Question

SELECT
 DATE_TRUNC('day', payment.paid_at) AS "DAILY",
 COUNT(CASE WHEN product.type = 'A' THEN 1 ELSE NULL END) AS "A count",
 SUM(CASE WHEN product.type = 'A' AND payment.price_currency_code='USD' THEN payment.price_amount*1080 ELSE payment.price_amount END) AS "A revenue",
 COUNT(CASE WHEN product.type = 'B' THEN 1 ELSE NULL END) AS "B count",
 SUM(CASE WHEN product.type = 'B' AND payment.price_currency_code='USD' THEN payment.price_amount*1080 ELSE payment.price_amount END) AS "B revenue"

您好，

我在使用＆＃34; CASE WHEN＆＃34;时遇到问题我已按上述方式编写了查询，但结果不准确。事实上，每个类别的收入总和计算的数量几乎相同，但实际上它应该是不同的（尽管计数是正确的）。

如果你能指出我错过的或做错了，我们将不胜感激！

谢谢：）

Answer 1

您的ELSE语句中存在一个错误，因为它包含了其他所有内容，而实际上您可能意图使用其他条件：

SELECT
 DATE_TRUNC('day', payment.paid_at) AS "DAILY",
 COUNT(CASE WHEN product.type = 'A' THEN 1 END) AS "A count",
 SUM(CASE WHEN product.type = 'A' AND payment.price_currency_code='USD' THEN payment.price_amount*1080 WHEN product.type = 'A' THEN payment.price_amount END) AS "A revenue",
 COUNT(CASE WHEN product.type = 'B' THEN 1 END) AS "B count",
 SUM(CASE WHEN product.type = 'B' AND payment.price_currency_code='USD' THEN payment.price_amount*1080 WHEN product.type = 'B' THEN payment.price_amount END) AS "B revenue"

此外，在case语句中，ELSE NULL不是必需的，因为NULL是非匹配的默认结果。

结果不准确的情况

1 个答案: