = 0 ms, elapsed time = 39 ms.UNION
(1172 row(s) affected)
SQL Server Execution Times:
CPU time = 10 ms, elapsed time = 25 ms.因此在performance-wise中,联合比使用Distinct的联合All要好得多
发布于 2017-08-28 23:59:11
另一个现场示例说明了四种可能的情况:
/* with each case we should expect a return set:
(1) DISTINCT UNION {1,2,3,4,5} - is redundant with case (2)
(2) UNION {1,2,3,4,5} - more efficient?
(3) DISTINCT UNION ALL {1,2,2,3,3,4,4,5}
(4) UNION ALL {1,1,2,2,2,3,3,4,4,5}
*/
declare @t1 table (c1 varchar(15));
declare @t2 table (c2 varchar(15));
insert into @t1 values ('1'),('1'),('2'),('3'),('4');
insert into @t2 values ('2'),('2'),('3'),('4'),('5');
select DISTINCT * from @t1 --case (1)
UNION
select DISTINCT * from @t2 order by c1
select * from @t1 --case (2)
UNION
select * from @t2 order by c1
select DISTINCT * from @t1 --case (3)
UNION ALL
select DISTINCT * from @t2 order by c1
select * from @t1 --case (4)
UNION ALL
select * from @t2 order by c1发布于 2020-06-15 13:46:29
UNION DISTINCT与UNION ALL的比较
此查询用于为下游系统创建具有附加备用ID的扩展employee表。此示例来自mySQL 8.0.20环境。
对于下面显示的数据和查询,测试产生了显著的差异:
UNION ALL 8.983 sec
UNION DISTINCT 15.344 sec为了显示此示例的规模和复杂性,表大小和查询代码如下所示
hqsource 600K records
accountingemppos 180K
accountingposld 200K
emp_no_accountingnumeric 20
First UNION block is approx 550K records, second approx 50K
SELECT a.`emp_no_imported` AS `emp_no`,
a.`supervisor_emp_no`,
a.`first name`,
a.`middle name`,
a.`last name`,
a.`jobtitle`,
a.`status`,
CASE WHEN rida.`accounting_emp_no` IS NOT NULL THEN
rida.`accounting_emp_no`
ELSE
a.`emp_no_imported`
END AS `accounting_id`,
CASE WHEN epfp.`emp_no` IS NOT NULL THEN
CASE WHEN `sridf`.`emp_no` IS NOT NULL THEN
`sridf`.`accounting_emp_no`
ELSE
epfp.`emp_no`
END
ELSE
CASE WHEN epp.`emp_no` IS NOT NULL THEN
CASE WHEN `srids`.`emp_no` IS NOT NULL THEN
`srids`.`accounting_emp_no`
ELSE
epp.`emp_no`
END
ELSE
CASE WHEN `srida`.`emp_no` IS NOT NULL THEN
`srida`.`accounting_emp_no`
ELSE
a.`supervisor_emp_no`
END
END
END AS `accounting_s_emp_no`,
ep.`emp_no` AS `traas_emp_no`,
epp.`emp_no` AS `traas_parent_emp_no`
FROM `hqsource`.hq_people a
LEFT OUTER JOIN `hqsource`.`emp_no_accountingnumeric` `rida` ON `rida`.emp_no = a.`emp_no_imported`
LEFT OUTER JOIN `hqsource`.`emp_no_accountingnumeric` `srida` ON `srida`.emp_no = a.`supervisor_emp_no`
LEFT OUTER JOIN `traas`.`accountingemppos_data_extract` ep ON ep.`emp_no` = a.`emp_no_imported` AND ep.`End` = '2899-12-31' AND ep.`Primary` = 'Y'
LEFT OUTER JOIN `epe`.`accountingposld_data_extract` p ON p.`RangeGID` = ep.`GID`
LEFT OUTER JOIN `traas`.`accountingemppos_data_extract` epp ON epp.`GID` = p.`ParentGID` AND epp.`End` = '2899-12-31' AND epp.`Primary` = 'Y'
LEFT OUTER JOIN `hqsource`.`emp_no_accountingnumeric` `rids` ON `rids`.emp_no = ep.`emp_no`
LEFT OUTER JOIN `hqsource`.`emp_no_accountingnumeric` `srids` ON `srids`.emp_no = epp.`emp_no` AND epp.`End` = '2899-12-31' AND epp.`Primary` = 'Y'
LEFT OUTER JOIN `epe`.`accountingemppos_data_extract_filtered` epf ON epf.`emp_no` = a.`emp_no_imported` AND epf.`End` = '2899-12-31' AND epf.`Primary` = 'Y'
LEFT OUTER JOIN `epe`.`accountingposld_data_extract` pf ON pf.`RangeGID` = epf.`GID`
LEFT OUTER JOIN `epe`.`accountingemppos_data_extract_filtered` epfp ON epfp.`GID` = pf.`ParentGID` AND epfp.`End` = '2899-12-31' AND epfp.`Primary` = 'Y'
LEFT OUTER JOIN `hqsource`.`emp_no_accountingnumeric` `ridf` ON `ridf`.emp_no = epf.`emp_no` AND epf.`End` = '2899-12-31' AND epf.`Primary` = 'Y'
LEFT OUTER JOIN `hqsource`.`emp_no_accountingnumeric` `sridf` ON `sridf`.emp_no = epfp.`emp_no` AND epfp.`End` = '2899-12-31' AND epfp.`Primary` = 'Y'
WHERE a.`emp_no_imported` REGEXP ('^[a-z]{2}\\d{5}.$')
UNION ALL
-- UNION DISTINCT
SELECT a.`emp_no_imported` AS `emp_no`, a.`supervisor_emp_no` AS `s_emp_no`, u.`First_Name`, 'ƒ' AS `MI`, u.`Last_Name`, u.`Job_Title`, NULL AS `status`,
CASE WHEN rid.`accounting_emp_no` IS NULL THEN
ep.`emp_no`
ELSE
rid.`accounting_emp_no`
END AS `accounting_emp_no`,
CASE WHEN `srid`.`accounting_emp_no` IS NULL THEN
epp.`emp_no`
ELSE
`srid`.`accounting_emp_no`
END AS `accounting_s_emp_no`,
ep.`emp_no` AS `traas_emp_no`,
epp.`emp_no` AS `traas_parent_emp_no`
FROM `epe`.`accountingemppos_data_extract_filtered` ep
LEFT OUTER JOIN `hqsource`.`hq_people` a ON a.`emp_no_imported` = ep.`emp_no`
LEFT OUTER JOIN `epe`.`accountingposld_data_extract` p ON p.`RangeGID` = ep.`GID`
LEFT OUTER JOIN `epe`.`accountingemppos_data_extract_filtered` epp ON epp.`GID` = p.`ParentGID`
LEFT OUTER JOIN `siebel`.`users_all_output` u ON u.`LOGIN` = ep.`emp_no`
LEFT OUTER JOIN `hqsource`.`emp_no_accountingnumeric` `rid` ON `rid`.emp_no = ep.`emp_no`
LEFT OUTER JOIN `hqsource`.`emp_no_accountingnumeric` `srid` ON `srid`.emp_no = epp.`emp_no`
WHERE
ep.`End` = '2899-12-31' AND
epp.`End` = '2899-12-31' AND
p.`End` = '2899-12-31' AND
ep.emp_no REGEXP ('^F\\d{8}$|^V[0-3]\\d{5}$')
ORDER BY LENGTH(accounting_emp_no) ASC, accounting_emp_no ASC;
两个UNION块中每个块中的WHERE子句保证结果是唯一的。(此查询已存在很久,每天都会运行。我希望我能早点尝试一下)。字段名称已被混淆
https://stackoverflow.com/questions/35627923
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