Question

我有两张桌子：

CREATE TABLE huge (
    id INT PRIMARY KEY,
    name VARCHAR2(100),
    day INT,
    errno INT,
    error_message VARCHAR2(100)
);

CREATE TABLE smallish (
   id INT PRIMARY KEY,
   name VARCHAR2(100),
   day int
);

-- Note the lack of a foreign key between huge and smallish on name
-- This is intentional

我想做三件事：

更新huge中的所有行，将day设置为smallish中huge中的行name中只有smallish的{{1}} }}。
更新huge中的所有行，使errno为1，error_message为:name not in smallish huge中的行在smallish。
更新huge中的所有行，使errno 2和error_message :name has multiple rows in smallish huge中的行name smallish

我使用下面的DML似乎可以正常工作，但其中两个DML在输出中提供了全表扫描，否则看起来不直观。

此外，在一个陈述中完成所有这一切将非常酷，而不是三个。

更新

以下似乎有效，看起来有些直观，但解释计划显示了巨大的全表扫描：

UPDATE huge h_out
SET (day, errno, error_message) = (
    select CASE WHEN DAY IS NOT NULL AND count = 1 THEN day ELSE NULL END as bill_day,
           CASE WHEN day IS NULL THEN 1 WHEN count > 1 THEN 2 ELSE NULL END AS errno,
           CASE WHEN day IS NULL THEN name || ' not in smallish' WHEN count > 1 THEN name || ' has multiple rows in smallish' ELSE NULL END as error_message
    FROM (
        select dhuge.name, max(smallish.day) as day, count(dhuge.name) as count
        from (select distinct huge.name from huge) dhuge left join smallish on dhuge.name = smallish.name
        group by dhuge.name
    ) h_in
    WHERE h_out.name = h_in.name
);

原件：

-- Problem #1
UPDATE huge h
SET (day) = (
    SELECT MIN(day)
    FROM smallish s
    WHERE h.name = s.name
    GROUP BY s.name
    HAVING count(1) = 1
) WHERE EXISTS (
    SELECT null
    FROM smallish s
    WHERE s.name = h.name
);

-- Problem #2 Explain plan shows a full table scan on huge
UPDATE huge h_out
SET (errno, error_message) = (
    select 1, h_out.name || ' not in smallish' AS error_message FROM DUAL
) WHERE NOT EXISTS (
    SELECT NULL
    FROM smallish s
    WHERE s.name = h_out.name
);


-- Problem #3 Explain plan shows a full table scan on huge
UPDATE huge h
SET (errno, error_message) = (
    SELECT 2, h.name || ' has multiple rows' FROM dual
) WHERE EXISTS (
    SELECT s.name
    FROM smallish s
    WHERE h.name = s.name
    GROUP BY s.name
    HAVING count(1) > 1
);

要复制：


DROP TABLE huge;
DROP TABLE smallish;

CREATE TABLE huge (
    id INT PRIMARY KEY,
    name VARCHAR2(100),
    day INT,
    errno INT,
    error_message VARCHAR2(100)
);


CREATE TABLE smallish (
   id INT PRIMARY KEY,
   name VARCHAR2(100),
   day int
);

create index huge_name_indx ON huge (name);
create index smallish_name_indx ON smallish (name);

insert into huge values (1, 'good1', null, 0, null);
insert into huge values (2, 'good1', null, 0, null);
insert into huge values (3, 'good1', null, 0, null);
insert into huge values (4, 'good1', null, 0, null);
insert into huge values (5, 'good2', null, 0, null);
insert into huge values (6, 'good2', null, 0, null);
insert into huge values (7, 'good2', null, 0, null);
insert into huge values (8, 'good2', null, 0, null);
insert into huge values (9, 'double1', null, 0, null);
insert into huge values (10, 'double1', null, 0, null);
insert into huge values (11, 'double1', null, 0, null);
insert into huge values (12, 'double1', null, 0, null);
insert into huge values (13, 'double2', null, 0, null);
insert into huge values (14, 'double2', null, 0, null);
insert into huge values (15, 'double2', null, 0, null);
insert into huge values (16, 'double2', null, 0, null);
insert into huge values(17, 'notin1', null, 0, null);
insert into huge values(18, 'notin1', null, 0, null);
insert into huge values(19, 'notin1', null, 0, null);
insert into huge values(20, 'notin1', null, 0, null);
insert into huge values(21, 'notin2', null, 0, null);
insert into huge values(22, 'notin2', null, 0, null);
insert into huge values(23, 'notin2', null, 0, null);
insert into huge values(24, 'notin2', null, 0, null);

insert into smallish values (1, 'good1', 1);
insert into smallish values (2, 'good2', 2);
insert into smallish values (3, 'double1', 3);
insert into smallish values (4, 'double1', 4);
insert into smallish values (5, 'double2', 5);
insert into smallish values (6, 'double2', 6);

commit;

Answer 1

要在一个语句中执行此操作，您可以使用merge：

merge into huge h
    using (select name, count(*) as cnt, max(day) as day
           from smallish
           group by name
          ) s
    on h.name = s.name
when matched then update
    set day = (case when s.cnt = 1 then s.day else h.day end),
        errno = (case when s.cnt > 1 then 2 else h.errno end),
        error_message = (case when s.cnt > 1 then s.name || ' has multiple rows in smallish' else error_message end)
when not matched then update
    set errno = 1,
        error_message = h.name || ' not in smallish';

Answer 2

上面的合并看起来很有希望，但是您不能将其用于名称不匹配（语法限制）-为此需要单独的UPDATE，并且可能无法避免全表扫描。我喜欢使用“ IN（子查询）”，因为它通常会提供很好的解释计划。试试这些：

UPDATE huge SET ...
WHERE name IN (SELECT name FROM smallish GROUP BY name HAVING count(*) = 1);

UPDATE huge SET ...
WHERE name IN (SELECT name FROM smallish GROUP BY name HAVING count(*) > 1);

UPDATE huge SET ...
WHERE name NOT IN (SELECT name FROM smallish);

为huge_name_indx设置BITMAP索引也可能有所帮助。

顺便说一句，汤姆·凯特（Tom Kyte）建议用更新的数据创建一个新表，而不是更新现有的大表（https://asktom.oracle.com/pls/asktom/asktom.search?tag=how-to-update-millions-or-records-in-a-table-200211）。

更新一个表中具有1个匹配项，多个匹配项或其他表中的0个匹配项的所有行

2 个答案: