Question

我有这个查询

SELECT 
    COUNT(
            CASE 
                WHEN restricted LIKE '%us%' THEN 2 
                ELSE 1 
            END) AS totalcount 
FROM 
    reviews 
WHERE 
    restricted LIKE '%us%'

这会计算出现在列（限制）中的“我们”等值的总数。此列由muultiselect复选框中的值以“不寻常”的方式填充：

* us * ca * uk *等。

这很有效。没问题。

现在，我需要计算这个值（我们）没有出现的地方。我试图这样做，一个“经典”。

SELECT 
    COUNT(
            CASE 
                WHEN restricted NOT LIKE '%us%' THEN 2 
                ELSE 1 
            END
    ) AS totalcount 
FROM 
    reviews 
WHERE 
    restricted NOT LIKE '%us%'

我已声明状态不喜欢但现在......问题......它还计算“限制”列未填充的行（某些列表不使用此列）。伯爵错了。

有人可以帮忙吗？

Answer 1

这个怎么样？

SELECT 
    COUNT(*) AS totalcount 
FROM 
    reviews 
WHERE 
    restricted IS NOT NULL 
-- In case you expect restricted to contain spaces, the following line should be 
-- LTRIM(RTRIM(restricted)) != ''
AND restricted != '' 
AND restricted NOT LIKE '%us%'

这会过滤掉限制为null或空的行。

Answer 2

不需要CASE声明。你的WHERE子句意味着ELSE永远不会被击中。

匹配“我们”：

SELECT 
  COUNT(restricted) AS 'Count matching *us*' 
FROM 
  reviews 
WHERE 
  restricted IS NOT NULL 
  AND restricted LIKE '%us%'

不匹配“我们”（包括null）：

SELECT 
  COUNT(restricted) AS 'Count not matching *us*' 
FROM 
  reviews 
WHERE 
  restricted IS NULL 
  OR restricted NOT LIKE '%us%'

Answer 3

限制＆＃39;空值或空值列将通过＆＃34;限制NOT LIKE＆＃39;％us％＆＃39; ＆＃34;声明。尝试改为：

SELECT 
    COUNT(
            CASE 
                WHEN restricted NOT LIKE '%us%' THEN 2 
                ELSE 1 
            END
    ) AS totalcount 
FROM 
    reviews 
WHERE 
    restricted NOT LIKE '%us%' 
    AND restricted != '' 
    AND restricted IS NOT NULL

Answer 4

SELECT 
    COUNT(*) AS totalcount 
FROM 
    reviews 
WHERE 
    restricted IS NOT NULL

如果您希望限制包含空格，则以下行应为

LTRIM(RTRIM(restricted)) != ''
AND restricted != '' 
AND restricted NOT LIKE '%us%'

MySql SELECT COUNT ...不喜欢

4 个答案: