I have a table with column a having not necessarily distinct values and column b having for each value of a a number of distinct values. I want to get a result having each value of a appearing only once and getting the first found value of b for that value of a. How do I do this in sql server 2000?
example table:
a b
1 aa
1 bb
2 zz
3 aa
3 zz
3 bb
4 bb
4 aa
Wanted result:
a b
1 aa
2 zz
3 aa
4 bb
In addition, I must add that the values in column b are all text values. I updated the example to reflect this.
Thanks
;with cte as
(
select *,
row_number() over(partition by a order by a) as rn
from yourtablename
)
select
a,b
from cte
where rn = 1
SQL does not know about ordering by table rows. You need to introduce order in the table structure (usually using an id column). That said, once you have an id column, it's rather easy:
SELECT a, b FROM test WHERE id in (SELECT MIN(id) FROM test GROUP BY a)
There might be a way to do this, using internal SQL Server functions. But this solution is portable and more easily understood by anyone who knows SQL.
Related
I work in healthcare. In a Postgres database, we have a table member IDs and dates. I'm trying to pull the latest two dates for each member ID.
Simplified sample data:
A 1
B 1
B 2
C 1
C 5
C 7
D 1
D 2
D 3
D 4
Desired result:
A 1
B 1
B 2
C 1
C 5
D 1
D 2
I get a strong feeling this is for a homework assignment and would recommend that you look into partitioning and specifically rank() function by yourself first before looking at my solution.
Moreover, you have not specified how you received the initial result you provided, so I'll have to assume you just did select letter_column, number_column from my_table; to achieve the result.
So, what you actually want here is partition the initial query result into groups by the letter_column and select the first two rows in each. rank() function lets you assign each row a number, counting within groups:
select letter_column,
number_column,
rank() over (partition by letter_column order by number_column) as rank
from my_table;
Since it's a function, you can't use it in a predicate in the same query, so you'll have to build another query around this one, this time filtering the results where rank is over 2:
with ranked_results as (select letter_column,
number_column,
rank() over (partition by letter_column order by number_column asc) as rank
from my_table mt)
select letter_column,
number_column
from ranked_results
where rank < 3;
Here's an SQLFiddle to play around: http://sqlfiddle.com/#!15/e90744/1/0
Hope this helps!
I have the task to find out if all columns in a SQL Server table have exact the same value. The table content is created by a stored procedure and can vary in the number of columns. The first column is an ID, the second and the following columns must be compared if the all columns have exact the same value.
At the moment I do not have a clue how to achieve this.
The best solution would be to display only the rows, which have different values in one or multiple columns except the first column with ID.
Thank you so much for your help!!
--> Edit: The table looks this:
ID Instance1 Instance2 Instance3 Instance4 Instance5
=====================================================
A 1 1 1 1 1
B 1 1 0 1 1
C 55 55 55 55 55
D Driver Driver Driver Co-driver Driver
E 90 0 90 0 50
F On On On On On
The result should look like this, only the rows with one or multiple different column values should be display.
ID Instance1 Instance2 Instance3 Instance4 Instance5
=====================================================
B 1 1 0 1 1
D Driver Driver Driver Co-driver Driver
E 90 0 90 0 50
My table has more than 1000 rows and 40 columns
you can achieve this by using row_number()
Try the following code
With c as(
Select id
,field_1
,field_2
,field_3
,field_n
,row_number() over(partition by field_1,field_2,field_3,field_n order by id asc) as rn
From Table
)
Select *
From c
Where rn = 1
row_number with partition is going to show you if the field is repeated by assigning a number to a row based on field_1,field_2,field_3,field_n, for example if you have 2 rows with same field values the inner query is going to show you
rn field_1 field_2 field_3 field_n id
1 x y z a 5
2 x y z a 9
After that on the outer part of the query pick rn = 1 and you are going to obtain a query without repetitions based on fields.
Also if you want to delete repeated numbers from your table you can apply
With c as(
Select id
,field_1
,field_2
,field_3
,field_n
,row_number() over(partition by field_1,field_2,field_3,field_n order by id asc) as rn
From Table
)
delete
From c
Where rn > 1
The best solution would be to display only the rows, which have different values in one or multiple columns except the first column with ID.
You may be looking for a the following simple query, whose WHERE clause filters out rows where all fields have the same value (I assumed 5 fields - id not included).
SELECT *
FROM mytable t
WHERE NOT (
field1 = field2
AND field1 = field3
AND field1 = field4
AND field1 = field5
);
Fair warning: I'm new to using SQL. I do so on an Oracle server either via AQT or with SQL Developer.
As I haven't been able to think or search my way to an answer, I put myself in your able hands...
I'd like to combine data from table A (high quality data) with data from table B (fresh data) such that the entries from B are only included when the date stamp are later than those available from table A.
Both tables include entries from multiple entities, and the latest date stamp varies with those entities.
On the 4th of january, the tables may look something like:
A____________________________ B_____________________________
entity date type value entity date type value
X 1.jan 1 1 X 1.jan 1 2
X 1.jan 0 1 X 1.jan 0 2
X 2.jan 1 1 X 2.jan 1 2
Y 1.jan 1 1 (new entry)X 3.jan 1 1
Y 3.jan 1 1 Y 1.jan 1 2
Y 3.jan 1 2
(new entry)Y 4.jan 1 1
I have made an attempt at some code that I hope clarify my need:
WITH
AA AS
(SELECT entity, date, SUM(value)
FROM table_A
GROUP BY
entity,
date),
BB AS
(SELECT entity, date, SUM(value)
FROM table_B
WHERE date > ALL (SELECT date FROM AA)
GROUP BY
entity,
date
)
SELECT * FROM (SELECT * FROM AA UNION ALL SELECT * FROM BB)
Now, if the WHERE date > ALL (SELECT date FROM AA)would work seperately for each entity, I think have what I need.
That is, for each entity I want all entries from A, and only newer entries from B.
As the data in table A often differ from that of B (values are often corrected) I dont think I can use something like: table A UNION ALL (table B MINUS table A)?
Thanks
Essentially you are looking for entries in BB which do not exist in AA. When you are doing date > ALL (SELECT date FROM AA) this will not take into consideration the entity in question and you will not get the correct records.
Alternative is to use the JOIN and filter out all matching entries with AA.
Something like below.
WITH
AA AS
(SELECT entity, date, SUM(value)
FROM table_A
GROUP BY
entity,
date),
BB AS
(SELECT entity, date, SUM(value)
FROM table_B
LEFT OUTER JOIN AA
ON AA.entity = BB.entity
AND AA.DATE = BB.date
WHERE AA.date == null
GROUP BY
entity,
date
)
SELECT * FROM (SELECT * FROM AA UNION ALL SELECT * FROM BB)
I find your question confusing, because I don't know where the aggregation is coming from.
The basic idea on getting newer rows from table_b uses conditions in the where clause, something like this:
select . . .
from table_a a
union all
select . . .
from table_b b
where b.date > (select max(a.date) from a where a.entity = b.entity);
You can, of course, run this on your CTEs, if those are what you really want to combine.
Use UNION instead of UNION ALL , it will remove the duplicate records
SELECT * FROM (
SELECT *
FROM AA
UNION
SELECT *
FROM BB )
I want to write a query to achieve the following. I have a table xyz in which there are multiple row with same column value(1) in say column a.
I want to find in column b doesn't have a particular value for the set of rows with value 1 in column a.
Table xyz
---------
a b
1 te
1 we
1 re
2 te
2 re
3 ge
4 re
So basically I want to find if the column b does not have the value 'te' for a set of values from column a
when i do
Select a from xyz where b <> 'te'
group by a
I will get 1,2,3 and 4 both for the result.
But I want the result should only contain 1 and 2. Please help.
Select a from xyz where (b<>'te') and ((a=1) or (a=2))
or as variant
select a from xyz where (b<>'te') and (a in (1, 2))
select a from xyz
where b! = 'tz' and
a in (select a from xyz where b = 'tz')
Is this what you are looking for?
Try this for you:
Select a from xyz where b = 'te'
group by a
I just realized I didn't and still don't understand what you are asking. Could you try and restate it? The only non-trivial interpretation I can come up with that would return 1 and 2 based on this data would be:
What are the values of a such that there is a row with both a and
'te' and a row with both a and a value other than 'te'
in which case a query would be:
SELECT DISTINCT q1.a FROM (SELECT a FROM xyz WHERE b='te') q1
JOIN (SELECT a FROM xyz WHERE b!='te') q2 ON
q1.a=q2.a
The interpretation which corresponds with returning 3 and 4 in your example or with returning 1 and 2 in your geo example would be:
What are the values of a for which a te row does not exist?
in which case a query would be:
SELECT DISTINCT a FROM xyz WHERE a NOT IN (SELECT a FROM xyx WHERE b='te')
as shown here (sqlfiddle is acting up, so I used ideone)
My Table: table1
ID Name Family
1 A AA
2 B BB
3 A AB
4 D DD
5 E EE
6 A AC
SQL command on Access:
select count(*) from table1
Output: ------------> True
6 row(s)
I tried to count unique names:
Expected output: 4 row(s)
select count(distinct Name) from table1
Output on Access: ------------> Error
What changes do I need to make to my query?
Try this
SELECT Count(*) AS N
FROM
(SELECT DISTINCT Name FROM table1) AS T;
Read this for more info.
Access-Engine does not support
SELECT count(DISTINCT....) FROM ...
You have to do it like this:
SELECT count(*)
FROM
(SELECT DISTINCT Name FROM table1)
Its a little workaround... you're counting a DISTINCT selection.
A quick trick to use for me is using the find duplicates query SQL and changing 1 to 0 in Having expression. Like this:
SELECT COUNT([UniqueField]) AS DistinctCNT FROM
(
SELECT First([FieldName]) AS [UniqueField]
FROM TableName
GROUP BY [FieldName]
HAVING (((Count([FieldName]))>0))
);
Hope this helps, not the best way I am sure, and Access should have had this built in.