Question

我编写了一个程序，其中一个语句没有正确执行：

SELECT thumb_image into v_thumb_image FROM RESTAURANT_IMAGE WHERE 
   RESTAURANT_ID = v_restaurant_id

我调查的原因是，如果在任何时间点结果集为空，则程序不会进一步运行语句。

请注意我在循环中调用它。

我担心的是，如果任何v_restaurant_id，结果集为空，则不会停止执行。

完整程序：

-- --------------------------------------------------------------------------------
-- Routine DDL
-- Note: comments before and after the routine body will not be stored by the server
-- --------------------------------------------------------------------------------
DELIMITER $$

CREATE DEFINER=`root`@`localhost` PROCEDURE `populate_restaurant_details`()
BEGIN
 DECLARE v_finished_cuisines, 
         v_finished, 
         v_restaurant_id, 
         v_count_discount
 INT DEFAULT 0;

 DECLARE v_cuisines, 
         v_thumb_image 
 varchar(200) DEFAULT "";

 DECLARE cuisine_title varchar(50) DEFAULT "";
 -- Fetch all restaurant id
 DECLARE restaurant_cursor CURSOR FOR
   SELECT id FROM delhifoodonline.restaurant order by id desc;

 DECLARE CONTINUE HANDLER 
  FOR NOT FOUND SET v_finished = 1;

 OPEN restaurant_cursor;

 get_restaurant: LOOP   

   FETCH restaurant_cursor INTO v_restaurant_id;
   IF v_finished = 1 THEN 
    LEAVE get_restaurant;
   END IF;

  SET v_finished_cuisines =""; 
  SET v_thumb_image = "";
begin
  DECLARE CONTINUE HANDLER FOR NOT FOUND SET v_thumb_image = NULL;

  SELECT thumb_image into v_thumb_image 
  FROM restaurant_image 
  WHERE restaurant_id = v_restaurant_id
  ORDER BY id
  LIMIT 1;
end;

  SELECT count(*) into v_count_discount FROM restaurant_discount WHERE 
    restaurant_id = v_restaurant_id;

BLOCK2: BEGIN

  DECLARE cuisines_cursor CURSOR FOR 
   SELECT cuisine.title FROM restaurant_cuisine INNER JOIN cuisine 
    ON restaurant_cuisine.cuisine_id = cuisine.id
    WHERE 
    restaurant_cuisine.restaurant_id = v_restaurant_id
    LIMIT 0,5;

  DECLARE CONTINUE HANDLER 
    FOR NOT FOUND SET v_finished_cuisines = 1;
  SET v_cuisines = "";
  OPEN cuisines_cursor;

  get_cuisine: LOOP
   FETCH cuisines_cursor INTO cuisine_title;

   IF v_finished_cuisines = 1 THEN 
    LEAVE get_cuisine;
   END IF;

   SET v_cuisines = CONCAT(cuisine_title,", ",v_cuisines);

   END LOOP get_cuisine;
  CLOSE cuisines_cursor;

END BLOCK2;

  SET v_cuisines = TRIM(BOTH ", " FROM v_cuisines);

  IF v_count_discount > 0 THEN
   SET v_count_discount = 1;
  ELSE
   SET v_count_discount = 0;
  END IF;

  UPDATE restaurant SET 
                        thumb_image = v_thumb_image,
                        cuisines_list = v_cuisines,
                        discount_available = v_count_discount
                   WHERE id= v_restaurant_id;
 END LOOP get_restaurant;

CLOSE restaurant_cursor;

END

Answer 1

来自documentation：

NOT FOUND是开始的SQLSTATE值类的简写与＆＃39; 02＆＃39;。这在游标的上下文中是相关的并且被使用控制光标到达数据集末尾时发生的情况。如果没有更多行可用，则会出现No Data条件 SQLSTATE值＆＃39; 02000＆＃39;。要检测此情况，您可以设置一个它的处理程序（或NOT FOUND条件）。有关示例，请参阅第13.6.6节“游标”。 SELECT也会出现这种情况...... INTO var_list语句，不检索任何行。

因此，当restaurant_image表中的select不返回任何行时，它也会满足NOT FOUND状态，并调用定义的处理程序，这会导致离开循环。

一种解决方案是通过将其放入BEGIN...END块内来为该选择声明另一个处理程序：

begin
  DECLARE CONTINUE HANDLER FOR NOT FOUND SET v_thumb_image = NULL;

  SELECT thumb_image into v_thumb_image 
  FROM restaurant_image 
  WHERE restaurant_id = v_restaurant_id
  ORDER BY id
  LIMIT 1;
end;

毕竟，为什么使用存储过程和游标这样做会很慢。您可以实现执行单个语句的相同功能：

UPDATE restaurant 
SET thumb_image = (
    SELECT thumb_image 
    FROM restaurant_image 
    WHERE restaurant_id = restaurant.id
    ORDER BY id
    LIMIT 1),
discount_available = IF(EXISTS(
    SELECT 1
    FROM restaurant_discount 
    WHERE restaurant_id = restaurant.id), 1, 0), 
cuisines_list = (
    SELECT group_concat(cuisine.title separator ', ')
    FROM restaurant_cuisine
    INNER JOIN cuisine ON restaurant_cuisine.cuisine_id = cuisine.id
    WHERE restaurant_cuisine.restaurant_id = restaurant.id
    LIMIT 0,5)

或者通过消除每一行的子查询使其更快：

UPDATE restaurant r
LEFT JOIN 
    (SELECT restaurant_id, count(*) AS discount_available
    FROM restaurant_discount 
    GROUP BY restaurant_id) d ON r.id = d.restaurant_id
LEFT JOIN 
    (SELECT restaurant_id, thumb_image 
    FROM restaurant_image r1
    WHERE NOT EXISTS (
        SELECT 1 FROM restaurant_image r2 WHERE r2.restaurant_id = r1.restaurant_id AND r2.id < r1.id
    )) t ON r.id = t.restaurant_id
LEFT JOIN
    (SELECT rc.restaurant_id, SUBSTRING_INDEX(GROUP_CONCAT(c.title SEPARATOR ', '), ',', 5) AS cuisines_list
    FROM restaurant_cuisine rc
    INNER JOIN cuisine c ON rc.cuisine_id = c.id
    GROUP BY rc.restaurant_id
    ) rc ON r.id = rc.restaurant_id
SET r.discount_available = IF(d.discount_available = 0, 0, 1),
r.thumb_image = t.thumb_image,
r.cuisines_list = rc.cuisines_list

分别尝试这些子查询以更好地理解。

Mysql存储过程：如何处理空结果集

1 个答案: