如何在 Cognos Report Studio 中创建每个重复值的不同计数？
How do I create a distinct count of every duplicate values in Cognos Report Studio?
这些是我目前的数据
ACCOUNT NUMBER PRODUCT STATUS RANK DESIGN
1 530033 Wealth Services Closed 1 Manual
2 530033 Wealth Services Closed 2 Manual
3 530033 Wealth Services Closed 3 Manual
4 530033 Wealth Services Closed 4 Manual
5 534656 Initial Escrow Open 1 Manual
6 534656 Initial Escrow Open 2 Manual
7 535161 Markets Integrity Closed 1 Manual
8 538379 Prepaid Cards Closed 1 Manual
9 538379 Prepaid Cards Closed 2 Manual
10 538379 Prepaid Cards Closed 3 Manual
11 538379 Prepaid Cards Closed 4 Manual
12 538915 Uploaded Cards Open 1 Manual
13 538915 Uploaded Cards Open 2 Manual
14 538915 Uploaded Cards Open 3 Manual
我想创建一个列来唯一地计算每个重复的帐号。比如账号530033出现了四次，我想每530033算一个，同理另一个账号534656出现两次算一个。
我已经尝试了以下但 none 这些工作。
count(distinct(Account Number))
count(distinct(Account Number) for Account Number)
聚合列等于 TOTAL、COUNT 或 AUTOMATIC。
我将使用数据创建交叉表。我的交叉表给了我 14 的总计数，因为它计算了所有重复的帐号。因为所有帐号都有十四行。但实际上只有5个帐号（530033、534656、535161、538379和53891）
TOTAL Open Closed
Wealth Services 4 4
Initial Escrow 2 2
Markets Integrity 1 1
Prepaid Cards 4 4
Uploaded Cards 3 3
TOTAL 14
理想情况下应该是这样的。
TOTAL Open Closed
Wealth Services 1 1
Initial Escrow 1 1
Markets Integrity 1 1
Prepaid Cards 1 1
Uploaded Cards 1 1
TOTAL 5
我应该创建一个名为唯一帐号计数的新列吗？但如何单独计算帐号？
我是否必须创建另一个具有唯一帐号的 table，然后内部加入两个 table？
我相信你需要的表达方式如下：
总计
COUNT(DISTINCT [ACCOUNT NUMBER] for [PRODUCT])
打开
CASE [STATUS]
WHEN 'Open' THEN COUNT(DISTINCT [ACCOUNT NUMBER] for [PRODUCT],[STATUS])
ELSE NULL
END
关闭
CASE [STATUS]
WHEN 'Closed' THEN COUNT(DISTINCT [ACCOUNT NUMBER] for [PRODUCT],[STATUS])
ELSE NULL
END
确保将每个数据项的聚合函数设置为 'Calculated'，因为我们手动指定聚合和汇总。
这些是我目前的数据
ACCOUNT NUMBER PRODUCT STATUS RANK DESIGN
1 530033 Wealth Services Closed 1 Manual
2 530033 Wealth Services Closed 2 Manual
3 530033 Wealth Services Closed 3 Manual
4 530033 Wealth Services Closed 4 Manual
5 534656 Initial Escrow Open 1 Manual
6 534656 Initial Escrow Open 2 Manual
7 535161 Markets Integrity Closed 1 Manual
8 538379 Prepaid Cards Closed 1 Manual
9 538379 Prepaid Cards Closed 2 Manual
10 538379 Prepaid Cards Closed 3 Manual
11 538379 Prepaid Cards Closed 4 Manual
12 538915 Uploaded Cards Open 1 Manual
13 538915 Uploaded Cards Open 2 Manual
14 538915 Uploaded Cards Open 3 Manual
我想创建一个列来唯一地计算每个重复的帐号。比如账号530033出现了四次，我想每530033算一个，同理另一个账号534656出现两次算一个。
我已经尝试了以下但 none 这些工作。
count(distinct(Account Number))
count(distinct(Account Number) for Account Number)
聚合列等于 TOTAL、COUNT 或 AUTOMATIC。
我将使用数据创建交叉表。我的交叉表给了我 14 的总计数，因为它计算了所有重复的帐号。因为所有帐号都有十四行。但实际上只有5个帐号（530033、534656、535161、538379和53891）
TOTAL Open Closed
Wealth Services 4 4
Initial Escrow 2 2
Markets Integrity 1 1
Prepaid Cards 4 4
Uploaded Cards 3 3
TOTAL 14
理想情况下应该是这样的。
TOTAL Open Closed
Wealth Services 1 1
Initial Escrow 1 1
Markets Integrity 1 1
Prepaid Cards 1 1
Uploaded Cards 1 1
TOTAL 5
我应该创建一个名为唯一帐号计数的新列吗？但如何单独计算帐号？
我是否必须创建另一个具有唯一帐号的 table，然后内部加入两个 table？
我相信你需要的表达方式如下：
总计
COUNT(DISTINCT [ACCOUNT NUMBER] for [PRODUCT])
打开
CASE [STATUS]
WHEN 'Open' THEN COUNT(DISTINCT [ACCOUNT NUMBER] for [PRODUCT],[STATUS])
ELSE NULL
END
关闭
CASE [STATUS]
WHEN 'Closed' THEN COUNT(DISTINCT [ACCOUNT NUMBER] for [PRODUCT],[STATUS])
ELSE NULL
END
确保将每个数据项的聚合函数设置为 'Calculated'，因为我们手动指定聚合和汇总。