获取具有值的列的百分比

Question

我试着这样做：

 SELECT (SELECT COUNT(*)
 FROM users WHERE email != '') / (SELECT COUNT(*) FROM users)

但它在php Operand should contain 1 column(s)中给了我这个错误，在mysql中一切正常。还有其他解决办法吗？

  编辑：它不完全像我几个月也有组，上面的查询有效，但接受的答案解决了我的错误

Answer 1

您可以在一个查询中执行此操作：

SELECT SUM(email != '') / COUNT(*) 
    AS ratio
  FROM users

......或者，如果你需要百分比：

SELECT 100 * SUM(email != '') / COUNT(*)
    AS percentage
  FROM users

Answer 2

试试这个..

SELECT (SELECT COUNT(your_primary_key) FROM users WHERE email != '') / (SELECT COUNT(your_primary_key) FROM users) ...

由于两个表中的字段数量不同而发生这种情况......

Answer 3

试试这个：

 SELECT COUNT(*) as counts
 FROM users 
 WHERE email != '' 

Answer 4

你想做什么？如果您可以在SQL查询中计算百分比，则可以点击链接Calculating Percentage within MYSQL query based on conditions