SELECT expression1, expression2, ... expression_n
FROM tables
[WHERE conditions]
UNION [ALL | DISTINCT]
SELECT expression1, expression2, ... expression_n
FROM tables
[WHERE conditions];

mysql> SELECT * FROM Websites;
+----+--------------+---------------------------+-------+---------+
| id | name     | url            | alexa | country |
+----+--------------+---------------------------+-------+---------+
| 1 | Google    | https://www.google.cm/  | 1   | USA   |
| 2 | 淘宝     | https://www.taobao.com/  | 13  | CN   |
| 3 | 菜鸟教程   | http://www.runoob.com/  | 4689 | CN   |
| 4 | 微博     | http://weibo.com/     | 20  | CN   |
| 5 | Facebook   | https://www.facebook.com/ | 3   | USA   |
| 7 | stackoverflow | http://stackoverflow.com/ |  0 | IND   |
+----+---------------+---------------------------+-------+---------+

mysql> SELECT * FROM apps;
+----+------------+-------------------------+---------+
| id | app_name  | url           | country |
+----+------------+-------------------------+---------+
| 1 | QQ APP   | http://im.qq.com/    | CN   |
| 2 | 微博 APP | http://weibo.com/    | CN   |
| 3 | 淘宝 APP | https://www.taobao.com/ | CN   |
+----+------------+-------------------------+---------+
3 rows in set (0.00 sec)

SELECT country FROM Websites
UNION
SELECT country FROM apps
ORDER BY country;

SELECT country FROM Websites
UNION ALL
SELECT country FROM apps
ORDER BY country;

SELECT country, name FROM Websites
WHERE country='CN'
UNION ALL
SELECT country, app_name FROM apps
WHERE country='CN'
ORDER BY country;