【发布时间】:2016-05-09 16:05:40
【问题描述】:
我正在处理一个比较查询,它将比较针对不同信用卡类型处理的内容与我们的 POS 服务实际处理的内容。我遇到了一个问题,如果某些卡类型在给定日期没有处理任何付款但可能已经退款,我看不到它们的行。 IE:没有使用 VISA 购买任何东西,但获得了 VISA 退款,但由于没有购买,即使该类型有退款,该行也不会显示。
这是我的查询:
WITH tran_total AS (
SELECT CONCAT(c.id_str, ' - ', c.clinic_str) AS Clinics,
t.clinic,
c.xcharge_mid,
p.pay_desc,
CAST(t.time_ran AS date) AS tran_date,
CASE WHEN SUM(t.amount) <> 0 THEN SUM(t.amount) * - 1
ELSE 0.00 END AS collection
FROM dbo.transactions AS t INNER JOIN
dbo.clinic_master AS c ON t.clinic = c.clinic INNER JOIN
dbo.paytype AS p ON t.clinic = p.clinic AND t.paytype_id = p.paytype_id
WHERE (t.time_ran > GETDATE() - 10)
AND (t.paytype_id IS NOT NULL)
AND (p.pay_desc = 'Visa' OR
p.pay_desc = 'MasterCard' OR
p.pay_desc = 'American Express' OR
p.pay_desc = 'Discover')
GROUP BY c.id_str, c.clinic_str, t.clinic, c.xcharge_mid, p.pay_desc, CAST(t.time_ran AS date))
SELECT w.Clinics,
w.pay_desc,
w.tran_date,
w.collection,
CASE WHEN w.pay_desc = 'Visa' THEN (SUM(VI_pur) - SUM(VI_ref))
WHEN w.pay_desc = 'MasterCard' THEN (SUM(MC_pur) - SUM(MC_ref))
WHEN w.pay_desc = 'American Express' THEN (SUM(AX_pur) - SUM(AX_ref))
WHEN w.pay_desc = 'Discover' THEN (SUM(DI_pur) - SUM(DI_ref)) END AS xcharge
FROM tran_total AS w LEFT OUTER JOIN
dbo.xcharge AS x ON w.xcharge_mid = x.xcharge_mid AND w.tran_date = x.settle_date
GROUP BY w.Clinics, w.pay_desc, w.tran_date, w.collection
ORDER BY w.Clinics, w.tran_date
我曾考虑过切换它并从xcharge 表中的比较开始,但没有pay_desc 可以轻松链接它们,并且如果某些内容被处理为错误卡(选择了VISA,但Discover 是扫描)然后我觉得这些行仍然会丢失。
我想要的是,即使 SUM(t.amount) 没有值,所有 pay_desc 也会出现在每个 tran_date 上。我已经尝试了一个案例语句来完成这个没有任何运气。
编辑:我觉得问题更多是由于t.time_ran 变量。如果给定的pay_desc 没有付款,那么也不会有t.time_ran,我可以做一个函数来列出 GETDATE()-10 和 GETDATE() 之间的日期吗?只是让t.time_ran 和x.settle_date 与该变量匹配?
对此有什么想法吗?提前致谢
更新:
我创建了一个包含各个日期的日历表,并尝试了以下查询:
WITH tran_test AS(
SELECT cal.calendardate AS cd,
p.clinic,
p.pay_desc,
p.paytype_id
FROM calendar cal, paytype p
WHERE cal.calendardate BETWEEN GETDATE()-10 AND GETDATE()+1
AND (p.pay_desc = 'Visa' OR
p.pay_desc = 'MasterCard' OR
p.pay_desc = 'American Express' OR
p.pay_desc = 'Discover'))
SELECT w.cd,
CONCAT(c.id_str, ' - ', c.clinic_str) AS Clinics,
w.pay_desc,
ISNULL(SUM(t.amount)*-1, 0) AS collection,
CASE WHEN w.pay_desc = 'Visa' THEN (SUM(x.VI_pur) - SUM(x.VI_ref))
WHEN w.pay_desc = 'MasterCard' THEN (SUM(x.MC_pur) - SUM(x.MC_ref))
WHEN w.pay_desc = 'American Express' THEN (SUM(x.AX_pur) - SUM(x.AX_ref))
WHEN w.pay_desc = 'Discover' THEN (SUM(x.DI_pur) - SUM(x.DI_ref)) END AS xcharge
FROM tran_test w
INNER JOIN clinic_master c
ON (w.clinic=c.clinic)
LEFT OUTER JOIN transactions t
ON (t.clinic=w.clinic AND t.paytype_id=w.paytype_id AND CAST(t.time_ran AS date) = w.cd)
LEFT OUTER JOIN xcharge x
ON (w.cd=x.settle_date AND c.xcharge_mid=x.xcharge_mid)
GROUP BY w.cd, c.id_str, c.clinic_str, w.pay_desc
ORDER BY w.cd
但是,我没有遇到问题,我的 xcharge 列似乎将自身乘以随机间隔,我不知道为什么。
【问题讨论】:
-
感谢您提供链接,但我也尝试了
ISNULL (SUM(t.amount)*-1, 0) AS collections,结果与不显示行的结果相同 -
不,它只是一个有
paytype_id键的表,pay_desc是ID 的描述,例如VISA、Cash、Check、Discover 等。 -
你有两个选择,不要透露哪些列和表有什么
标签: sql sql-server tsql