mysql - 找不到结果时返回 '0'

Question

如果没有找到记录，是否可以修改此查询以在成绩列中返回“0”？现在，当什么都找不到时，我得到一个空白行。

SELECT  COALESCE(Score, 0) AS Score, COALESCE(Grade,0) AS Grade, LAST_UPDATE, '' as  dummy
FROM (
SELECT (CASE ScanName WHEN '%=APP%' THEN '= Network System' END) AS   System, 
(sum(CASE severity WHEN 1 THEN 10 WHEN 2 THEN 9 WHEN 3 THEN 6 WHEN 4 THEN 3    
END)/COUNT(severity )) AS Score,
(sum(CASE severity WHEN 1 THEN 10 WHEN 2 THEN 9 WHEN 3 THEN 6 WHEN 4 THEN 3   
END)/COUNT(severity )) AS Grade, LAST_UPDATE 

FROM missing_internal

WHERE ScanName like '%APP%'

and LAST_UPDATE >= SUBDATE((SELECT MAX(LAST_UPDATE) from missing_internal), INTERVAL 1 day) 
HAVING System IS NOT NULL  
LIMIT 1

UNION ALL 

SELECT NULL AS Score, NULL AS Grade, 0 AS LAST_UPDATE, '' as dummy

ORDER BY LAST_UPDATE DESC

)  as T2
LIMIT 1;

Score 和 Grade 是故意的，因为我使用单独的脚本对分数应用字母等级。

Answer 1

看看 COALESCE功能...

如果我很好地理解了您的问题，那么下面的查询应该完全按照您的要求执行(现在返回行时得到零):

SELECT COALESCE(Score, 0) AS Score, COALESCE(Grade,0) AS Grade, LAST_UPDATE 
FROM (
SELECT (CASE ScanName WHEN 'APP' THEN 'Network System' END) AS System, 
(sum(CASE severity WHEN 1 THEN 10 WHEN 2 THEN 9 WHEN 3 THEN 6 WHEN 4 THEN 3    
END)/COUNT(severity )) AS Score,
(sum(CASE severity WHEN 1 THEN 10 WHEN 2 THEN 9 WHEN 3 THEN 6 WHEN 4 THEN 3   
END)/COUNT(severity )) AS Grade, LAST_UPDATE

FROM missing_internal

WHERE ScanName like '%APP%'

and LAST_UPDATE >= SUBDATE((SELECT MAX(LAST_UPDATE) from missing_internal), INTERVAL 1 day)
HAVING System IS NOT NULL  
LIMIT 1

UNION ALL 

SELECT NULL AS Score, NULL AS Grade, 0 AS LAST_UPDATE 

ORDER BY LAST_UPDATE DESC
) AS T2
LIMIT 1;

我还没有测试查询，所以你可能需要安排它......但是接受这个想法......

如果你只想在列返回空结果时得到零，你可以使用COALESCE与上面的例子一样:

COALESCE(sum(CASE severity WHEN 1 THEN 10 WHEN 2 THEN 9 WHEN 3 THEN 6 WHEN 4 THEN 3   
END)/COUNT(severity ), 0) AS Grade