I want to compare data of a table with its other records. The count of rows with a specific condition has to match the count of rows without the where clause but on the same grouping.
Below is the table
-------------
id name time status
1 John 10 C
2 Alex 10 R
3 Dan 10 C
4 Tim 11 C
5 Tom 11 C
Output should be time = 11 as the count for grouping on time column is different when a where clause is added on status = 'C'
SELECT q1.time
FROM (SELECT time,
Count(id)
FROM table
GROUP BY time) AS q1
INNER JOIN (SELECT time,
Count(id)
FROM table
WHERE status = 'C'
GROUP BY time) AS q2
ON q1.time = q2.time
WHERE q1.count = q2.count
This is giving the desired output but is there a better and efficient way to get the desired result?
Are you looking for this :
select t.*
from table t
where not exists (select 1 from table t1 where t1.time = t.time and t1.status <> 'C');
However you can do :
select time
from table t
group by time
having sum (case when status <> 'c' then 1 else 0 end ) = 0;
If you want the times where the rows all satisfy the where clause, then in Postgres, you can express this as:
select time
from t
group by time
having count(*) = count(*) filter (where status = 'C');
Related
I have a simple table of values:
column1
-------
2
5
7
5
8
7
and this simple query of number count:
SELECT column1, count(column1) as counter
FROM table
GROUP BY column1
ORDER BY count(column1) DESC
The question how can I add rows with values 0 when I have a number range for example from 1 to 8.
I want to get the result like this:
column1 Counter
-------- -------
5 2
7 2
2 1
8 1
1 0 <-- Row Add
2 0 <-- Row Add
3 0 <-- Row Add
4 0 <-- Row Add
6 0 <-- Row Add
Thanks very much.
If you are willing to use more than 1 query and to put your range of numbers in a table at least temporarily then you can do this with a full outer join. To do a Full outer join in access you combine a left join and a right join using either UNION OR UNION ALL.
copy the sql from the right and left joins into a union query then add the union and the order by statements to the union query.
SELECT SpecialNumbers.Numbers, mytable.Column1, Count(mytable.Column1) AS CountOfColumn1
FROM SpecialNumbers LEFT JOIN mytable ON SpecialNumbers.Numbers = mytable.Column1
GROUP BY SpecialNumbers.Numbers, mytable.Column1
UNION
SELECT SpecialNumbers.Numbers, mytable.Column1, Count(mytable.Column1) AS CountOfColumn1
FROM SpecialNumbers RIGHT JOIN mytable ON SpecialNumbers.Numbers = mytable.Column1
GROUP BY SpecialNumbers.Numbers, mytable.Column1
ORDER BY CountOfColumn1 DESC;
Given my special numbers where 1 to 10 This turns
to
Assuming that you only wish the four missing rows added, this is not that difficult in Access, as it is easy to create a small query generating numbers from a system table:
SELECT
T.Column1,
Count(Table.[Column1]) AS [Counter]
FROM
(SELECT DISTINCT Abs([id] Mod 10) AS Column1
FROM MSysObjects
WHERE (Abs([id] Mod 10)) Between 1 And 8) As T
LEFT JOIN
[Table]
ON T.Column1 = Table.Column1
GROUP BY
T.Column1
ORDER BY
Count(Table.[Column1]) DESC,
T.Column1;
Output:
My goal is something like following table:
Key | Count since date X | Count total
1 | 4 | 28
With two simple selects I could gain this values: (the key of the table consists of 3 columns [t$ncmp, t$trav, t$seqn])
1. SELECT COUNT(*) FROM db.table WHERE t$date >= sysdate-2 GROUP BY t$ncmp, t$trav, t$seqn
2. SELECT COUNT(*) FROM db.table GROUP BY t$ncmp, t$trav, t$seqn
How can I join these statements?
What I tried:
SELECT n.t$trav, COUNT(n.t$trav), m.total FROM db.table n
LEFT JOIN (SELECT t$ncmp, t$trav, t$seqn, COUNT(*) as total FROM db.table
GROUP BY t$ncmp, t$trav, t$seqn) m
ON (n.t$ncmp = m.t$ncmp AND n.t$trav = m.t$trav AND n.t$seqn = m.t$seqn)
WHERE n.t$date >= sysdate-2
GROUP BY n.t$ncmp, n.t$trav, n.t$seqn
I tried different variantes, but always got errors like 'group by is missing' or 'unknown qualifier'.
Now this at least executes, but total is always 2.
T$TRAV COUNT(N.T$TRAV) TOTAL
4 2 2
29 3 2
51 1 2
62 2 2
16 1 2
....
If it matter, I will run this as an OPENQUERY from MSSQLSERVER to Oracle-DB.
I'd try
GROUP BY n.t$trav, m.total
You typically GROUP BY the same columns as you SELECT - except those who are arguments to set functions.
My goal is something like following table:
If so, you seem to want conditional aggregation:
select key, count(*) as total,
sum(case when datecol >= date 'xxxx-xx-xx' then 1 else 0 end) as total_since_x
from t
group by key;
I'm not sure how this relates to your sample queries. I simply don't see the relationship between that code and your question.
I'm looking to filter out rows in the database (PostgreSQL) if one of the values in the status column occurs. The idea is to sum the amount column if the unique reference only has a status equals to 1. The query should not SELECT the reference at all if it has also a status of 2 or any other status for that matter. status refers to the state of the transaction.
Current data table:
reference | amount | status
1 100 1
2 120 1
2 -120 2
3 200 1
3 -200 2
4 450 1
Result:
amount | status
550 1
I've simplified the data example but I think it gives a good idea of what I'm looking for.
I'm unsuccessful in selecting only references that only have status 1.
I've tried sub-queries, using the HAVING clause and other methods without success.
Thanks
Here's a way using not exists to sum all rows where the status is 1 and other rows with the same reference and a non 1 status do not exist.
select sum(amount) from mytable t1
where status = 1
and not exists (
select 1 from mytable t2
where t2.reference = t1.reference
and t2.status <> 1
)
SELECT SUM(amount)
FROM table
WHERE reference NOT IN (
SELECT reference
FROM table
WHERE status<>1
)
The subquery SELECTs all references that must be excluded, then the main query sums everything except them
select sum (amount) as amount
from (
select sum(amount) as amount
from t
group by reference
having not bool_or(status <> 1)
) s;
amount
--------
550
You could use windowed functions to count occurences of status different than 1 per each group:
SELECT SUM(amount) AS amount
FROM (SELECT *,COUNT(*) FILTER(WHERE status<>1) OVER(PARTITION BY reference) cnt
FROM tc) AS sub
WHERE cnt = 0;
Rextester Demo
I have a table that has the below columns.
I need to find out those people that has More than 2 ApplicantRowid with same jobcategoryrowid and AssessmentTest should have atleast one row NULL with Different Appstatusrowid's.
The result should look exeactly like the below table.
Rowid ApplicantRowid JobCategoryRowid AssessmentTestRowid AppstatusRowid
10770598 6952346 157 3 5
11619676 6952346 157 NULL 6
select t.*
from
(
select ApplicantRowid, JobCategoryRowid
from tbl
group by ApplicantRowid, JobCategoryRowid
having count(AssessmentTestRowid) < count(*)
and count(distinct AppstatusRowid) > 1
) x
inner join t on t.ApplicantRowid = x.ApplicantRowid
and t.JobCategoryRowid = x.JobCategoryRowid
COUNT does not include NULLs, so count(AssessmentTestRowid) < count(*) ensures there is at least a NULL
count(distinct AppstatusRowid) > 1 ensure there are different AppstatusRowids
Suppose I have a table with 2 columns (status and date) like the following:
status: U T U U L
date: 12 14 15 16 17
Can I (using only 1 SQL statement) count the number of distinct values in the status? That is:
count(U)=3
count(T)=1
count(L)=2
count(P)=0
Can I do this with 1 SQL query?
Note: I have static values in status. I can only have (U-T-L-P)
You need to use Group By:
SELECT Status, Count(Status)
FROM table
GROUP BY Status
This will not return P = 0 if P is not populated in the table. In your application logic you will need to check and if a certain status is not returned, it means there are no entries (i.e. 0).
SQL cannot query records that are not there.
This will return a row for every status and the count in the second column:
SELECT Status, COUNT(*) Cnt
FROM Tbl
GROUP BY Status
So it would return
Status Cnt
U 3
T 1
L 1
for your example (in no defined order). Use ORDER BY if you want to sort the results.
You can do this with a query which groups on your status column, e.g.
SELECT COUNT(*) as StatusCount, Status
FROM MyTable
GROUP BY Status
To get the zero for the status P, you have to do some devious stuff using a table that lists all the possible statuses.
SELECT COUNT(A.Status), B.Status
FROM AnonymousTable AS A RIGHT OUTER JOIN
(SELECT 'P' AS Status FROM Dual
UNION
SELECT 'U' AS Status FROM Dual
UNION
SELECT 'L' AS Status FROM Dual
UNION
SELECT 'T' AS Status FROM Dual
) AS B ON A.Status = B.Status
GROUP BY B.Status;
The 4-way UNION is one way of generating a list of values; your DBMS may provide more compact alternatives. I'm assuming that the table Dual contains just one row (as found in Oracle).
The COUNT(A.Status) counts the number of non-null values in A.Status. The RIGHT OUTER JOIN lists the row from B with Status = 'P' and joins it with a single NULL for the A.Status, which the COUNT(A.Status) therefore counts as zero. If you used COUNT(*), you'd get a 1 for the count.