獲得百分比SELECT查詢在PHP SQL行工作

Question

我終於得到這個代碼 MySQL的內部工作作爲一個單一的查詢，而不是2 SELECT查詢一起在PHP代碼的一個$ SQL行：

SELECT count(*) from mytable into @AgentCount; 

SELECT area 
    , COUNT(area) AS thecount 
    , ROUND ((COUNT(*)/(@AgentCount) * 100), 1) AS percentage 
FROM mytable 
GROUP BY area 
ORDER BY thecount DESC LIMIT 50; 

此外，仍然無法弄清楚如何在ROUND中使用％符號，以使百分比出現爲53.3％（等等）。我比我開始時更近，但仍然沒有雪茄。

我很可能在這裏錯過了一些簡單的事情，但頭腦凍結了。

感謝您的解凍幫助。

-stucko

Answer 1

使用concat函數添加％符號並使用交叉連接。

SELECT area 
    , COUNT(area) AS thecount 
    , concat(ROUND((COUNT(*)/AgentCount * 100), 1), '%') AS percentage 
FROM mytable 
    cross join 
     (SELECT count(*) AgentCount from mytable) t 
GROUP BY area 
ORDER BY thecount DESC LIMIT 50; 

Answer 2

SELECT area 
    , COUNT(area) AS thecount 
    , concat(ROUND ((COUNT(*)/(total_count) * 100), 1),"%") AS percentage 
FROM mytable,(SELECT count(*) as total_count from mytable)temp 
GROUP BY area 
ORDER BY thecount DESC LIMIT 50;