Re: SQl help to build a result with custom aliased bool column

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От Arup Rakshit
Тема Re: SQl help to build a result with custom aliased bool column
Дата
Msg-id DA14E90F-BA29-42F4-AA3B-8D17D889348B@zeit.io
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Ответ на Re: SQl help to build a result with custom aliased bool column  (Szymon Lipiński <mabewlun@gmail.com>)
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I knew that will be more compact way. Thanks for showing it. One thing I still would like to handle is that, to make
surethe column contains only True/False. But right now sometimes it shows NULL. How can I fix this? 

id|name|active|
--|----|------|
 1|f1  |true  |
 2|f2  |true  |
 3|f3  |false |
 4|f4  |      |


Thanks,

Arup Rakshit
ar@zeit.io



> On 08-Apr-2019, at 3:28 PM, Szymon Lipiński <mabewlun@gmail.com> wrote:
>
> Hey,
> you could just use
>
> SELECT
>         features.id,
>         features.name,
>         company_features.company_id = 1 as active
>
> regards,
> Szymon
>
> On Mon, 8 Apr 2019 at 09:55, Arup Rakshit <ar@zeit.io> wrote:
> I have 2 tables Company and Feature. They are connected via a join table called CompanyFeature. I want to build a
resultset where it will have id, name and a custom boolean column. This boolean column is there to say if the feature
ispresent for the company or not. 
>
> Company table:
>
>     | id | name |
>     |----|------|
>     | 1  | c1   |
>     | 2  | c2   |
>     | 3  | c3   |
>
> Feature table:
>
>     | id | name |
>     |----|------|
>     | 1  | f1   |
>     | 2  | f2   |
>     | 3  | f3   |
>
> Company Feature table:
>
>     | id | feature_id | company_id |
>     |----|------------|------------|
>     | 1  | 2          | 1          |
>     | 2  | 1          | 1          |
>     | 3  | 3          | 2          |
>
> The result should look like for company `c1`:
>
>     | id | name | active |
>     |----|------|--------|
>     | 1  | f1   | t      |
>     | 2  | f2   | t      |
>     | 3  | f3   | f      |
>
> I tried something like:
>
> SELECT
>         features.id,
>         features.name,
>         CASE WHEN company_features.company_id = 1 THEN
>                 TRUE
>         ELSE
>                 FALSE
>         END AS active
> FROM
>         features
>         LEFT JOIN company_features ON company_features.feature_id = features.id
>
> It works. But is there any better way to achieve this?
>
>
> Thanks,
>
> Arup Rakshit
> ar@zeit.io
>
>
>
>
>




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