MySQL:UNION+JOIN

我正试图弄明白如何在相同的连接中使用UNION，而不是陷入#2014-命令不同步

我已经创建了四个简单的表，并为它们编写了简化的代码。主要的想法是要得到每一把剑，枪和帽子，其中有'最好的'的名字+如果这些所有者很酷

SELECT pirate_id, sword, null AS gun, null AS hat, legs = 1 AS is_cool FROM swords as t1
LEFT JOIN
(SELECT legs, pirate_id as pid FROM body_parts) AS t2
ON t1.pirate_id = t2.pid
WHERE sword LIKE '%best%'

UNION
SELECT pirate_id, null AS sword, gun, null AS hat, legs = 1 AS is_cool FROM guns as t3
LEFT JOIN
(SELECT legs, pirate_id as pid FROM body_parts) AS t4
ON t3.pirate_id = t4.pid
WHERE gun LIKE '%best%' 

UNION
SELECT pirate_id, null AS sword, null AS gun, hat, legs = 1 AS is_cool FROM hats as t5
LEFT JOIN
(SELECT legs, pirate_id as pid FROM body_parts) AS t6
ON t5.pirate_id = t6.pid
WHERE hat LIKE '%best%' 

ORDER BY is_cool DESC, pirate_id ASC

目前，我必须重复加入每次我必须联盟。我希望有更好的方法

表格创建：

CREATE TABLE IF NOT EXISTS `body_parts` (
  `pirate_id` int(11) NOT NULL,
  `legs` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `body_parts` (`pirate_id`, `legs`) VALUES
(1, 0),
(2, 1),
(3, 2);

CREATE TABLE IF NOT EXISTS `guns` (
  `pirate_id` int(11) NOT NULL,
  `gun` varchar(100) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `guns` (`pirate_id`, `gun`) VALUES
(1, 'best 1');

CREATE TABLE IF NOT EXISTS `hats` (
  `pirate_id` int(11) NOT NULL,
  `hat` varchar(100) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `hats` (`pirate_id`, `hat`) VALUES
(2, 'best 2');

CREATE TABLE IF NOT EXISTS `swords` (
  `pirate_id` int(11) NOT NULL,
  `sword` varchar(100) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `swords` (`pirate_id`, `sword`) VALUES
(3, 'best 3');

---

您可以使用单个左连接，通过子查询更改

    select t1.pirate_id, t1.sword, t1.gun, t1.hat, t2.legs=1  is_cool

    from  (
        SELECT pirate_id, sword, null AS gun, null AS hat 
        FROM swords 
        WHERE sword LIKE '%best%'

        UNION 

        SELECT pirate_id, null, gun, null
        FROM guns 
        WHERE gun LIKE '%best%' 

        UNION 
        SELECT pirate_id, null, null, hat
        FROM hats 
        WHERE hat LIKE '%best%'
    ) t1  
    LEFT JOIN (
        SELECT legs, pirate_id 
        FROM body_parts
    ) t2 ON t2.pirate_id = t1.pirate_id
    ORDER BY is_cool DESC, pirate_id ASC

---

#1054-字段列表中的未知列“legs”#1222-使用的SELECT语句有不同数量的列SANSWER更新。。由于每个select UNION中有4列，因此缺少一个字符。无论如何，谢谢你的回答！为什么不写“选择海盗身份证、剑、枪、帽子、腿=1，因为它很酷”，而不写表名？我缺少的字符是什么“为什么不写‘选择海盗身份证、剑、枪、帽子、腿=1，因为它很酷’，而不写桌名？”这句话是什么意思。。我不清楚。。更好地解释