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Getting a BIT value as to whether a non-Aggregated Value is in a constant list


Getting a BIT value as to whether a non-Aggregated Value is in a constant list

By : GautiGotam
Date : November 22 2020, 04:01 AM
fixed the issue. Will look into that further The Logs and SuperLogs tables have a one-to-many relationship. This means that for each record in SuperLogs there can be many records in Logs. This is why your query doesn't make sense and can't work as is.
The way I understand the question is this: If Any Log that belongs to the current SuperLog have a ScanMode that is either one of the values in the list, you want to get 1. If none of them have a ScanMode that fits the list, you want to get 0.
code :
SELECT Name,
CAST
(
    MAX(
        CASE WHEN
        Logs.ScanMode IN
        (
            'Navigational',
            'Locatable',
            'Automatic'
        )
    THEN
        1
    ELSE
        0
    END
    )
    AS BIT
) AS HasLogsWithGpsData
FROM SuperLogs
INNER JOIN Logs ON Logs.SuperLogId = SuperLogs.Id
GROUP BY SuperLogs.Id


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Pivot columns with aggregated function and calculate total of aggregated value in last column

Pivot columns with aggregated function and calculate total of aggregated value in last column


By : Raymond Colletti
Date : March 29 2020, 07:55 AM
it helps some times Hi my source data is like, , Try this query
code :
select a.error 'Error Type',a.Total,sum(b.Total) 'Cumulative',convert(varchar(25),a.Total)+'/'+convert(varchar(25),sum(a.total) over()) from(
select 1 as id, 'E1' as Error ,sum(E1) as total
from tablename
Union all
select 2 as id, 'E2' as Error ,sum(E2) as total
from tablename
Union all
select 3 as id, 'E3' as Error ,sum(E3) as total
from tablename
Union all
select 4 as id, 'E4' as Error ,sum(E4) as total
from tablename)a
join (select 1 as id, 'E1' as Error ,sum(E1) as total
from tablename
Union all
select 2 as id, 'E2' as Error ,sum(E2) as total
from tablename
Union all
select 3 as id, 'E3' as Error ,sum(E3) as total
from tablename
Union all
select 4 as id, 'E4' as Error ,sum(E4) as total
from tablename)b on b.id<=a.id
how to access different aggregated columns of aggregated data frame r

how to access different aggregated columns of aggregated data frame r


By : Salam Dulaimi
Date : March 29 2020, 07:55 AM
wish help you to fix your issue The answer is here: R: results from aggregate with multiple functions not usable in further calculations. WHY?
Once I aggregate, I convert to a list and then a data.frame
code :
> as.data.frame(as.list(a))
  Group.1 x.Min. x.1st.Qu. x.Median x.Mean x.3rd.Qu. x.Max. x.V7      x.V8
1       a  29.76     30.17    30.56  30.38     30.60  30.83    5 0.4244106
2       b  27.54     28.79    29.48  29.27     29.97  30.60    4 1.2884043
> dim(as.data.frame(as.list(a)))
[1] 2 9
> 
how to get distinct non-aggregated columns along with aggregated column when group by clause is applied in oracle?

how to get distinct non-aggregated columns along with aggregated column when group by clause is applied in oracle?


By : TAC
Date : March 29 2020, 07:55 AM
may help you . DISTINCT is not an operation that is applied to a single column - it applies to all columns and does not make sense in the context of an aggregation.
code :
col1(sum) col2
2         '...'
SELECT SUM( col1 ),
       CASE
       WHEN COUNT( DISTINCT col2 ) = 1
       THEN TO_CHAR( MAX( col2 ) )
       ELSE '...'
       END
FROM   your_table
GROUP BY col1
Compare aggregated value for groups to the aggregated value for the entire table

Compare aggregated value for groups to the aggregated value for the entire table


By : user2094107
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Let's say I have one table that has E_name, DepartmentName, Salary , You may use GROUP BY with a HAVING clause
code :
select DepartmentName  
  from employees
 group by DepartmentName
 having avg(Salary) > ( select avg(salary) from employees );
Keep the aggregated column constant in SQL Server?

Keep the aggregated column constant in SQL Server?


By : Del
Date : March 29 2020, 07:55 AM
it should still fix some issue I am trying to do count of client locations that are count > 1, which seems to work fine as indicated in the first query. , You could use windowed SUM per clientid
code :
select reveneueid,clientID,
  SUM(count(clientlocations)) OVER(PARITION BY clientid) AS clientcount
from db.schema.table
group by revenueid,clientid
having count(clientid) > 1
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