I have a table with the following columns:
bkng_date
bkng_id (varchar)
villa_id (varchar)
This query
select bkng_date,count(*) as cnt
from tab_bkng_det
group by bkng_date;
returns the no.of records for each date as count.
Now I need to find dates in the resultset of this query where cnt = 2.
I tried a couple of subqueries but I'm not getting the desired results.
The simplest, correct and safe solution is adding having count(*) = 2 clause as Gordon said.
For completeness, if you were curious how to solve it using subqueries (you didn't provide your db vendor though it's very likely your vendor supports having clause), it would be:
select x.bkng_date, x.cnt from (
select bkng_date,count(*) as cnt
from tab_bkng_det
group by bkng_date
) x
where x.cnt = 2
or
with x as (
select bkng_date,count(*) as cnt
from tab_bkng_det
group by bkng_date
)
select * from x where cnt = 2
Best Option is to use the Having Clause as follows,
select bkng_date,count(*) as cnt
from tab_bkng_det
group by bkng_date
having count(*) = 2
Related
I have this table called item:
| PERSON_id | ITEM_id |
|------------------|----------------|
|------CP2---------|-----A03--------|
|------CP2---------|-----A02--------|
|------HB3---------|-----A02--------|
|------BW4---------|-----A01--------|
I need an SQL statement that would output the person with the most Items. Not really sure where to start either.
I advice you to use inner query for this purpose. the inner query is going to include group by and order by statement. and outer query will select the first statement which has the most items.
SELECT * FROM
(
SELECT PERSON_ID, COUNT(*) FROM TABLE1
GROUP BY PERSON_ID
ORDER BY 2 DESC
)
WHERE ROWNUM = 1
here is the fiddler link : http://sqlfiddle.com/#!4/4c4228/5
Locating the maximum of an aggregated column requires more than a single calculation, so here you can use a "common table expression" (cte) to hold the result and then re-use that result in a where clause:
with cte as (
select
person_id
, count(item_id) count_items
from mytable
group by
person_id
)
select
*
from cte
where count_items = (select max(count_items) from cte)
Note, if more than one person shares the same maximum count; more than one row will be returned bu this query.
I am trying to find a query that would give me a count of another table in the query. The problem is that I have no idea what to set where in the count part to. As it is now it will just give back a count of all the values in that table.
Select
ID as Num,
(select Count(*) from TASK where ID=ID(Also tried Num)) as Total
from ORDER
The goal is to have a result that reads like
Num Total
_________________
1 13
2 5
3 22
You need table aliases. So I think you want:
Select ID as Num,
(select Count(*) from TASK t where t.ID = o.ID) as Total
from ORDER o;
By the way, ORDER is a terrible name for a table because it is a reserved work in SQL.
You can do it as a sub query or a join (or an OVER statement.)
I think the join is clearest when you are first learning SQL
Select
ID as Num, count(TASK.ID) AS Total
from ORDER
left join TASK ON ORDER.ID=TASK.ID
GROUP BY ORDER.ID
How can one filter a grouped resultset for only those groups that meet some criterion compared against the other groups? For example, only those groups that have the maximum number of constituent records?
I had thought that a subquery as follows should do the trick:
SELECT * FROM (
SELECT *, COUNT(*) AS Records
FROM T
GROUP BY X
) t HAVING Records = MAX(Records);
However the addition of the final HAVING clause results in an empty recordset... what's going on?
In MySQL (Which I assume you are using since you have posted SELECT *, COUNT(*) FROM T GROUP BY X Which would fail in all RDBMS that I know of). You can use:
SELECT T.*
FROM T
INNER JOIN
( SELECT X, COUNT(*) AS Records
FROM T
GROUP BY X
ORDER BY Records DESC
LIMIT 1
) T2
ON T2.X = T.X
This has been tested in MySQL and removes the implicit grouping/aggregation.
If you can use windowed functions and one of TOP/LIMIT with Ties or Common Table expressions it becomes even shorter:
Windowed function + CTE: (MS SQL-Server & PostgreSQL Tested)
WITH CTE AS
( SELECT *, COUNT(*) OVER(PARTITION BY X) AS Records
FROM T
)
SELECT *
FROM CTE
WHERE Records = (SELECT MAX(Records) FROM CTE)
Windowed Function with TOP (MS SQL-Server Tested)
SELECT TOP 1 WITH TIES *
FROM ( SELECT *, COUNT(*) OVER(PARTITION BY X) [Records]
FROM T
)
ORDER BY Records DESC
Lastly, I have never used oracle so apolgies for not adding a solution that works on oracle...
EDIT
My Solution for MySQL did not take into account ties, and my suggestion for a solution to this kind of steps on the toes of what you have said you want to avoid (duplicate subqueries) so I am not sure I can help after all, however just in case it is preferable here is a version that will work as required on your fiddle:
SELECT T.*
FROM T
INNER JOIN
( SELECT X
FROM T
GROUP BY X
HAVING COUNT(*) =
( SELECT COUNT(*) AS Records
FROM T
GROUP BY X
ORDER BY Records DESC
LIMIT 1
)
) T2
ON T2.X = T.X
For the exact question you give, one way to look at it is that you want the group of records where there is no other group that has more records. So if you say
SELECT taxid, COUNT(*) as howMany
GROUP by taxid
You get all counties and their counts
Then you can treat that expressions as a table by making it a subquery, and give it an alias. Below I assign two "copies" of the query the names X and Y and ask for taxids that don't have any more in one table. If there are two with the same number I'd get two or more. Different databases have proprietary syntax, notably TOP and LIMIT, that make this kind of query simpler, easier to understand.
SELECT taxid FROM
(select taxid, count(*) as HowMany from flats
GROUP by taxid) as X
WHERE NOT EXISTS
(
SELECT * from
(
SELECT taxid, count(*) as HowMany FROM
flats
GROUP by taxid
) AS Y
WHERE Y.howmany > X.howmany
)
Try this:
SELECT * FROM (
SELECT *, MAX(Records) as max_records FROM (
SELECT *, COUNT(*) AS Records
FROM T
GROUP BY X
) t
) WHERE Records = max_records
I'm sorry that I can't test the validity of this query right now.
I'm not sure if this is even a good question or not.
I have a complex query with lot's of unions that searches multiple tables for a certain keyword (user input). All tables in which there is searched are related to the table book.
There is paging on the resultset using LIMIT, so there's always a maximum of 10 results that get withdrawn.
I want an extra column in the resultset displaying the total amount of results found however. I do not want to do this using a separate query. Is it possible to add a count() column to the resultset that counts every result found?
the output would look like this:
ID Title Author Count(...)
1 book_1 auth_1 23
2 book_2 auth_2 23
4 book_4 auth_.. 23
...
Thanks!
This won't add the count to each row, but one way to get the total count without running a second query is to run your first query using the SQL_CALC_FOUND_ROWS option and then select FOUND_ROWS(). This is sometimes useful if you want to know how many total results there are so you can calculate the page count.
Example:
select SQL_CALC_FOUND_ROWS ID, Title, Author
from yourtable
limit 0, 10;
SELECT FOUND_ROWS();
From the manual:
http://dev.mysql.com/doc/refman/5.1/en/information-functions.html#function_found-rows
The usual way of counting in a query is to group on the fields that are returned:
select ID, Title, Author, count(*) as Cnt
from ...
group by ID, Title, Author
order by Title
limit 1, 10
The Cnt column will contain the number of records in each group, i.e. for each title.
Regarding second query:
select tbl.id, tbl.title, tbl.author, x.cnt
from tbl
cross join (select count(*) as cnt from tbl) as x
If you will not join to other table(s):
select tbl.id, tbl.title, tbl.author, x.cnt
from tbl, (select count(*) as cnt from tbl) as x
My Solution:
SELECT COUNT(1) over(partition BY text) totalRecordNumber
FROM (SELECT 'a' text, id_consult_req
FROM consult_req cr);
If your problem is simply the speed/cost of doing a second (complex) query I would suggest you simply select the resultset into a hash-table and then count the rows from there while returning, or even more efficiently use the rowcount of the previous resultset, then you do not even have to recount
This will add the total count on each row:
select count(*) over (order by (select 1)) as Cnt,*
from yourtable
Here is your answare:
SELECT *, #cnt count_rows FROM (
SELECT *, (#cnt := #cnt + 1) row_number FROM your_table
CROSS JOIN (SELECT #cnt := 0 AS variable) t
) t;
You simply cannot do this, you'll have to use a second query.
If I have a table with columns id, name, score, date
and I wanted to run a sql query to get the record where id = 2 with the earliest date in the data set.
Can you do this within the query or do you need to loop after the fact?
I want to get all of the fields of that record..
If you just want the date:
SELECT MIN(date) as EarliestDate
FROM YourTable
WHERE id = 2
If you want all of the information:
SELECT TOP 1 id, name, score, date
FROM YourTable
WHERE id = 2
ORDER BY Date
Prevent loops when you can. Loops often lead to cursors, and cursors are almost never necessary and very often really inefficient.
SELECT TOP 1 ID, Name, Score, [Date]
FROM myTable
WHERE ID = 2
Order BY [Date]
While using TOP or a sub-query both work, I would break the problem into steps:
Find target record
SELECT MIN( date ) AS date, id
FROM myTable
WHERE id = 2
GROUP BY id
Join to get other fields
SELECT mt.id, mt.name, mt.score, mt.date
FROM myTable mt
INNER JOIN
(
SELECT MIN( date ) AS date, id
FROM myTable
WHERE id = 2
GROUP BY id
) x ON x.date = mt.date AND x.id = mt.id
While this solution, using derived tables, is longer, it is:
Easier to test
Self documenting
Extendable
It is easier to test as parts of the query can be run standalone.
It is self documenting as the query directly reflects the requirement
ie the derived table lists the row where id = 2 with the earliest date.
It is extendable as if another condition is required, this can be easily added to the derived table.
Try
select * from dataset
where id = 2
order by date limit 1
Been a while since I did sql, so this might need some tweaking.
Using "limit" and "top" will not work with all SQL servers (for example with Oracle).
You can try a more complex query in pure sql:
select mt1.id, mt1."name", mt1.score, mt1."date" from mytable mt1
where mt1.id=2
and mt1."date"= (select min(mt2."date") from mytable mt2 where mt2.id=2)