Assuming there is a table with 100 columns, how can I select all columns with a sum without having to type out all the columns?
For example something like this:
select *, sum(price) as sales
from table
group by *
order by date
try this
select table.* , t.sales from table
inner join (select id, sum(price) as sales from table group by id ) t
on table.id=t.id
order by date
But in general it is not recommended to use an stare in a select statement,
for example dont use * in table-valued function
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'm fairly new to Oracle SQL, but already it's logic is beginning to confuse me. I'm trying to select all columns from a table where a particular column PRICE has the minimum value.
This works:
SELECT MIN(PRICE) FROM my_tab;
This returns me the minimum value. But why can't I select all the columns in that row? The following won't work:
SELECT * FROM my_tab WHERE PRICE = MIN( PRICE );
What am I missing here? Cheers folks!
*EDIT*
Here is the full code I'm having trouble with:
SELECT * FROM ( SELECT c.NAME, o.* FROM customers c JOIN customer_orders o ON c.CUST_NBR = o.CUST_NBR ) AS t WHERE t.PRICE = ( SELECT MIN( t.PRICE) FROM t );
SELECT * FROM TABLE WHERE PRICE = (SELECT MIN(PRICE) FROM TABLE)
--Edited
WITH
TABLE AS
(QUERY)
SELECT * FROM TABLE
WHERE PRICE = (SELECT MIN(PRICE) FROM TABLE)
You can also use a subquery to get the result:
select t1.*
from my_tab t1
inner join
(
SELECT MIN(PRICE) MinPrice
FROM my_tab
) t2
on t1.price = t2.minprice
See previous SO question, and especially answer by "Vash" which is best for your purposes. Note that you probably want to avoid a subselect since Oracle may be smart enough to use an index on the price if available to look at only one record.
Most databases, but apparently not Oracle, have either TOP 1 or LIMIT clauses for questions like these.
I have been trying to find some info on how to select a non-aggregate column that is not contained in the Group By statement in SQL, but nothing I've found so far seems to answer my question. I have a table with three columns that I want from it. One is a create date, one is a ID that groups the records by a particular Claim ID, and the final is the PK. I want to find the record that has the max creation date in each group of claim IDs. I am selecting the MAX(creation date), and Claim ID (cpe.fmgcms_cpeclaimid), and grouping by the Claim ID. But I need the PK from these records (cpe.fmgcms_claimid), and if I try to add it to my select clause, I get an error. And I can't add it to my group by clause because then it will throw off my intended grouping. Does anyone know any workarounds for this? Here is a sample of my code:
Select MAX(cpe.createdon) As MaxDate, cpe.fmgcms_cpeclaimid
from Filteredfmgcms_claimpaymentestimate cpe
where cpe.createdon < 'reportstartdate'
group by cpe.fmgcms_cpeclaimid
This is the result I'd like to get:
Select MAX(cpe.createdon) As MaxDate, cpe.fmgcms_cpeclaimid, cpe.fmgcms_claimid
from Filteredfmgcms_claimpaymentestimate cpe
where cpe.createdon < 'reportstartdate'
group by cpe.fmgcms_cpeclaimid
The columns in the result set of a select query with group by clause must be:
an expression used as one of the group by criteria , or ...
an aggregate function , or ...
a literal value
So, you can't do what you want to do in a single, simple query. The first thing to do is state your problem statement in a clear way, something like:
I want to find the individual claim row bearing the most recent
creation date within each group in my claims table
Given
create table dbo.some_claims_table
(
claim_id int not null ,
group_id int not null ,
date_created datetime not null ,
constraint some_table_PK primary key ( claim_id ) ,
constraint some_table_AK01 unique ( group_id , claim_id ) ,
constraint some_Table_AK02 unique ( group_id , date_created ) ,
)
The first thing to do is identify the most recent creation date for each group:
select group_id ,
date_created = max( date_created )
from dbo.claims_table
group by group_id
That gives you the selection criteria you need (1 row per group, with 2 columns: group_id and the highwater created date) to fullfill the 1st part of the requirement (selecting the individual row from each group. That needs to be a virtual table in your final select query:
select *
from dbo.claims_table t
join ( select group_id ,
date_created = max( date_created )
from dbo.claims_table
group by group_id
) x on x.group_id = t.group_id
and x.date_created = t.date_created
If the table is not unique by date_created within group_id (AK02), you you can get duplicate rows for a given group.
You can do this with PARTITION and RANK:
select * from
(
select MyPK, fmgcms_cpeclaimid, createdon,
Rank() over (Partition BY fmgcms_cpeclaimid order by createdon DESC) as Rank
from Filteredfmgcms_claimpaymentestimate
where createdon < 'reportstartdate'
) tmp
where Rank = 1
The direct answer is that you can't. You must select either an aggregate or something that you are grouping by.
So, you need an alternative approach.
1). Take you current query and join the base data back on it
SELECT
cpe.*
FROM
Filteredfmgcms_claimpaymentestimate cpe
INNER JOIN
(yourQuery) AS lookup
ON lookup.MaxData = cpe.createdOn
AND lookup.fmgcms_cpeclaimid = cpe.fmgcms_cpeclaimid
2). Use a CTE to do it all in one go...
WITH
sequenced_data AS
(
SELECT
*,
ROW_NUMBER() OVER (PARITION BY fmgcms_cpeclaimid ORDER BY CreatedOn DESC) AS sequence_id
FROM
Filteredfmgcms_claimpaymentestimate
WHERE
createdon < 'reportstartdate'
)
SELECT
*
FROM
sequenced_data
WHERE
sequence_id = 1
NOTE: Using ROW_NUMBER() will ensure just one record per fmgcms_cpeclaimid. Even if multiple records are tied with the exact same createdon value. If you can have ties, and want all records with the same createdon value, use RANK() instead.
You can join the table on itself to get the PK:
Select cpe1.PK, cpe2.MaxDate, cpe1.fmgcms_cpeclaimid
from Filteredfmgcms_claimpaymentestimate cpe1
INNER JOIN
(
select MAX(createdon) As MaxDate, fmgcms_cpeclaimid
from Filteredfmgcms_claimpaymentestimate
group by fmgcms_cpeclaimid
) cpe2
on cpe1.fmgcms_cpeclaimid = cpe2.fmgcms_cpeclaimid
and cpe1.createdon = cpe2.MaxDate
where cpe1.createdon < 'reportstartdate'
Thing I like to do is to wrap addition columns in aggregate function, like max().
It works very good when you don't expect duplicate values.
Select MAX(cpe.createdon) As MaxDate, cpe.fmgcms_cpeclaimid, MAX(cpe.fmgcms_claimid) As fmgcms_claimid
from Filteredfmgcms_claimpaymentestimate cpe
where cpe.createdon < 'reportstartdate'
group by cpe.fmgcms_cpeclaimid
What you are asking, Sir, is as the answer of RedFilter.
This answer as well helps in understanding why group by is somehow a simpler version or partition over:
SQL Server: Difference between PARTITION BY and GROUP BY
since it changes the way the returned value is calculated and therefore you could (somehow) return columns group by can not return.
You can use as below,
Select X.a, X.b, Y.c from (
Select X.a as a, sum (b) as sum_b from name_table X
group by X.a)X
left join from name_table Y on Y.a = X.a
Example;
CREATE TABLE #products (
product_name VARCHAR(MAX),
code varchar(3),
list_price [numeric](8, 2) NOT NULL
);
INSERT INTO #products VALUES ('paku', 'ACE', 2000)
INSERT INTO #products VALUES ('paku', 'ACE', 2000)
INSERT INTO #products VALUES ('Dinding', 'ADE', 2000)
INSERT INTO #products VALUES ('Kaca', 'AKB', 2000)
INSERT INTO #products VALUES ('paku', 'ACE', 2000)
--SELECT * FROM #products
SELECT distinct x.code, x.SUM_PRICE, product_name FROM (SELECT code, SUM(list_price) as SUM_PRICE From #products
group by code)x
left join #products y on y.code=x.code
DROP TABLE #products
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.