Python 如果列的值为零值，如何从SQL查询中获取值？

基本上我有一个训练数据集，但我希望它显示为零，而不是根本不显示

下面的节目都是不完整和完整的，但我希望它显示不完整，并且来自于零不完整为0的部门

这就是我到目前为止所拥有的，我需要它来展示一切，我的同事试图帮助我，但不得不离开，所以我请求你们

Select distinct Department, Total FROM
(
Select Department, Total FROM

(
SELECT 'Incomplete' AS Status, department, count(*) as Total
FROM MyTable
WHERE CompletedTraining = 'Incomplete'
GROUP BY department
union all
Select 'Complete' AS Status, department, Count(*) as Total
FROM MyTable
WHERE CompletedTraining = ' Complete'
Group By Department
)
WHERE Status = 'Incomplete'
)
union all
SELECT DISTINCT Department, NULL AS Total
FROM MyTable

当我运行python脚本时，我得到了这个

Incomplete Training
[6, 8, 2, 3, 6, 4, 4, 5, 2, 4]

但是如果有人接受了我的训练，我希望在名单上显示为零 所以我的预期输出是这样的

Incomplete Training
[6, 8, 2, 3, 6, 4, 4, 5, 2, 4, 0, 0]

示例文档X 1500行，包含多个用户

Curriculum Name     Department          Employee Name           Employee Email  Employee Status Date Assigned   Completion Date CompletedTraining
Security Training   OIS John Doe        john.doe@email.org      Active  7/18/2017 21:47 8/2/2017 21:31                          Complete
Support Training    OIS Home Simpson    home.simpson@email.gov  Active  4/20/2017 15:33 5/3/2017 22:18                          Complete
Security Training   ASD bart simpson    bart.simpson@email.gove Active  5/5/2017 20:22  11/30/2017 19:43                        incomplete
Security Training   CO  jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   ECARS    jack johnson   jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   EO  jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   ISD jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   MSCD    jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   RD  jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   TTD jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   DP  jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   MLD jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   OIS jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   TTD jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   TTD jack johnson    jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete
Security Training   AQPSD jack johnson  jack.johnson@email.gov  Active  5/9/2017 21:15  5/10/2017 20:23                         incomplete

---

这会让你得到不完整的

在同一行上获得不完整和完成

select distinct 
    Department
    , (select count(*) from MyTable t2 where t2.Department = t1.Department and t2.CompletedTraining='Incomplete') as TotalIncomplete
    , (select count(*) from MyTable t2 where t2.Department = t1.Department and t2.CompletedTraining='Complete') as TotalComplete
from MyTable t1

或者类似的

select distinct 'Incomplete' as Src
    , Department
    , (select count(*) from MyTable t2 where t2.Department = t1.Department and t2.CompletedTraining='Incomplete') as Total
from MyTable t1
union
select distinct 'Complete' as Src
    , Department
    , (select count(*) from MyTable t2 where t2.Department = t1.Department and t2.CompletedTraining='Completeomplete') as Total
from MyTable t1

---

这会让你得到不完整的

在同一行上获得不完整和完成

select distinct 
    Department
    , (select count(*) from MyTable t2 where t2.Department = t1.Department and t2.CompletedTraining='Incomplete') as TotalIncomplete
    , (select count(*) from MyTable t2 where t2.Department = t1.Department and t2.CompletedTraining='Complete') as TotalComplete
from MyTable t1

或者类似的

select distinct 'Incomplete' as Src
    , Department
    , (select count(*) from MyTable t2 where t2.Department = t1.Department and t2.CompletedTraining='Incomplete') as Total
from MyTable t1
union
select distinct 'Complete' as Src
    , Department
    , (select count(*) from MyTable t2 where t2.Department = t1.Department and t2.CompletedTraining='Completeomplete') as Total
from MyTable t1

---

---

你能给出示例表吗？更新后，基本上我需要我的查询来输出所有部门，总共有13个部门。从这4行的示例数据中，你会期望OIS 0、ASD 1、TTD1？我总共有13个部门，我应该在示例中编辑吗？检查我的答案是否有不完整之处。我不确定它需要所有的工会。我会注意的。你能给出样本表吗。更新后，基本上我需要我的查询来输出所有部门，总共有13个部门。从这4行的样本数据中，你会期望OIS 0，ASD 1，TTD1？我总共有13个部门，我应该在示例中编辑吗？检查我的答案是否有不完整之处。我不确定它需要所有的工会。我会留意的。如果可能的话，我不想依赖部门名称，我会试试这个很棒的，我想我也可以用它来完成训练，只需更改completedtraining='complete'？那么completedtraining呢？我试着将“不完整”改为“完整”，但没有成功…更新了答案，gl.Hmm…，对“完整”没有任何结果我不想依赖部门名称，如果可能的话，我会试试这个很棒的，我想我也可以用它来完成训练，只需更改completedtraining='complete'？那么completedtraining呢？我试着将“不完整”改为“完整”，但它不起作用…更新了答案，gl.Hmm…，它没有为“完整”生成任何结果