Table data and expected data as below
I wanted to select the distinct ID and Name, irrespective of branch. But I want the branch name to be displayed.
If branch need not be displayed I can use substring. Can the result be achieved using CTE.
As you don't appear to care about which 'branch' needs to be returned, you can simply use a row_number within a CTE to return just one result per ID value:
declare #t table(ID int,Name varchar(20));
insert into #t values
(10,'Manoj (CS)')
,(10,'Manoj (IS)')
,(20,'Ajay (CS)')
,(20,'AJAY (IS)')
,(30,'Sunjay(EC)')
,(40,'Lina(IS)')
,(40,'Lina(CS)')
,(40,'Lina(EC)')
,(50,'Mary(IS)')
,(50,'Mary(EC)');
with d as
(
select ID
,Name
,row_number() over (partition by ID order by Name) as rn
from #t
)
select ID
,Name
from d
where rn = 1;
Output:
+----+------------+
| ID | Name |
+----+------------+
| 10 | Manoj (CS) |
| 20 | Ajay (CS) |
| 30 | Sunjay(EC) |
| 40 | Lina(CS) |
| 50 | Mary(EC) |
+----+------------+
If you do have a preference for the (CS) branch however, you would need to alter the row_number slightly:
with d as
(
select ID
,Name
,row_number() over (partition by ID
order by case when right(Name,4) = '(CS)'
then 1
else 2
end
,Name
) as rn
from #t
)
select ID
,Name
from d
where rn = 1;
You can use row_number() function with TIES :
select top (1) with ties *
from table t
order by row_number() over (partition by id order by name);
As mentioned in the comments: Change your datamodel.
If you must live with the table as is, all you want is: all student IDs, each with one branhc/subject name arbitrarily picked. This can be achieved with a simple aggregation:
select id, min(name) from mytable group by id;
Related
I have two rows with the same id but different values. I want a query to get the second value and display it in the first row.
There are only two rows for each productId and 2 different values.
I've tried looking for this for the solution everywhere.
What I have, example:
+-----+-------+
| ID | Value |
+-----+-------+
| 123 | 1 |
| 123 | 2 |
+-----+-------+
What I want
+------+-------+---------+
| ID | Value | Value 1 |
+------+-------+---------+
| 123 | 1 | 2 |
+------+-------+---------+
Not sure whether order matters to you. Here is one way:
SELECT MIN(Value), MAX(Value), ID
FROM Table
GROUP BY ID;
This is a self-join:
SELECT a.ID, a.Value, b.Value
FROM table a
JOIN table b on a.ID = b.ID
and a.Value <> b.Value
You can use a LEFT JOIN instead if there are IDs that only have one value and would be lost by the above JOIN
May be you may try this
DECLARE #T TABLE
(
Id INT,
Val INT
)
INSERT INTO #T
VALUES(123,1),(123,2),
(456,1),(789,1),(789,2)
;WITH CTE
AS
(
SELECT
RN = ROW_NUMBER() OVER(PARTITION BY Id ORDER BY Val),
*
FROM #T
)
SELECT
*
FROM CTE
PIVOT
(
MAX(Val)
FOR
RN IN
(
[1],[2]--Add More Numbers here if there are more values
)
)Q
I don't really know how to describe it. I have a table:
ID | Name | Date
-------------------------
1 | Mike | 01.01.2016
1 | Michael | 02.03.2016
2 | Samuel | 23.12.2015
2 | Sam | 05.03.2015
3 | Tony | 02.04.2012
I want to select pairs of IDs and Names with latest dates in each pair. The result here should be:
ID | Name | Date
-------------------------
1 | Michael | 02.03.2016
2 | Samuel | 23.12.2015
3 | Tony | 02.04.2012
How do I achieve this?
Oracle Database 11g
You can do it using the ROW_NUMBER() analytic function:
SELECT id, name, "date"
FROM (
SELECT t.*,
ROW_NUMBER() OVER ( PARTITION BY id ORDER BY "date" DESC ) rn
FROM table_name t
)
WHERE rn = 1
This requires only a single table scan (it does not have a self-join or correlated sub-query - i.e. IN (...) or EXISTS(...)).
Have a sub-select that returns each id and it's max date:
select * from table
where (id, date) in (select id, max(date) from table group by id)
You can use NOT EXISTS() :
SELECT * FROM YourTable t
WHERE NOT EXISTS(SELECT 1 FROM YourTable s
WHERE t.id = s.id and s.date > t.date)
Possibly the most efficient method is:
select t.*
from table t
where t.date = (select max(date) from table t2 where t2.id = t.id);
along with an index on table(id, date).
This version should scan the table and look up the correct value in the index.
Or, if there are only three columns, you can use keep:
select id, max(date) as date,
max(name) keep (dense_rank first order by date desc) as name
from table
group by id;
I have found that this version works very well in Oracle.
I have the following data structure
ID | REFID | NAME
1 | 100 | A
2 | 101 | B
3 | 101 | C
With
SELECT DISTINCT REFID, ID, NAME
FROM my_table
ORDER BY ID
I would like to have the following result:
1 | 100 | A
2 | 101 | B
Colum NAME and ID should contain the MIN or FIRST value.
But actually I get stuck at using MIN/FIRST here.
I welcome every tipps :-)
select id,
refid,
name
from (select id,
refid,
name,
row_number() over(partition by refid order by name) as rn
from my_table)
where rn = 1
order by id
You can use a subquery to do this.
WITH Q AS
( SELECT MIN(NAME) AS NAME, REFID FROM T GROUP BY REFID )
SELECT T.ID, T.REFID, T.NAME
FROM T
JOIN Q
ON (T.NAME = Q.NAME)
Also, note that SQL tables have no order. So there's no "First" value.
I am fighting with the distinct keyword in sql.
I just want to display all row numbers of unique (distinct) values in a column & so I tried:
SELECT DISTINCT id, ROW_NUMBER() OVER (ORDER BY id) AS RowNum
FROM table
WHERE fid = 64
however the below code giving me the distinct values:
SELECT distinct id FROM table WHERE fid = 64
but when tried it with Row_Number.
then it is not working.
This can be done very simple, you were pretty close already
SELECT distinct id, DENSE_RANK() OVER (ORDER BY id) AS RowNum
FROM table
WHERE fid = 64
Use this:
SELECT *, ROW_NUMBER() OVER (ORDER BY id) AS RowNum FROM
(SELECT DISTINCT id FROM table WHERE fid = 64) Base
and put the "output" of a query as the "input" of another.
Using CTE:
; WITH Base AS (
SELECT DISTINCT id FROM table WHERE fid = 64
)
SELECT *, ROW_NUMBER() OVER (ORDER BY id) AS RowNum FROM Base
The two queries should be equivalent.
Technically you could
SELECT DISTINCT id, ROW_NUMBER() OVER (PARTITION BY id ORDER BY id) AS RowNum
FROM table
WHERE fid = 64
but if you increase the number of DISTINCT fields, you have to put all these fields in the PARTITION BY, so for example
SELECT DISTINCT id, description,
ROW_NUMBER() OVER (PARTITION BY id, description ORDER BY id) AS RowNum
FROM table
WHERE fid = 64
I even hope you comprehend that you are going against standard naming conventions here, id should probably be a primary key, so unique by definition, so a DISTINCT would be useless on it, unless you coupled the query with some JOINs/UNION ALL...
This article covers an interesting relationship between ROW_NUMBER() and DENSE_RANK() (the RANK() function is not treated specifically). When you need a generated ROW_NUMBER() on a SELECT DISTINCT statement, the ROW_NUMBER() will produce distinct values before they are removed by the DISTINCT keyword. E.g. this query
SELECT DISTINCT
v,
ROW_NUMBER() OVER (ORDER BY v) row_number
FROM t
ORDER BY v, row_number
... might produce this result (DISTINCT has no effect):
+---+------------+
| V | ROW_NUMBER |
+---+------------+
| a | 1 |
| a | 2 |
| a | 3 |
| b | 4 |
| c | 5 |
| c | 6 |
| d | 7 |
| e | 8 |
+---+------------+
Whereas this query:
SELECT DISTINCT
v,
DENSE_RANK() OVER (ORDER BY v) row_number
FROM t
ORDER BY v, row_number
... produces what you probably want in this case:
+---+------------+
| V | ROW_NUMBER |
+---+------------+
| a | 1 |
| b | 2 |
| c | 3 |
| d | 4 |
| e | 5 |
+---+------------+
Note that the ORDER BY clause of the DENSE_RANK() function will need all other columns from the SELECT DISTINCT clause to work properly.
All three functions in comparison
Using PostgreSQL / Sybase / SQL standard syntax (WINDOW clause):
SELECT
v,
ROW_NUMBER() OVER (window) row_number,
RANK() OVER (window) rank,
DENSE_RANK() OVER (window) dense_rank
FROM t
WINDOW window AS (ORDER BY v)
ORDER BY v
... you'll get:
+---+------------+------+------------+
| V | ROW_NUMBER | RANK | DENSE_RANK |
+---+------------+------+------------+
| a | 1 | 1 | 1 |
| a | 2 | 1 | 1 |
| a | 3 | 1 | 1 |
| b | 4 | 4 | 2 |
| c | 5 | 5 | 3 |
| c | 6 | 5 | 3 |
| d | 7 | 7 | 4 |
| e | 8 | 8 | 5 |
+---+------------+------+------------+
Using DISTINCT causes issues as you add fields and it can also mask problems in your select. Use GROUP BY as an alternative like this:
SELECT id
,ROW_NUMBER() OVER (ORDER BY id) AS RowNum
FROM table
where fid = 64
group by id
Then you can add other interesting information from your select like this:
,count(*) as thecount
or
,max(description) as description
How about something like
;WITH DistinctVals AS (
SELECT distinct id
FROM table
where fid = 64
)
SELECT id,
ROW_NUMBER() OVER (ORDER BY id) AS RowNum
FROM DistinctVals
SQL Fiddle DEMO
You could also try
SELECT distinct id, DENSE_RANK() OVER (ORDER BY id) AS RowNum
FROM #mytable
where fid = 64
SQL Fiddle DEMO
Try this:
;WITH CTE AS (
SELECT DISTINCT id FROM table WHERE fid = 64
)
SELECT id, ROW_NUMBER() OVER (ORDER BY id) AS RowNum
FROM cte
WHERE fid = 64
Try this
SELECT distinct id
FROM (SELECT id, ROW_NUMBER() OVER (ORDER BY id) AS RowNum
FROM table
WHERE fid = 64) t
Or use RANK() instead of row number and select records DISTINCT rank
SELECT id
FROM (SELECT id, ROW_NUMBER() OVER (PARTITION BY id ORDER BY id) AS RowNum
FROM table
WHERE fid = 64) t
WHERE t.RowNum=1
This also returns the distinct ids
Question is too old and my answer might not add much but here are my two cents for making query a little useful:
;WITH DistinctRecords AS (
SELECT DISTINCT [col1,col2,col3,..]
FROM tableName
where [my condition]
),
serialize AS (
SELECT
ROW_NUMBER() OVER (PARTITION BY [colNameAsNeeded] ORDER BY [colNameNeeded]) AS Sr,*
FROM DistinctRecords
)
SELECT * FROM serialize
Usefulness of using two cte's lies in the fact that now you can use serialized record much easily in your query and do count(*) etc very easily.
DistinctRecords will select all distinct records and serialize apply serial numbers to distinct records. after wards you can use final serialized result for your purposes without clutter.
Partition By might not be needed in most cases
table indexed on the field name
for given value of name "name1" give me that row as well as N rows before and N rows after (alphabetically)
Did it in two select statements replace the number 5 with whatever you want you N to be and change the table name and this will do it. Also replace the asterisk with correct column names. Let me know if you have any problems with this.
select * from
(
Select *
,row_number() over (order by firstname desc) as 'rowNumber'
from attendees
) as temp
where rowNumber between
(
select rownumber-1
from
(
Select *, row_number() over (order by firstname desc) as 'rowNumber'
from attendees
) as temp
where firstname = 'name1') AND (
select rownumber+1
from
(
Select *, row_number() over (order by firstname desc) as 'rowNumber'
from attendees
) as temp
where firstname = 'name1')
The following gets you the row with name = 'name4', the two rows before that, and the two rows after that.
drop table t;
create table t(
name varchar(20)
,primary key(name)
);
insert into t(name) values('name1');
insert into t(name) values('name2');
insert into t(name) values('name3');
insert into t(name) values('name4');
insert into t(name) values('name5');
insert into t(name) values('name6');
insert into t(name) values('name7');
commit;
(select name from t where name = 'name4')
union all
(select name from t where name > 'name4' order by name asc limit 2)
union all
(select name from t where name < 'name4' order by name desc limit 2);
+-------+
| name |
+-------+
| name1 |
| name2 |
| name4 |
| name5 |
| name6 |
+-------+
Edit:
Added descending order by as pointed out by cyberkiwi (otherwise I would have gotten the "first" 2 items on the wrong end).