Question

我有这个SQL

   SELECT COUNT( DISTINCT `ct`.`content_type_id`) AS `type_count`
        , COUNT(`content_id`) AS `content_count`
     FROM `user_to_content` `uc`
     JOIN `content` `c` USING (`content_id`)
     JOIN `content_type` `ct` USING (`content_type_id`)
    WHERE `user_id`= 1;

我想在同一张桌子上同时获得2个数字。

此SQLfiddle文件显示上述SQL正在运行...

但是我无法让SP版本正常工作。我似乎无法让SQLfiddle与SP合作（也许有些人可以告诉我如何），所以here是在pastebin的SP ......

无论如何，SP不会用第二个COUNT命令“编译”......

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 24

这是第24行

DECLARE $type_count    TINYINT(1) UNSIGNED DEFAULT 0;

我在这方面的工作也是两次调用桌子上的COUNT。我不愿意。

Answer 1

这是一个有一些修改的例子：

DELIMITER $$

CREATE PROCEDURE `SP_system_content_badges`
(  IN `$user_id` INT(10) UNSIGNED
 , IN `$content_id` INT(10) UNSIGNED
 , IN `$content_type_id` INT(10) UNSIGNED
)
BEGIN
  -- DECLARE $type_count    TINYINT(1) UNSIGNED DEFAULT 0;
  -- DECLARE $content_count TINYINT(1) UNSIGNED DEFAULT 0;

  DECLARE `$type_count`    INT UNSIGNED DEFAULT 0;
  DECLARE `$content_count` INT UNSIGNED DEFAULT 0;

  -- How many pieces of content has this person consumed
  -- Is there 4 unique kinds as well

  -- SELECT COUNT( DISTINCT `ct`.`content_type_id`) INTO $type_count
  --      , COUNT(`content_id`) INTO $content_count

  SELECT COUNT(DISTINCT `ct`.`content_type_id`), COUNT(`c`.`content_id`)
    INTO `$type_count`, `$content_count`
  FROM `user_to_content` `uc`
    JOIN `content` `c` USING (`content_id`)
    JOIN `content_type` `ct` USING (`content_type_id`)
  WHERE `user_id`= `$user_id`;
  SELECT `$type_count` 'TYPE_COUNT', `$content_count` 'CONTENT_COUNT';
END$$

DELIMITER ;

SQL Fiddle example

mySQL，存储过程，双COUNT不起作用（带有SQLfiddle）

1 个答案: