Having touble getting only values that appear once. I currently have some sql code that gets out the all the entries that have 0 percent. The problem is that two rows can contain the same person With different percentages. If one of these is above 0 then i dont want it to come out in the Query
abridged table:
Name - Percent
steve 0
dan 0
mike 100
harold 50
steve 80
carl 0
carl 0
Result:
dan - 0
Carl - 0
Here is how far ive gotten, but not managed to make any variation of Count() or having or Group by working.
select person, Value2, Value3, Value4, percent
from
Table1
INNER JOIN Table1 ON Table2.valueNum = Table1.valueNum
INNER JOINTable1 ON Table3.valueNum = Table1.valueNum
INNER JOIN Table1 ON Table4.valueNum = Table1.valueNum
WHERE
(#date BETWEEN table1.FROMDATE AND table1.todate)
AND table1.percent = 0
AND table1.varchar IN ('T', 'X')
This is one method
select name,0 as percent from abridged
group by name
having min(percent)=0 and max(percent)=0
Your example SQL and abridged table don't match. However, this looks like the basic idea you are after:
select
*
from
dbo.table a
where
a.percent = 0 and
not exists (
select
'x'
from
dbo.table b
where
a.Name = b.Name and
b.percent > 0
);
I would just use window functions:
with t as (
<your query here>
)
select t.*
from (select t.*, count(*) over (partition by name) as cnt
from t
) t
where cnt = 1;
To resolve your problem you can use this query:
select [name], 0 as [percent] from [abridged]
group by [name]
having sum([percent])=0
This should solve your problem right?
select name,sum([percent]) from abridged
group by name
having SUM([percent]) = 0
Query
SELECT distinct name, min(percentage) from a
group by name
having min(percentage) = 0
and count(*) > 1;
Related
I am writing a large SQL query and have narrowed down the problem to the following (simple example).
id
status
1
-1
2
1
3
-1
3
1
Now I want to filter out all ids which never have a status of 1. In this example, I only want to return id 1 since id 3 both can have status 1 and -1. How can I do that?
Not tested but this should do the trick.
select *
from yourTable T
where not exists(
select 1
from yourTable X
where X.id = T.id
and X.status = 1
)
(sorry, edited: you don't want the records having -1 but 1)
This might help
select id from my_table
group by id
having sum(status) = -count(id)
You can use sub query to filter:
SELECT *
FROM YourTable YT
WHERE id NOT IN (
SELECT DISTINCT X.id FROM YourTable X WHERE X.status = 1
)
I have a table called tbl_A with the following schema:
After insert, I have the following data in tbl_A:
Now the question is how to write a query for the following scenario:
Put (1) in front of any employee who was present three days consecutively
Put (0) in front of employee who was not present three days consecutively
The output screen shoot:
I think we should use case statement, but I am not able to check three consecutive days from date. I hope I am helped in this
Thank you
select name, case when max(cons_days) >= 3 then 1 else 0 end as presence
from (
select name, count(*) as cons_days
from tbl_A, (values (0),(1),(2)) as a(dd)
group by name, adate + dd
)x
group by name
With a self-join on name and available = 'Y', we create an inner table with different combinations of dates for a given name and take a count of those entries in which the dates of the two instances of the table are less than 2 units apart i.e. for each value of a date adate, it will check for entries with its own value adate as well as adate + 1 and adate + 2. If all 3 entries are present, the count will be 3 and you will have a flag with value 1 for such names(this is done in the outer query). Try the below query:
SELECT Z.NAME,
CASE WHEN Z.CONSEQ_AVAIL >= 3 THEN 1 ELSE 0 END AS YOUR_FLAG
FROM
(
SELECT A.NAME,
SUM(CASE WHEN B.ADATE >= A.ADATE AND B.ADATE <= A.ADATE + 2 THEN 1 ELSE 0 END) AS CONSEQ_AVAIL
FROM
TABL_A A INNER JOIN TABL_A B
ON A.NAME = B.NAME AND A.AVAILABLE = 'Y' AND B.AVAILABLE = 'Y'
GROUP BY A.NAME
) Z;
Due to the complexity of the problem, I have not been able to test it out. If something is really wrong, please let me know and I will be happy to take down my answer.
--Below is My Approch
select Name,
Case WHen Max_Count>=3 Then 1 else 0 end as Presence
from
(
Select Name,MAx(Coun) as Max_Count
from
(
select Name, (count(*) over (partition by Name,Ref_Date)) as Coun from
(
select Name,adate + row_number() over (partition by Name order by Adate desc) as Ref_Date
from temp
where available='Y'
)
) group by Name
);
select name as employee , case when sum(diff) > =3 then 1 else 0 end as presence
from
(select id, name, Available,Adate, lead(Adate,1) over(order by name) as lead,
case when datediff(day, Adate,lead(Adate,1) over(order by name)) = 1 then 1 else 0 end as diff
from table_A
where Available = 'Y') A
group by name;
I'll cut right to the chase: I have a select I'm currently writing with a rather lengthy where clause, what I want to do is calculate percentages.
So what I need is the count of all results and then my each distinct counts.
SELECT distinct count(*)
FROM mytable
WHERE mywhereclause
ORDER BY columnIuseInWhereClause
works fine for getting each individual values, but I want to avoid doing something like
Select (Select count(*) from mytable WHERE mywhereclause),
distinct count(*)
FROM mytable
WHERE mywhereclause
because I'd be using the same where-clause twice which just seems unnecessary.
This is for OracleDB but I'm only using standard SQL syntax, nothing database specific if I can help it.
Thanks for any ideas.
Edit:
Sample Data
__ID__,__someValue__
1 | A
2 | A
3 | B
4 | C
I want the occurances of A, B, C as numbers as well as the overall count.
__CountAll__,__ACounts__,__BCounts__,__CCounts__
4 | 2 | 1 | 1
So I can get to
100% | 50% | 25% | 25%
That last part I can probably figure out on my own. Excuse my lack of experience or even logic thinking, it's early in the morning. ;)
Edit2:
I do have written a query that works but is clumsy and long as all holy heck, this one is for trying with group by.
Try:
select count(*) as CountAll,
count(distinct SomeColumn) as CoundDistinct -- The DISTINCT goes inside the brackets
from myTable
where SomeOtherColumn = 'Something'
Use case expressions to do conditional counting:
select count(*) as CountAll,
count(case when someValue = 'A' then 1 end) as ACounts,
count(case when someValue = 'B' then 1 end) as BCounts,
count(case when someValue = 'C' then 1 end) as CCounts
FROM mytable
WHERE mywhereclause
Wrap it up in a derived table to do the % part easy:
select 100,
ACounts * 100 / CountAll,
BCounts * 100 / CountAll,
CCounts * 100 / CountAll
from
(
select count(*) as CountAll,
count(case when someValue = 'A' then 1 end) as ACounts,
count(case when someValue = 'B' then 1 end) as BCounts,
count(case when someValue = 'C' then 1 end) as CCounts
FROM mytable
WHERE mywhereclause
) dt
Here's an alternative using window function:
with data_table(ID, some_value)
AS
(SELECT 1,'A' UNION ALL
SELECT 2,'A' UNION ALL
SELECT 3,'B' UNION ALL
SELECT 4,'C'
)
SELECT DISTINCT [some_value],
COUNT([some_value]) OVER () AS Count_All,
COUNT([some_value]) OVER (PARTITION BY [some_value]) AS 'Counts' FROM [data_table]
ORDER BY [some_value]
The advantage is that you don't have to hard-code your [some_value]
I've the following table with 3 columns: Id, FeatureName and Value:
Id FeatureName Value
-- ----------- -----
1 AAA 10
1 ABB 12
1 BBB 12
2 AAA 15
2 ABB 12
2 ACD 7
3 AAA 10
3 ABB 12
3 CCC 12
.............
Each Id has different features and each Feature has a value for that Id.
I need to write a query which gives me the Ids that have exactly the same features and values than a given one, but only taking into account those whose name starts with 'A'. For example, in the top table, I can use that query to search for all the Ids that have the same features. For example, features with values where Id=1 would result Id=3 with same features starting with 'A' and same values for these features.
I found a couple of different ways to do this, but all of them go very slow when the table has lots of rows (more than hundred of thousands)
The way I obtain the best performance is using the next query:
select a2.Id
from (select a.FeatureName, a.Value
from Table1 a
where a.Id = 1) a1,
(select a.Id, a.FeatureName, a.Value
from Table1 a
where a.FeatureName like 'A%') a2
where a1.FeatureName = a2.FeatureName
and a1.value = a2.value
group by a2.Id
having count(*) = 2
intersect
select a.Id
from Table1 a
where a.FeatureName like 'A%'
group by a.Id
having count(*)= 2
where #nFeatures is the number of features starting by 'A' in Id=1. I counted them before calling this query. I make the intersection to avoid results that have the same parameters than Id=1 but also some others whose name starts with 'A'.
I think that the slowest part is the second subquery:
select a.Id, a.FeaureName, a.Value
from MyTable a
where a.FeatureName = 'A%'
but I don't know how to make it faster. Maybe I will have to play with the indexes.
Any idea of how could I write a fast query for this purpose?
So you want all rows where the combination of FeatureName and Value is not unique? You can use EXISTS:
SELECT t.*
FROM dbo.Table1 t
WHERE t.FeatureName LIKE 'A%'
AND EXISTS(SELECT 1 FROM dbo.Table1 t2
WHERE t.Id <> t2.ID
AND t.FeatureName = t2.FeatureName
AND t.Value = t2.Value)
Demo
how could I write a fast query for this purpose?
If it's not fast enough create an index on FeatureName + Value.
I tried to eliminate the join with MyTable again to select the data for the ID's that have matching FeatureName and Value values. Here's the query:
with joined_set as
(
SELECT
mt1.*, mt2.id as mt2_id, mt2.featurename as mt2_FeatureName, mt2.value as mt2_value
from
(
select *
from mytable
where featurename like 'A%'
) mt1
left join
(
select *
from mytable
where featurename like 'A%'
) mt2
on mt2.id <> mt1.id and mt2.FeatureName = mt1.featurename and mt2.value = mt1.value
)
select distinct id
from joined_set
where id not in
(select id
from joined_set
group by id
having SUM(
CASE
WHEN mt2_id is null THEN 1
ELSE 0
END
) <> 0
);
Here is the SQL Fiddle demo. It has an extra condition in the inline view mt2, to perform this search only for id = 1.
I'm a little dense this morning, I'm not sure if you wanted just the ID's or...
Here's my take on it...
You could probably move the where FeatureName like 'A%' into the inner query to filter the data on the initial table scan.
with dupFeatures (FeatureName, Value, dupCount)
as
(
select FeatureName, Value, count(*) as dupCount from MyTable
group by FeatureName, Value
having count(*) > 1
)
select MyTable.Id, dupFeatures.FeatureName,dupFeatures.Value
from dupFeatures
join MyTable on (MyTable.FeatureName = dupFeatures.FeatureName and
MyTable.Value = dupFeatures.Value )
where dupFeatures.FeatureName like 'A%'
order by FeatureName, Value, Id
A general solution is
With Rows As (
select id
, FeatureName
, Value
, rows = Count(id) OVER (PARTITION BY id)
FROM test
WHERE FeatureName LIKE 'A%')
SELECT a.id aID, b.id bID
FROM Rows a
INNER JOIN Rows b ON a.id < b.id and a.FeatureName = b.FeatureName
and a.rows = b.rows
GROUP BY a.id, b.id
ORDER BY a.id, b.id
to limit the solution to a group just add a WHERE condition on the main query for a.ID. The CTE is needed to get the correct number of rows for each id
SQLFiddle demo, in the demo I changed little the test data to have a another couple of ID with only one of the FeatureName of 1 and 3
I don't quite know how I should describe the problem for title, but here's my question.
I have a table named hello with two columns named time and state.
Time | State
Here's an example of the data I have
1 DC
1 VA
1 VA
2 DC
2 MD
3 MD
3 MD
3 VA
3 DC
I would like to get all the possible time and the count of "VA" (0 if "VA" doesn't appear at the time)
The output would look like this
Time Number
1 2
2 0
3 1
I tried to do
SELECT DISTINCT time,
COUNT(state) as Number
FROM hello
WHERE state = 'VA'
GROUP BY time
but it doesn't seem to work.
This is a conditional aggregation:
select time, sum(case when state = 'VA' then 1 else 0 end) as NumVA
from hello
group by time
I want to add that you should never use distinct when you have a group by. The two are redundant. Distinct as a keyword is not even needed in the SQL language; semantically, it is just shorthand for grouping by all the columns.
SELECT TIME,
SUM(CASE WHEN State = 'VA' THEN 1 ELSE 0 END)
FROm tableName
GROUP BY Time
SQLFiddle Demo
One rule of thumb is to get your counts first and put them into a temp for use later.
See below:
Create table temp(Num int, [state] varchar(2))
Insert into temp(Num,[state])
Select 1,'DC'
UNION ALL
Select 1,'VA'
UNION ALL
Select 1,'VA'
UNION ALL
Select 2,'DC'
UNION ALL
Select 2,'MD'
UNION ALL
Select 3,'MD'
UNION All
Select 3,'MD'
UNION ALL
Select 3,'VA'
UNION ALL
Select 3,'DC'
Select t.Num [Time],t.[State]
, CASE WHEN t.[state] = 'VA' THEN Count(t.[State]) ELSE 0 END [Number]
INTO #temp2
From temp t
Group by t.Num, t.[state]
--drop table #temp2
Select
t2.[time]
,SUM(t2.[Number])
From #temp2 t2
group by t2.[time]