In SQL Server I am able to create a query that uses both Top and Distinct in the Select clause, such as this one:
Select Distinct Top 10 program_name
From sampleTable
Will the database return the distinct values from the top 10 results, or will it return the top 10 results of the distinct values? Is this behavior consistent in SQL or is it database dependent?
TOP is executed last, so your DISTINCT runs first then the TOP
http://blog.sqlauthority.com/2009/04/06/sql-server-logical-query-processing-phases-order-of-statement-execution/
Use
Select Top 10 program_name
From sampleTable group by program_name;
It will return you the top 10 distinct program_name.
Your query will also return the distinct 10 program_name.
Try this:
select distinct top 10 c from
(
select 1 c union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 2
) as T
order by c
Compare that result to these queries:
select distinct c from (
select top 10 c from
(
select 1 c union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 2
) as T
order by c
) as T2
select top 10 c from (
select distinct c from
(
select 1 c union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 2
) as T
) as T2
order by c
Related
I have this query that needs to be executed for oracle sql instead of mysql which is where it originally came from, but I have the ADDDATE() function which I don't see any other alternative than DateAdd since it needs more parameters than I really need..
Apart from that, if I try to execute it, it also indicates an error in the
SELECT 0 i UNION.................
part, saying the following ORA-00923: FROM keyword not found where expected
Maybe in oracle it is not allowed to do a select 0 union select 1 union...
Any suggestions or help I appreciate it, thanks
SELECT
ADDDATE('1970-01-01', t4.i * 10000 + t3.i * 1000 + t2.i * 100 + t1.i * 10 + t0.i) selected_date
FROM
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t0,
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t1,
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t2,
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t3,
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t4
In Oracle you must select from the one-row table dual in order to select one row. You cannot select without a from clause.
If you want to generate dates, you'll write a standard SQL recursive CTE. (And this is the typical approach now in MySQL, too, since version 8.0.)
Here is an example selecting all days for 1970:
with dates (dt) as
(
select date '1970-01-01' from dual
union all
select dt + interval '1' day from dates where dt < date '1970-12-31'
)
select dt from dates;
Here is another way to SELECT a list of dates for the year 1970. Adjust the starting and ending dates if you want different years or the INTERVAL if you want different periods like seconds, minutes, hours…
ALTER SESSION SET NLS_DATE_FORMAT = 'DD-MON-YYYY HH24:MI:SS';
with dt (dt, interv) as (
select date '1970-01-01', numtodsinterval(1,'DAY') from dual
union all
select dt.dt + interv, interv from dt
where dt.dt + interv <= date '1970-12-31')
select dt from dt;
/
In SQLite, if I type:
SELECT (SELECT 1 UNION SELECT 2 UNION SELECT 3) INTERSECT SELECT 3 UNION SELECT 4
I get the result 4. How is that possible?
SELECT 1 UNION SELECT 2 SELECT 3 is (1, 2, 3), right? And SELECT 3 UNION SELECT 4 is (3, 4). So, the intersect should be 3, right? What am I getting wrong?
EDIT: Saying that INTERSECT is evaluated first does not answer my question, as ((1,2,3) INTERSECT (3)) UNION (4) is (3,4), rather than 4.
If you write your statement like this:
SELECT (SELECT 1 UNION SELECT 2 UNION SELECT 3)
INTERSECT
SELECT 3
UNION
SELECT 4
you can see that you are combining 3 SELECT statements with the operators UNION and INTERSECT.
All 3 statements should return the same number of columns.
Your 1st statement:
SELECT (SELECT 1 UNION SELECT 2 UNION SELECT 3)
actually returns only 1 row with 1 column, try it, which is the 1st row and the result is 1.
So your code is equivalent to:
SELECT 1
INTERSECT
SELECT 3
UNION
SELECT 4
which returns nothing for INTERSECT and finally UNION returns 4.
If you meant to write:
SELECT * FROM (SELECT 1 UNION SELECT 2 UNION SELECT 3)
INTERSECT
SELECT 3
UNION
SELECT 4
then the result would be (3, 4).
Can HUE Impala create a column which shows all dates between a specified start and end dates?
I want to list a column with date values.
You can use this sql.
select a.Date_Range
from (
select date1 - INTERVAL (a.a + (10 * b.a) + (100 * c.a) + (1000 * d.a) ) DAY as Date_Range
from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as d
) a
where a.Date_Range <= date2
Explanation -
You first create a range of numbers. And then add it to the date1 to get a range. Then you can pick your date range less than date2.
Here is answer to request
The question is how to count by each selected_date e.x:
2012-02-10: 1
2012-02-15: 0
2012-02-14: 3
2012-02-11: 0
How to make this request
Here is the request to get above answer
select selected_date, date1 from
(select selected_date from
(select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) selected_date from
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
where selected_date between '2012-02-10' and '2012-02-15' ) vv left join clicker on clicker.date1=vv.selected_date
This might work:
SELECT selected_date, SUM(CASE WHEN date1 IS NULL THEN 0 ELSE 1 END) FROM table
GROUP BY selected_date
So , basically this?
SELECT t.selected_date, COUNT(t.date1)
FROM ( Your Query Here )
GROUP BY t.selected_date
COUNT() ignores NULL values by default, so it will count only matches .
For example, I have table:
ID | Value
1 hi
1 yo
2 foo
2 bar
2 hehe
3 ha
6 gaga
I want my query to get ID, Value; meanwhile the returned set should be in the order of frequency count of each ID.
I tried the query below but don't know how to get the ID and Value column at the same time:
SELECT COUNT(*) FROM TABLE group by ID order by COUNT(*) desc;
The count number doesn't matter to me, I just need the data to be in such order.
Desire Result:
ID | Value
2 foo
2 bar
2 hehe
1 hi
1 yo
3 ha
6 gaga
As you can see because ID:2 appears most times(3 times), it's first on the list,
then ID:1(2 times) etc.
you can try this -
select id, value, count(*) over (partition by id) freq_count
from
(
select 2 as ID, 'foo' as value
from dual
union all
select 2, 'bar'
from dual
union all
select 2, 'hehe'
from dual
union all
select 1 , 'hi'
from dual
union all
select 1 , 'yo'
from dual
union all
select 3 , 'ha'
from dual
union all
select 6 , 'gaga'
from dual
)
order by 3 desc;
select t.id, t.value
from TABLE t
inner join
(
SELECT id, count(*) as cnt
FROM TABLE
group by ID
)
x on x.id = t.id
order by x.cnt desc
How about something like
SELECT t.ID,
t.Value,
c.Cnt
FROM TABLE t INNER JOIN
(
SELECT ID,
COUNT(*) Cnt
FROM TABLE
GROUP BY ID
) c ON t.ID = c.ID
ORDER BY c.Cnt DESC
SQL Fiddle DEMO
I see the question is already answered, but since the most obvious and most simple solution is missing, I'm posting it anyway. It doesn't use self joins nor subqueries:
SQL> create table t (id,value)
2 as
3 select 1, 'hi' from dual union all
4 select 1, 'yo' from dual union all
5 select 2, 'foo' from dual union all
6 select 2, 'bar' from dual union all
7 select 2, 'hehe' from dual union all
8 select 3, 'ha' from dual union all
9 select 6, 'gaga' from dual
10 /
Table created.
SQL> select id
2 , value
3 from t
4 order by count(*) over (partition by id) desc
5 /
ID VALU
---------- ----
2 bar
2 hehe
2 foo
1 yo
1 hi
6 gaga
3 ha
7 rows selected.