如何在OracleSQLDeveloper上加速REGEXP级查询
我有一个INSERT INTO SELECT语句,它用一个函数解析的值填充一个表;在源表中:如何在OracleSQLDeveloper上加速REGEXP级查询,sql,regex,oracle,performance,insert-into,Sql,Regex,Oracle,Performance,Insert Into,我有一个INSERT INTO SELECT语句,它用一个函数解析的值填充一个表;在源表中: INSERT INTO PC_MATERIALS_BRIDGE (MATERIAL_BRIDGE_ID, VARIABLE_ID, MATERIAL_NAME) SELECT PC_VAR_MATERIALS_BRIDGE_SEQ.NEXTVAL, VARIABLE_ID, MATERIAL_NAME FROM (SELECT DISTINCT E.VARIABLE_ID, LOWER(TRIM(
INSERT INTO PC_MATERIALS_BRIDGE (MATERIAL_BRIDGE_ID, VARIABLE_ID, MATERIAL_NAME)
SELECT PC_VAR_MATERIALS_BRIDGE_SEQ.NEXTVAL, VARIABLE_ID, MATERIAL_NAME FROM (SELECT DISTINCT E.VARIABLE_ID, LOWER(TRIM(REGEXP_SUBSTR(e.MATERIALS, '[^;]+', 1, LEVEL))) MATERIAL_NAME
FROM (SELECT VARIABLE_ID, MATERIALS FROM SRC_VARS_OCEAN_ALL WHERE MATERIALS IS NOT NULL AND MATERIALS != 'N/A) e
CONNECT BY LOWER(TRIM(REGEXP_SUBSTR(e.MATERIALS, '[^;]+', 1, LEVEL))) IS NOT NULL);
因此,源表中的数据
ID MATERIAL_NAME
1 paper
2 paper; plastic
将显示为
MATERIAL_BRIDGE_ID MATERIAL_NAME
1 paper
2 paper
3 plastic
在目标表中
脚本运行良好;但是,它非常昂贵,因为源表有近40000条记录,有些记录有三个值,例如paper;塑料橡胶我知道水平仪很贵。我将MATERIAL_NAME设置为VARCHAR2255字节。除了编写另一种类型的查询(例如递归查询),不确定如何改进,但这可能很困难。是不是也导致了它的速度减慢?DISTINCT可能不再需要,因为e.VARIABLE_ID现在是主键。这是一种效率非常低的方法。在下面的简单演示中,您可以观察到当您删除DISTINCT时它会导致问题的原因:
create table SRC_VARS_OCEAN_ALL(
VARIABLE_ID int,
MATERIALS varchar2(200)
);
insert into SRC_VARS_OCEAN_ALL values( 1, 'ala;ma;kota' );
insert into SRC_VARS_OCEAN_ALL values( 2, 'as;to;pies' );
insert into SRC_VARS_OCEAN_ALL values( 3, 'baba;jaga' );
insert into SRC_VARS_OCEAN_ALL values( 4, 'zupa;obiad' );
以及:
此查询仅为4个输入行生成52条输出记录,其中包含10个值。你可以猜到4万美元会有多少。
该查询生成数百个thausand甚至数百万行,然后对这个巨大的结果集进行DISTINCT排序以消除重复项
下面的查询应该执行得更好,因为它只生成10条记录,不多也不少,与执行此任务所需的记录相同:
SELECT a.VARIABLE_ID, b.lev_el,
trim( regexp_substr( a.MATERIALS, '[^;]+', 1, b.lev_el )) as MATERIAL_NAME
FROM SRC_VARS_OCEAN_ALL a
JOIN (
SELECT level as lev_el
FROM dual CONNECT BY level <= 100
) b
ON b.lev_el <= regexp_count( a.MATERIALS, ';' ) + 1
VARIABLE_ID LEV_EL MATERIAL_NAME
----------- ---------- --------------
1 1 ala
2 1 as
3 1 baba
4 1 zupa
1 2 ma
2 2 to
3 2 jaga
4 2 obiad
1 3 kota
2 3 pies
10 rows selected.
我假设每个列表中的值不超过100个,每一行都有一个列表,其中的值不超过100个,所以这里有一个dual CONNECT BY level很棒,谢谢。我一直在使用DISTINCT,但忽略了一个事实,即在检索到所有可能的行之后对行进行排序。干杯
SELECT a.VARIABLE_ID, b.lev_el,
trim( regexp_substr( a.MATERIALS, '[^;]+', 1, b.lev_el )) as MATERIAL_NAME
FROM SRC_VARS_OCEAN_ALL a
JOIN (
SELECT level as lev_el
FROM dual CONNECT BY level <= 100
) b
ON b.lev_el <= regexp_count( a.MATERIALS, ';' ) + 1
VARIABLE_ID LEV_EL MATERIAL_NAME
----------- ---------- --------------
1 1 ala
2 1 as
3 1 baba
4 1 zupa
1 2 ma
2 2 to
3 2 jaga
4 2 obiad
1 3 kota
2 3 pies
10 rows selected.