Here
Please Find the About fiddle link.
There i have table with the data.But when tried to pivot the table i am getting some rows in alphabetically order.
I don't want the rows to be converted to Alphabetical how can i restrict that Please help me .
I want to see the output like
A|B|C|January|February|December
-----------------------------
100|DEF|OL|200|290|100
100|ABC|OL|300|360|260
200|DEF|OL|600|550|500
200|ABC|OL|600|680|570
I dont want to change the order Please help me.
I have searched but nothing worked for me.If i use order by case coulmn-name when '' then 1
gives me syntax error
SQL tables represent unordered sets. So your original data has no ordering. You can assign an ordering by providing an id column. Then you can do the following:
Calculate the minimum id for combinations of a/b/c.
Order by this minimum id
You can see this solution in this db<>fiddle.
The change in the pivot code is:
set #sql = '
select A,B,C,'+#col_list+'
from (select a, b, c, d, e, min(id) over (partition by a, b, c) as min_id
from t1
) t1
pivot (max([E]) for [D] in ('+#col_list+'))pv
order by min_id'
Related
I have a table test with fields - A (ID), B (Flag). I need to add a new column - C (Result) in this table and it's value will be derived based on B (Flag) field. If flag is false then keep checking previous rows till we get flag as true and then take value of A (ID) field and populate it in C (Result) column. So C will have the last value of A with B field as True.
I have the query in SQL but when I try to use it in Redshift I get following errors.
1st Query Option:
WITH
cte1 AS (
SELECT A, SUM(B='T') OVER (ORDER BY A) group_no
FROM test
),
cte2 AS (
SELECT A, MIN(A) OVER (PARTITION BY group_no) previous_T
FROM cte1
)
UPDATE test
JOIN cte2 USING (A)
SET test.C = cte2.previous_T;
I am getting errors in SUM and MIN function.
2nd Query Option:
UPDATE test
JOIN (
SELECT A,
#tmp := CASE WHEN B='T' THEN A ELSE #tmp END C
FROM test
JOIN (SELECT #tmp:=0) init
ORDER BY A
) data USING (A)
SET test.C = data.C;
Getting error in temporary table.
I am new to SQL with no experience in Redshift, appreciate any help I get. Thanks!
I only have a few min but I think I can get you started. Let’s stick to query #1.
SUM(B=‘T’) isn’t going to work in Redshift. Look at the function DECODE() as it will allow you to switch how a column is generated based on another column.
It looks like you want to do a rolling SUM() so you will need a frame clause (likely “rows unbounded preceding”).
It’s not clear why you want SUM() of ids. I’d think you would want MAX() as this will give you the highest preceding id. Advise if I’m missing something.
I’d think you would want something like:
DECODE(B=‘T’, true, A, MAX(A) OVER (order by A rows unbounded preceding) as prev_id
But I don’t know your data or the process you are trying to implement.
As for the UPDATE I suggest you look at the Redshift docs. This will look more like:
…
update test set c = cte.prev_id
from cte
where test.id = cte.id;
This assumes the id is unique and that I have half a clue what you are trying to do.
My query is to avoid duplicate in a particular column while selecting all columns. But DISTINCT is not working since seq.number column is also being selected.
Any idea to make the query work
In the below example query seq_num is unique key.
Edit: including sample data in picture
select DISTINCT(name), seq_num from table_1;
For two columns this query will be enough:
SELECT name, min(seq_num)
FROM table
GROUP BY name
For more column, use row_number analytic functon
SELECT name, col1, col2, .... col500, seq_num
FROM (
SELECT t.*, row_number() over (partition by name order by seq_num ) As rn
FROM table t
)
WHERE rn = 1
The above queries pick only one row with a given name and the smallest seq_num value for each name.
You cannot do what you want. Please read more about DISTINCT clause and query result set. You will understand that distinct is not suitable for your issue. If you provide some sample data for what you have and what should query show, when possible we will help you.
I have a table with two columns a and b where a is an ID and b is a timestamp.
I need to select all of the a's which are distinct but I only care about the most up to date row per ID.
I.e. I need a way of selecting distinct a's conditional on the b values.
Is there a way to do this using DISTINCT ON in postgres?
Cheers
Like #a_horse_with_no_name suggests, the solution is
SELECT DISTINCT ON (a) a, b FROM the_table ORDER BY a, b DESC
As the manual says,
Note that the "first row" of a set is unpredictable unless the query
is sorted on enough columns to guarantee a unique ordering of the rows
arriving at the DISTINCT filter. (DISTINCT ON processing occurs after
ORDER BY sorting.)
As posted by the upvoted answers, SELECT DISTINCT ON (a) a, b FROM the_table ORDER BY a, b DESC works on Postgre 12. However, I am posting this answer to highlight few important points:
The results will be sorted based on column a; not column b.
Within each result row, the most recent (highest value) for column b would be picked.
In case, someone wants to get the most recent value for column b on the entire result set, in sql, we can run : SELECT MAX(b) from (SELECT DISTINCT ON (a) a, b FROM the_table ORDER BY a, b DESC).
I have a scenario for a type2 table where I have to remove duplicates on total row level.
Lets consider below example as the data in table.
A|B|C|D|E
100|12-01-2016|2|3|4
100|13-01-2016|3|4|5
100|14-01-2016|2|3|4
100|15-01-2016|5|6|7
100|16-01-2016|5|6|7
If you consider A as key column, you know that last 2 rows are duplicates.
Generally to find duplicates, we use group by function.
select A,C,D,E,count(1)
from table
group by A,C,D,E
having count(*)>1
for this output would be 100|2|3|4 as duplicate and also 100|5|6|7.
However, only 100|5|6|7 is only duplicate as per type 2 and not 100|2|3|4 because this value has come back in 3rd run and not soon after 1st load.
If I add date field into group by 100|5|6|7 will not be considered as duplicate, but in reality it is.
Trying to figure out duplicates as explained above.
Duplicates should only be 100|5|6|7 and not 100|2|3|4.
can someone please help out with SQL for the same.
Regards
Raghav
Use row_number analytical function to get rid of duplicates.
delete from
(
select a,b,c,d,e,row_number() over (partition by a,b,c,d,e) as rownumb
from table
) as a
where rownumb > 1
if you want to see all duplicated rows, you need join table with your group by query or filter table using group query as subquery.
wITH CTE AS (select a, B, C,D,E, count(*)
from TABLE
group by 1,2,3,4,5
having count(*)>1)
sELECT * FROM cte
WHERE B <> B + 1
Try this query and see if it works. In case you are getting any errors then let me know.
I am assuming that your column B is in the Date format if not then cast it to date
If you can see the duplicate then just replace select * to delete
I have a table:
abc_test with columns n_num, k_str.
This query doesnt work:
select distinct(n_num) from abc_test order by(k_str)
But this one works:
select n_num from abc_test order by(k_str)
How do DISTINCT and ORDER BY keywords work internally that output of both the queries is changed?
As far as i understood from your question .
distinct :- means select a distinct(all selected values should be unique).
order By :- simply means to order the selected rows as per your requirement .
The problem in your first query is
For example :
I have a table
ID name
01 a
02 b
03 c
04 d
04 a
now the query select distinct(ID) from table order by (name) is confused which record it should take for ID - 04 (since two values are there,d and a in Name column). So the problem for the DB engine is here when you say
order by (name).........
You might think about using group by instead:
select n_num
from abc_test
group by n_num
order by min(k_str)
The first query is impossible.
Lets explain this by example. we have this test:
n_num k_str
2 a
2 c
1 b
select distinct (n_num) from abc_test is
2
1
Select n_num from abc_test order by k_str is
2
1
2
What do you want to return
select distinct (n_num) from abc_test order by k_str?
it should return only 1 and 2, but how to order them?
How do extended sort key columns
The logical order of operations in SQL for your first query, is (simplified):
FROM abc_test
SELECT n_num, k_str i.e. add a so called extended sort key column
ORDER BY k_str DESC
SELECT n_num i.e. remove the extended sort key column again from the result.
Thanks to the SQL standard extended sort key column feature, it is possible to order by something that is not in the SELECT clause, because it is being temporarily added to it behind the scenes prior to ordering, and then removed again after ordering.
So, why doesn't this work with DISTINCT?
If we add the DISTINCT operation, it would need to be added between SELECT and ORDER BY:
FROM abc_test
SELECT n_num, k_str i.e. add a so called extended sort key column
DISTINCT
ORDER BY k_str DESC
SELECT n_num i.e. remove the extended sort key column again from the result.
But now, with the extended sort key column k_str, the semantics of the DISTINCT operation has been changed, so the result will no longer be the same. This is not what we want, so both the SQL standard, and all reasonable databases forbid this usage.
Workarounds
PostgreSQL has the DISTINCT ON syntax, which can be used here for precisely this job:
SELECT DISTINCT ON (k_str) n_num
FROM abc_test
ORDER BY k_str DESC
It can be emulated with standard syntax as follows, if you're not using PostgreSQL
SELECT n_num
FROM (
SELECT n_num, MIN(k_str) AS k_str
FROM abc_test
GROUP BY n_num
) t
ORDER BY k_str
Or, just simply (in this case)
SELECT n_num, MIN(k_str) AS k_str
FROM abc_test
GROUP BY n_num
ORDER BY k_str
I have blogged about SQL DISTINCT and ORDER BY more in detail here.
You are selecting the collection distinct(n_num) from the resultset from your query. So there is no actual relation with the column k_str anymore. A n_num can be from two rows each having a different value for k_str. So you can't order the collection distinct(n_num) by k_str.
According to SQL Standards, a SELECT clause may refer either to as clauses ("aliases") in the top level SELECT clause or columns of the resultset by ordinal position, and therefore nether of your queries would be compliant.
It seems Oracle, in common with other SQL implemetations, allows you to refer to columns that existed (logically) immediately prior to being projected away in the SELECT clause. I'm not sure whether such flexibility is such a good thing: IMO it is good practice to expose the sort order to the calling application by including the column/expressions etc in the SELECT clause.
As ever, you need to apply dsicpline to get meaningful results. For your first query, the definition of order is potentially entirely arbitrary.You should be grateful for the error ;)
This approach is available in SQL server 2000, you can select distinct values from a table and order by different column which is not included in Distinct.
But in SQL 2012 this will through you an error
"ORDER BY items must appear in the select list if SELECT DISTINCT is specified."
So, still if you want to use the same feature as of SQL 2000 you can use the column number for ordering(its not recommended in best practice).
select distinct(n_num) from abc_test order by 1
This will order the first column after fetching the result. If you want the ordering should be done based on different column other than distinct then you have to add that column also in select statement and use column number to order by.
select distinct(n_num), k_str from abc_test order by 2
When I got same error, I got it resolved by changing it as
SELECT n_num
FROM(
SELECT DISTINCT(n_num) AS n_num, k_str
FROM abc_test
) as tbl
ORDER BY tbl.k_str
My query doesn't match yours exactly, but it's pretty close.
select distinct a.character_01 , (select top 1 b.sort_order from LookupData b where a.character_01 = b.character_01 )
from LookupData a
where
Dataset_Name = 'Sample' and status = 200
order by 2, 1
did you try this?
SELECT DISTINCT n_num as iResult
FROM abc_test
ORDER BY iResult
you can do
select distinct top 10000 (n_num) --assuming you won't have more than 10,000 rows
from abc_test order by(k_str)