原始查询
delete B from
TABLE_BASE B ,
TABLE_INC I
where B.ID = I.IDID and B.NUM = I.NUM;
上述查询的性能统计数据
+-------------------+---------+-----------+
| Response Time | SumCPU | ImpactCPU |
+-------------------+---------+-----------+
| 00:05:29.190000 | 2852 | 319672 |
+-------------------+---------+-----------+
优化查询1
DEL FROM TABLE_BASE WHERE (ID, NUM) IN
(SELECT ID, NUM FROM TABLE_INC);
上述查询的统计信息
+-----------------+--------+-----------+
| QryRespTime | SumCPU | ImpactCPU |
+-----------------+--------+-----------+
| 00:00:00.570000 | 15.42 | 49.92 |
+-----------------+--------+-----------+
优化查询2
DELETE FROM TABLE_BASE B WHERE EXISTS
(SELECT * FROM TABLE_INC I WHERE B.ID = I.ID AND B.NUM = I.NUM);
上述查询的统计信息
+-----------------+--------+-----------+
| QryRespTime | SumCPU | ImpactCPU |
+-----------------+--------+-----------+
| 00:00:00.400000 | 11.96 | 44.93 |
+-----------------+--------+-----------+
我的问题 -
- 优化查询 1 和 2 如何/为何对性能影响如此之大？
- 此类 DELETE 查询的最佳实践是什么？
- 我应该选择查询 1 还是查询 2？哪一个是理想的/更好的/可靠的？我觉得查询 1 是理想的，因为而不是
SELECT *我在用SELECT ID,NUM减少到只有两列，但查询 2 显示了更好的结果。
QUERY 1
This query is optimized using type 2 profile T2_Linux64, profileid 21.
1) First, we lock TEMP_DB.TABLE_BASE for write on a
reserved RowHash to prevent global deadlock.
2) Next, we lock TEMP_DB_T.TABLE_INC for access, and we
lock TEMP_DB.TABLE_BASE for write.
3) We execute the following steps in parallel.
1) We do an all-AMPs RETRIEVE step from
TEMP_DB.TABLE_BASE by way of an all-rows scan
with no residual conditions into Spool 2 (all_amps), which is
redistributed by the hash code of (
TEMP_DB.TABLE_BASE.NUM,
TEMP_DB.TABLE_BASE.ID) to all AMPs. Then
we do a SORT to order Spool 2 by row hash. The size of Spool
2 is estimated with low confidence to be 168,480 rows (
5,054,400 bytes). The estimated time for this step is 0.03
seconds.
2) We do an all-AMPs RETRIEVE step from
TEMP_DB_T.TABLE_INC by way of an all-rows scan
with no residual conditions into Spool 3 (all_amps), which is
redistributed by the hash code of (
TEMP_DB_T.TABLE_INC.NUM,
TEMP_DB_T.TABLE_INC.ID) to all AMPs. Then
we do a SORT to order Spool 3 by row hash and the sort key in
spool field1 eliminating duplicate rows. The size of Spool 3
is estimated with high confidence to be 5,640 rows (310,200
bytes). The estimated time for this step is 0.03 seconds.
4) We do an all-AMPs JOIN step from Spool 2 (Last Use) by way of an
all-rows scan, which is joined to Spool 3 (Last Use) by way of an
all-rows scan. Spool 2 and Spool 3 are joined using an inclusion
merge join, with a join condition of ("(ID = ID) AND
(NUM = NUM)"). The result goes into Spool 1 (all_amps),
which is redistributed by the hash code of (
TEMP_DB.TABLE_BASE.ROWID) to all AMPs. Then we do
a SORT to order Spool 1 by row hash and the sort key in spool
field1 eliminating duplicate rows. The size of Spool 1 is
estimated with no confidence to be 168,480 rows (3,032,640 bytes).
The estimated time for this step is 1.32 seconds.
5) We do an all-AMPs MERGE DELETE to
TEMP_DB.TABLE_BASE from Spool 1 (Last Use) via the
row id. The size is estimated with no confidence to be 168,480
rows. The estimated time for this step is 42.95 seconds.
6) We spoil the parser's dictionary cache for the table.
7) Finally, we send out an END TRANSACTION step to all AMPs involved
in processing the request.
-> No rows are returned to the user as the result of statement 1.
QUERY 2 EXPLAIN PLAN
This query is optimized using type 2 profile T2_Linux64, profileid 21.
1) First, we lock TEMP_DB.TABLE_BASE for write on a reserved RowHash to
prevent global deadlock.
2) Next, we lock TEMP_DB_T.TABLE_INC for access, and we
lock TEMP_DB.TABLE_BASE for write.
3) We execute the following steps in parallel.
1) We do an all-AMPs RETRIEVE step from TEMP_DB.TABLE_BASE by way of
an all-rows scan with no residual conditions into Spool 2
(all_amps), which is redistributed by the hash code of (
TEMP_DB.TABLE_BASE.NUM, TEMP_DB.TABLE_BASE.ID) to all AMPs.
Then we do a SORT to order Spool 2 by row hash. The size of
Spool 2 is estimated with low confidence to be 168,480 rows (
5,054,400 bytes). The estimated time for this step is 0.03
seconds.
2) We do an all-AMPs RETRIEVE step from
TEMP_DB_T.TABLE_INC by way of an all-rows scan
with no residual conditions into Spool 3 (all_amps), which is
redistributed by the hash code of (
TEMP_DB_T.TABLE_INC.NUM,
TEMP_DB_T.TABLE_INC.ID) to all AMPs. Then
we do a SORT to order Spool 3 by row hash and the sort key in
spool field1 eliminating duplicate rows. The size of Spool 3
is estimated with high confidence to be 5,640 rows (310,200
bytes). The estimated time for this step is 0.03 seconds.
4) We do an all-AMPs JOIN step from Spool 2 (Last Use) by way of an
all-rows scan, which is joined to Spool 3 (Last Use) by way of an
all-rows scan. Spool 2 and Spool 3 are joined using an inclusion
merge join, with a join condition of ("(NUM = NUM) AND
(ID = ID)"). The result goes into Spool 1 (all_amps), which
is redistributed by the hash code of (TEMP_DB.TABLE_BASE.ROWID) to all
AMPs. Then we do a SORT to order Spool 1 by row hash and the sort
key in spool field1 eliminating duplicate rows. The size of Spool
1 is estimated with no confidence to be 168,480 rows (3,032,640
bytes). The estimated time for this step is 1.32 seconds.
5) We do an all-AMPs MERGE DELETE to TEMP_DB.TABLE_BASE from Spool 1 (Last
Use) via the row id. The size is estimated with no confidence to
be 168,480 rows. The estimated time for this step is 42.95
seconds.
6) We spoil the parser's dictionary cache for the table.
7) Finally, we send out an END TRANSACTION step to all AMPs involved
in processing the request.
-> No rows are returned to the user as the result of statement 1.
对于 TABLE_BASE
+----------------+----------+
| table_bytes | skewness |
+----------------+----------+
| 16842085888.00 | 22.78 |
+----------------+----------+
对于 TABLE_INC
+-------------+----------+
| table_bytes | skewness |
+-------------+----------+
| 5317120.00 | 44.52 |
+-------------+----------+
