Oracle 使用另一个集合更新一个集合中的值

Question

我有一个使用 for 循环的解决方案，但这种方法效率低下且性能不佳。我是使用集合的新手，还不能在互联网上找到我的答案。谁能指出我正确的方向？ 这是我想要实现的简化版本。

DECLARE
TYPE rec_numbers IS RECORD
(
  DIGIT NUMBER,
  ENGLISH VARCHAR2 (10),
  FRENCH VARCHAR2(10),
  ITALIAN VARCHAR2(10)
);
TYPE tab_numbers IS TABLE OF rec_numbers;
blk_num tab_numbers;
blk_num2 tab_numbers;

BEGIN

SELECT 1 DIGIT, 'ONE' ENGLISH, NULL FRENCH, NULL ITALIAN 
BULK COLLECT INTO blk_num
FROM DUAL;

SELECT 1 DIGIT, NULL ENGLISH, 'UN' FRENCH, 'UNO' ITALIAN 
BULK COLLECT INTO blk_num2
FROM DUAL;    

FOR i IN blk_num.FIRST .. blk_num.LAST LOOP
    FOR j IN blk_num2.FIRST .. blk_num2.LAST LOOP
        IF blk_num(i).digit = blk_num2(j).digit THEN
            blk_num(i).french := blk_num2(j).french;
            blk_num(i).italian := blk_num2(j).italian;
        END IF;
    END LOOP;      
END LOOP;

END;    

Answer 1

你没有提到为什么你需要一个收藏品。如果您将 TYPE 和集合定义为模式对象，则可以使用 TABLE 函数执行简单的连接查询，以完全执行您的 for 循环所做的事情。

CREATE OR REPLACE
         TYPE rec_numbers AS OBJECT 
         ( digit      NUMBER,
         english    VARCHAR2(10),
         french     VARCHAR2(10),
         italian    VARCHAR2(10) );
         /

CREATE OR REPLACE    TYPE tab_numbers AS
              TABLE OF rec_numbers;
              /

代码

SET SERVEROUTPUT ON
DECLARE
     blk_num    tab_numbers;
     blk_num2   tab_numbers;
     blk_num3   tab_numbers;
BEGIN
     SELECT rec_numbers(1,'ONE',NULL,NULL) BULK COLLECT
       INTO blk_num
      FROM dual;

     SELECT rec_numbers(1,NULL,'UN','UNO') BULK COLLECT
       INTO blk_num2
     FROM dual;

SELECT rec_numbers
 ( a.digit,
   a.english, 
   COALESCE(b.french,a.french ),
   COALESCE(b.italian,a.italian)  
  ) BULK COLLECT
  INTO blk_num3
     FROM TABLE   ( blk_num  ) a
  LEFT JOIN TABLE ( blk_num2 ) b 
  ON a.digit = b.digit;
  blk_num := blk_num3;

  for i in 1..blk_num.count
  loop
  dbms_output.put_line(blk_num(i).digit ||','||blk_num(i).english
                ||','||blk_num(i).french||','||blk_num(i).italian
                );
  END LOOP;
END;
/

输出

1,ONE,UN,UNO

PL/SQL procedure successfully completed.

Demo

Answer 2

嵌套循环可能很慢。如果 blk_num 和 blk_num2 各有 1000 条记录，则您正在执行 1000x1000 = 100 万次迭代。这是 O(n^2) 的性能。

您可以在加载后花时间将blk_num2 转换为关联数组，从而将其降低到 O(n)。然后，您无需循环遍历 blk_num2 中的所有条目，只需通过索引值查询即可找到您要查找的条目。

这是我的意思的一个例子：

DECLARE
TYPE rec_numbers IS RECORD
(
  DIGIT NUMBER,
  ENGLISH VARCHAR2 (10),
  FRENCH VARCHAR2(10),
  ITALIAN VARCHAR2(10)
);
TYPE tab_numbers IS TABLE OF rec_numbers;
blk_num tab_numbers;
blk_num2 tab_numbers;
-- Define an associative array to copy blk_num2 into.
-- Note: I did not INDEX BY PLS_INTEGER because I do not know how big DIGIT can be.
-- if DIGIT will always fit into a PLS_INTEGER, you can use that instead.
TYPE assoc_numbers IS TABLE OF rec_numbers INDEX BY VARCHAR2(30);
blk_num2_aa assoc_numbers;


BEGIN

SELECT 1 DIGIT, 'ONE' ENGLISH, NULL FRENCH, NULL ITALIAN 
BULK COLLECT INTO blk_num
FROM DUAL;

SELECT 1 DIGIT, NULL ENGLISH, 'UN' FRENCH, 'UNO' ITALIAN 
BULK COLLECT INTO blk_num2
FROM DUAL;    

-- Copy blk_num2 into blk_num2_aa
FOR i in blk_num2.FIRST..blk_num2.LAST LOOP
  blk_num2_aa(to_char(blk_num2(i).digit)) := blk_num2(i);
END LOOP;

FOR i IN blk_num.FIRST .. blk_num.LAST LOOP
  -- Find the match based on the associative array's index value
  IF blk_num2_aa.exists(to_char(blk_num(i).digit)) THEN
    blk_num(i).french := blk_num2_aa(to_char(blk_num(i).digit)).french;
    blk_num(i).italian := blk_num2_aa(to_char(blk_num(i).digit)).italian;
  END IF;
END LOOP;

END;

Answer 3

这个怎么样？

update (select * from table(blk_num )) a
set (FRENCH, ITALIAN) = 
    (select FRENCH, ITALIAN 
     from table(blk_num2) b 
     where a.DIGIT = b.DIGIT);