如何在MySQL中显示GROUP BY计数大于1的所有结果？

Question

这是我的MySQL查询及其返回的内容：

SELECT email, COUNT(*) AS num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL    
GROUP BY email
ORDER BY dateadded DESC;

我只想返回num＆gt;的结果1.我试图像这样更改我的查询，但它没有说明num不是公认的列：

SELECT email, COUNT(*) AS num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL
AND num > 1
GROUP BY email
ORDER BY dateadded DESC;

如何在num＆gt;处返回结果？ 1？

Answer 1

在GROUP BY子句之后，在ORDER BY子句之前添加：

HAVING COUNT(*) > 1 

在HAVING子句之前准备好所有行之后，LIMIT子句几乎在执行计划中最后应用。这对于在访问行时无法检查的条件非常有用，但仅在访问了行之后才会检查，例如像COUNT（*）这样的聚合函数，尽管它可以用于非聚合。 / p>

Answer 2

使用HAVING子句。或者另一种方式：

SELECT * FROM
 (SELECT email, COUNT(*) AS num  
  FROM collectors_users
  WHERE subscribed != 'NO'
  AND lastLogin IS NULL    
  GROUP BY email
  ORDER BY dateadded DESC;
  ) T
WHERE num>1

Answer 3

我认为你应该使用HAVING

SELECT email, COUNT(*) AS num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL
GROUP BY email
ORDER BY dateadded DESC
HAVING COUNT(*) > 1; 

这个问题可以帮助您理解为什么HAVING而不是WHERE

WHERE vs HAVING

Answer 4

SELECT email, COUNT(*) as num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL
GROUP BY email
HAVING COUNT(*) > 1

Answer 5

您可以使用Having子句。

SELECT email, COUNT(*) AS num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL
GROUP BY email
HAVING count(*) > 1
ORDER BY dateadded DESC;

了解更多信息：http://www.w3schools.com/sql/sql_having.asp