Question

如果购买的桌子上已经有砖块，我需要更新砖块的价格，否则我必须插入它。我写了这段代码：

DECLARE 
    colour VARCHAR2(10) :='brown';
    shape VARCHAR2(10) :='rectangle';
    price NUMBER(10,2) :=21;

BEGIN 
    UPDATE purchased_bricks pb
    SET pb.price = price
    WHERE pb.colour = colour
    AND pb.shape = shape;

    IF sql%rowcount = 0 
        THEN
        INSERT INTO purchased_bricks
        VALUES(colour,shape,price);
    END IF;
END;

我第一次运行它，插入了记录。然后，当我尝试更新砖的价格时，更改未反映在表中。

Answer 1

我建议使用 MERGE 代替PL / SQL，也称为 UPSERT （UPdate和/或inSERT）。

这是一个例子：

SQL> CREATE TABLE purchased_bricks
  2  (
  3     colour   VARCHAR2 (10),
  4     shape    VARCHAR2 (20),
  5     price    NUMBER
  6  );

Table created.

SQL> MERGE INTO purchased_bricks p
  2       USING (SELECT 1 FROM DUAL) x
  3          ON (    p.colour = '&&par_colour'
  4              AND p.shape = '&&par_shape')
  5  WHEN MATCHED
  6  THEN
  7     UPDATE SET p.price = &&par_price
  8  WHEN NOT MATCHED
  9  THEN
 10     INSERT     (colour, shape, price)
 11         VALUES ( '&&par_colour', '&&par_shape', &&par_price);
Enter value for par_colour: brown
Enter value for par_shape: rectangular
Enter value for par_price: 21

1 row merged.

SQL> select * From purchased_bricks;

COLOUR     SHAPE                     PRICE
---------- -------------------- ----------
brown      rectangular                  21

让我们尝试其他价格：

SQL> undefine par_price
SQL> MERGE INTO purchased_bricks p
  2       USING (SELECT 1 FROM DUAL) x
  3          ON (    p.colour = '&&par_colour'
  4              AND p.shape = '&&par_shape')
  5  WHEN MATCHED
  6  THEN
  7     UPDATE SET p.price = &&par_price
  8  WHEN NOT MATCHED
  9  THEN
 10     INSERT     (colour, shape, price)
 11         VALUES ( '&&par_colour', '&&par_shape', &&par_price);
Enter value for par_price: 50

1 row merged.

SQL> select * from purchased_bricks;

COLOUR     SHAPE                     PRICE
---------- -------------------- ----------
brown      rectangular                  50

SQL>

调整了上面的示例，使其可以在SQL * Plus中使用。根据您使用的工具，参数的值可能会被不同地引用，例如

MERGE INTO purchased_bricks p
     USING (SELECT 1 FROM DUAL) x
        ON (    p.colour = :par_colour
            AND p.shape = :par_shape)
WHEN MATCHED
THEN
   UPDATE SET p.price = :par_price
WHEN NOT MATCHED
THEN
   INSERT     (colour, shape, price)
       VALUES ( :par_colour, :par_shape, :par_price);

Answer 2

我找到了。问题在于变量名与列名相同。因此set pb.price = price用现有表值而不是变量值更新了表。解决方案是通过使用前缀来使名称不同。

DECLARE 
    l_colour VARCHAR2(10) :='brown';
    l_shape VARCHAR2(10) :='rectangle';
    l_price NUMBER(10,2) :=21;

BEGIN 
    UPDATE purchased_bricks pb
    SET pb.price = l_price          
    WHERE pb.colour = l_colour                
    AND pb.shape = l_shape;

    IF sql%rowcount = 0 
        THEN
        INSERT INTO purchased_bricks
        VALUES(colour,shape,price);
    END IF;
END;

Answer 3

您可以对所需的过滤条件进行计数，如果大于0则不更新，否则插入新行；如下所示：

DECLARE 
        colour VARCHAR2(10) :='brown';
        shape VARCHAR2(10) :='rectangle';
        price NUMBER(10,2) :=21;
        count_row Number;

      BEGIN 
        Select count(*) into count_row from purchased_bricks pb
        WHERE pb.colour = colour
        AND pb.shape = shape;

        IF count_row > 0
        THEN
        UPDATE purchased_bricks pb
        SET pb.price = price
        WHERE pb.colour = colour
        AND pb.shape = shape;

         ELSE

            INSERT INTO purchased_bricks
            VALUES(colour,shape,price);
        END IF;
    END;

需要更新（如果存在）或插入值

3 个答案: