MySql:“字段列表”中的未知列 ID

我遵循这个问题的答案：MySQL：是否可以group_concat多行？

但是我在MySql错误中挣扎：

#1054 - Unknown column 'CM_Cocktail.id' in 'field list' .

当我删除这个 CM_Cocktail.id 的东西时，下一个错误应用者：#1054 - Unknown column 'CM_Zutat.name' in 'field list' .我尝试使用别名，但情况变得更糟...也许我是瞎子。

SELECT 
  CM_Cocktail.id, CM_Cocktail.name,
  GROUP_CONCAT(CM_Zutat.name SEPARATOR ', ')
FROM (
  SELECT CM_Cocktail.id, CM_Cocktail.name AS cname, CM_Zutat.name
              FROM CM_Cocktail, CM_CocktailHatZutat, CM_Zutat
              WHERE CM_Cocktail.id=CM_CocktailHatZutat.cocktail_id
              AND CM_CocktailHatZutat.zutat_id=CM_Zutat.id
) group_attr
GROUP BY CM_Cocktail.id;

CREATE TABLE IF NOT EXISTS `CM_Cocktail` (
  `id` int(3) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `name` (`name`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;

CREATE TABLE IF NOT EXISTS `CM_CocktailHatZutat` (
  `cocktail_id` int(3) NOT NULL,
  `zutat_id` int(3) NOT NULL,
  `zutat_menge` int(4) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

CREATE TABLE IF NOT EXISTS `CM_Zutat` (
  `id` int(3) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `name` (`name`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

SELECT 
CocktailID, cname,
GROUP_CONCAT(ZName SEPARATOR ', ')
FROM (
SELECT CM_Cocktail.id as CocktailID, CM_Cocktail.name AS cname, CM_Zutat.name as ZName
     FROM CM_Cocktail, CM_CocktailHatZutat, CM_Zutat
     WHERE CM_Cocktail.id=CM_CocktailHatZutat.cocktail_id
     AND CM_CocktailHatZutat.zutat_id=CM_Zutat.id
) group_attr
GROUP BY CocktailID;

SELECT 
id, name,
GROUP_CONCAT(CM_Zutat.name SEPARATOR ', ')
FROM (
SELECT CM_Cocktail.id, CM_Cocktail.name AS cname, CM_Zutat.name
          FROM CM_Cocktail, CM_CocktailHatZutat, CM_Zutat
          WHERE CM_Cocktail.id=CM_CocktailHatZutat.cocktail_id
          AND CM_CocktailHatZutat.zutat_id=CM_Zutat.id
) group_attr
GROUP BY id;