容易MySQL查詢問題

這裏是mysql的容易MySQL查詢問題

sent_to customer msg  read 
------- -------- ------ ----- 
45   3  bla  0 
34   4  bla  1 
34   6  bla  0 
45   3  bla  0 
56   7  bla  1 
45   8  bla  0

對於其ID號爲45級的日誌中例如用戶的「味精」表，

我希望他看到這一點，

you have 2 unread msg to your "number 3" customer 
you have 1 unread msg to your "number 8" customer

像新聞飼料

我應該使用什麼查詢？

THX

來源

2010-04-21 Utku Dalmaz

您可能需要使用下面的查詢。

SELECT CONCAT('You have ', 
       COUNT(`read`), 
       ' unread msg to your number ', 
       customer, 
       ' customer') AS news 
FROM  msg 
WHERE `read` = '0' AND `sent_to` = '45' 
GROUP BY customer;

注意read是在MySQL中的保留字，所以你必須把它們放在反引號。（Source）

測試用例：

CREATE TABLE msg (
    `sent_to` int, 
    `customer` int, 
    `msg`  varchar(10), 
    `read`  int 
); 

INSERT INTO msg VALUES(45, 3, 'bla', 0); 
INSERT INTO msg VALUES(34, 4, 'bla', 1); 
INSERT INTO msg VALUES(34, 6, 'bla', 0); 
INSERT INTO msg VALUES(45, 3, 'bla', 0); 
INSERT INTO msg VALUES(56, 7, 'bla', 1); 
INSERT INTO msg VALUES(45, 8, 'bla', 0);

查詢結果：

+-------------------------------------------------+ 
| news           | 
+-------------------------------------------------+ 
| You have 2 unread msg to your number 3 customer | 
| You have 1 unread msg to your number 8 customer | 
+-------------------------------------------------+ 
2 rows in set (0.00 sec)

來源

2010-04-21 17:21:19

感謝您的很好的回答，也給予關於「讀」的信息，我有一個問題，mysql_num_rows給出0，即使我有「讀= 0」在數據庫中？ – 2010-04-21 17:36:08

哦好吧我修好了thx – 2010-04-21 17:38:08

...

select count(*), customer from msg where read = 0 group by customer

來源

2010-04-21 17:19:34

和'sent_to'想必？最確定的是 – 2010-04-21 17:22:48

！ – 2010-04-22 00:11:19

SELECT COUNT(read), customer FROM msg WHERE read = 0 AND sent_to = '45' GROUP BY customer;

來源

2010-04-21 17:23:16 darcy

select count(sent_to), customer 
from msg 
where read < 1 
and sent_to = 45 
group by customer

來源

2010-06-14 19:00:35 geeko

容易MySQL查詢問題

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