計數和排序，並加入

Question

我previous question給我，我可以採取

mysql> describe taps; 
+------------+-----------+------+-----+-------------------+-------+ 
| Field  | Type  | Null | Key | Default   | Extra | 
+------------+-----------+------+-----+-------------------+-------+ 
| tag  | int(11) | NO |  | NULL    |  | 
| station | int(11) | NO |  | NULL    |  | 
| time_Stamp | timestamp | NO |  | CURRENT_TIMESTAMP |  | 
+------------+-----------+------+-----+-------------------+-------+ 
3 rows in set (0.00 sec) 

，並使用查詢

SELECT tag 
    , COUNT(DISTINCT station) as `visit_count` 
    FROM taps 
GROUP 
    BY tag 
ORDER 
    BY COUNT(DISTINCT station) DESC 

得到由他們訪問站點的數量下令遊客答案。

現在我想添加

mysql> describe visitors; 
+--------+---------+------+-----+---------+-------+ 
| Field | Type | Null | Key | Default | Extra | 
+--------+---------+------+-----+---------+-------+ 
| tag_id | int(11) | NO |  | NULL |  | 
| name | text | NO |  | NULL |  | 
| email | text | NO |  | NULL |  | 
| phone | text | NO |  | NULL |  | 
+--------+---------+------+-----+---------+-------+ 
4 rows in set (0.00 sec) 

而且，而不是讓遊客tag_id，我希望得到他的name，email和phone。我知道，它涉及到一個JOIN，但就是無法弄清楚:-(

---

[更新]只是要清楚，我想輸出HTML表格，由誰訪問了大多數電視臺訂購，顯示的姓名，電子郵件&電話

Answer 1

SELECT tag 
    ,v.email, COUNT(DISTINCT station) as `visit_count` 
    FROM taps as t JOIN visitors as v ON t.tag = v.tag_id 
GROUP 
    BY v.email 
ORDER 
    BY COUNT(DISTINCT station) DESC