mysql - 返回表中的所有字段或返回一个查询中不存在的字段

Question

我想在给出id后从行中获取所有字段，但如果表中没有指定id的行，那么我想返回“不存在”或其他内容。我在一个查询中需要这个。

http://sqlfiddle.com/#!9/0afa50

Answer 1

我只能想到这个解决方案:

(SELECT * FROM `new_table` WHERE id = 1)
UNION 
(SELECT 'not exists!' AS id, 'not exists!' AS name, 'not exists!' AS surname, 'not exists!' AS anotherfield 
 FROM new_table 
 WHERE NOT EXISTS (SELECT * FROM `new_table` WHERE id = 1)
);
+----+------+---------+--------------+
| id | name | surname | anotherfield |
+----+------+---------+--------------+
| 1  | aaa  | ssss    | adad         |
+----+------+---------+--------------+
1 row in set (0,00 sec)

(SELECT * FROM `new_table` WHERE id = 0) 
UNION 
(SELECT 'not exists!' AS id, 'not exists!' AS name, 'not exists!' AS surname, 'not exists!' AS anotherfield         
 FROM new_table         
 WHERE NOT EXISTS (SELECT * FROM `new_table` WHERE id = 0)        
);
+-------------+-------------+-------------+--------------+
| id          | name        | surname     | anotherfield |
+-------------+-------------+-------------+--------------+
| not exists! | not exists! | not exists! | not exists!  |
+-------------+-------------+-------------+--------------+
1 row in set (0,00 sec)

更新:如您所见，它非常丑陋。为什么需要通过 SQL 来解决这个任务？用您使用的编程语言替换默认值不是更容易吗(我希望)？