Mysql存儲過程：如何處理空結果集

Question

我編寫了一個程序，其中一個語句沒有正確執行：

SELECT thumb_image into v_thumb_image FROM RESTAURANT_IMAGE WHERE 
   RESTAURANT_ID = v_restaurant_id

我調查的原因是，如果在任何時候結果集為空，則程序不會進一步運行語句。

請注意，我在一個循環中調用它。

我關心的是如果對於任何v_restaurant_id ，結果集為空，則不會停止執行。

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

從文檔 ：

NOT FOUND是以'02'開頭的SQLSTATE值類的簡寫。 這與游標的上下文相關，用於控制光標到達數據集末尾時發生的情況。 如果沒有更多行可用，則SQLSTATE值為“02000”時會出現“無數據”條件。 要檢測此情況，可以為其設置處理程序（或對於NOT FOUND條件）。 有關示例，請參見第13.6.6節“游標”。 對於不檢索任何行的SELECT ... INTO var_list語句，也會出現這種情況。

因此，當restaurant_image表中沒有返回任何行時，它也會遇到NOT FOUND狀態，並調用定義的處理程序，這會導致離開循環。

一種解決方案是通過將其放在BEGIN...END塊中來為該選擇聲明另一個處理程序：

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;

---

畢竟，為什么使用存儲過程和游標這樣做會很慢。 您可以實現執行單個語句的相同功能：

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)

或者通過消除每行的子查詢使其更快：

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

分別嘗試這些子查詢以找到更好的理解。