Question

嗨，这是我的疑问..

SELECT u.url,u.url_id,@mailid=wu.email FROM urls_uptime_log u INNER JOIN website_suburls wd ON u.url_id=wd.id
 INNER JOIN website_to_sms wu ON wd.websiteId=wu.website_id WHERE  u.time_stamp>=DATE_SUB(curdate(),INTERVAL 1 DAY) AND u.status<>200

修改：我得到的输出表是:(此输出包含网站的状态！= 200的网址）

url     |  url_id    |@mailid=wc.email
+-------+------------+----------------
http:/dfg   1616          aaa@gmail.com
+--------+------------+----------------
http:dfff   1621           aaa@gmail.com
+--------+------------+----------------
http:dfff   1619           bbb@ymail.com

我需要查看特定电子邮件的网站数量是多少？我可以使用以下查询吗？

SELECT COUNT(u.url) FROM urls_uptime_log u INNER JOIN website_suburls wd ON u.url_id=wd.id 
INNER JOIN website_to_sms wu ON wd.websiteId=wu.website_id WHERE wu.email=@mailid

Answer 1

要为用户变量赋值，必须使用赋值运算符:=而不是比较运算符=。如果你的意思是尝试

SELECT u.url,u.url_id, @mailid := wu.email 
  FROM urls_uptime_log u JOIN website_suburls wd 
    ON u.url_id = wd.id JOIN website_to_sms wu 
    ON wd.websiteId = wu.website_id 
 WHERE u.time_stamp >= CURDATE() - INTERVAL 1 DAY 
   AND u.status <> 200

另一方面，如果你只想给列做一个替代名称（别名）

SELECT u.url, u.url_id, wu.email mailid ...

更新：如果没有看到您的表架构，示例数据和所需的输出，很难说，但您可以尝试这样的事情

SELECT COUNT(DISTINCT u.url) 
  FROM urls_uptime_log u JOIN website_suburls wd 
    ON u.url_id = wd.id JOIN website_to_sms wu 
    ON wd.websiteId = wu.website_id
 WHERE wu.email IN
(
    SELECT DISTINCT wu.email 
      FROM urls_uptime_log u JOIN website_suburls wd 
        ON u.url_id = wd.id JOIN website_to_sms wu 
        ON wd.websiteId = wu.website_id 
     WHERE u.time_stamp >= CURDATE() - INTERVAL 1 DAY 
       AND u.status <> 200
)

执行查询时，mysql变量返回null

1 个答案: