I want to delete the duplicated data in a table , I know there is a way use
SELECT
fruit,
COUNT( fruit )
FROM
basket
GROUP BY
fruit
HAVING
COUNT( fruit )> 1
ORDER BY
fruit;
to find them , buy I need to determine every column's value is equal , which means tableA.* = tableA.* (except id , id is the auto-increment primary key )
and I tried this:
SELECT
*,
COUNT( * )
FROM
myTable
GROUP BY
*
HAVING
COUNT( * )> 1
ORDER BY
id;
but it says I can't use GROUP BY * , so how can I find & delete the duplicated data(need every column's value is equal except id)?
using
SELECT * DISTINCT
DISTINCT remove duplicated result
You need to try something similar to be below query. You apply PARTITION BY for the columns other than Id (as it is incrementing unique value). PARTITION BY should be applied for columns, for which you want to check duplicates.
Also refer to Row_Number in Postgres & Common Table expression in Postgres
WITH DuplicateTableRows AS
(
SELECT Id, Row_Number() OVER (PARTITION BY col1, col2... ORDER BY Id)
FROM
Table1
)
DELETE FROM Table1
WHERE Id IN (SELECT Id FROM Table1 WHERE row_number > 1)
You can do this using JSON:
select (to_jsonb(b) - 'id')
from basket b
group by 1
having count(*) > 1;
The result is as JSON. Unfortunately, to extract the values back into a record, you need to list the columns individually.
Related
I'm trying to make an SQL query that returns the greatest number from a column and its respective id.
For more information I have two columns ID and NUMBER. Both of them have 2 entries and I want to get the highest number with the ID next to it. This is what I tried but didn't success.
SELECT ID, MAX(NUMBER) AS MAXNUMB
FROM TABLE1
GROUP BY ID, MAXNUMB;
The problem I'm experiencing is that it just shows ALL the entries and if I add a "where" expression it just shows the same (all entries [ids+numbers]).
Pd.: Yes, I got what I wanted but only with one column (number) if I add another column (ID) to select it "brokes".
Try:
SELECT
ID,
A_NUMBER
FROM TABLE1
WHERE A_NUMBER = (
SELECT MAX(A_NUMBER)
FROM TABLE1);
Presuming you want the IDs* of the row with the highest number (and not, instead, the highest number for each ID -- if IDs were not unique in your table, for example).
* there may be more than one ID returned if there are two or more IDs with equal maximum numbers
you can try this
Select ID,maxNumber
From
(
SELECT
ID,
(Select Max(NUMBER) from Tmp where Id = t.Id) maxNumber
FROM
Tmp t
)T1
Group By ID,maxNumber
The query you posted has an illegal column name (number) and is group by the alias for the max value, which is illegal and also doesn't make sense; and you can't include the unaliased max() within the group-by either. So it's likely you're actually doing something like:
select id, max(numb) as maxnumb
from table1
group by id;
which will give one row per ID, with the maximum numb (which is the new name I've made up for your numeric column) for each ID. Or as you said you get "ALL the entries" you might have group by id, numb, which would show all rows from the table (unless there are duplicate combinations).
To get the maximum numb and the corresponding id you could group by id only, order by descending maxnumb, and then return the first row only:
select id, max(numb) as maxnumb
from table1
group by id
order by maxnumb desc
fetch first 1 row only
If there are two ID with the same maxnumb then you would only get one of them - and which one is indeterminate unless you modify the order by - but in that case you might prefer to use first 1 row with ties to see them all.
You could achieve the same thing with a subquery and analytic function to generating a ranking, and have the outer query return the highest-ranking row(s):
select id, numb as maxnumb
from (
select id, numb, dense_rank() over (order by numb desc) as rnk
from table1
)
where rnk = 1
You could also use keep to get the same result as first 1 row only:
select max(id) keep (dense_rank last order by numb) as id, max(numb) as maxnumb
from table1
fiddle
(select ID,EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE,ROW_NUMBER() OVER(PARTITION BY EXTERNAL_TRANSACTION_ID ORDER BY ID ) AS SEQNUM
from AC_POS_TRANSACTION_TRK aptt WHERE [RESULT] ='Success'
GROUP BY ID, EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE )
Hello,
On above query, I want to get rows of transaction id's which has seqnum=1 and seqnum=2
But if that transaction id has no second row (seqnum=2), I dont want to get any row for that transaction id.
Thanks!!
Something like this
Not 100% sure if this is correct without you table definition, but my understanding is that you want to EXCLUDE records if that record has an entry with seqnum=2 -- you can't use a where clause alone because that would still return seqnum = 1.
You can use an exists /not exists or in/not in clause like this
(select ID,EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE,ROW_NUMBER() OVER(PARTITION BY EXTERNAL_TRANSACTION_ID ORDER BY ID ) AS SEQNUM
from AC_POS_TRANSACTION_TRK aptt WHERE [RESULT] ='Success'
and not exists ( select 1 from AC_POS_TRANSACTION_TRK a where a.id = aptt.id
and a.seqnum = 2)
GROUP BY ID, EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE )
basically what this does is it excludes records if a record exists as specified in the NOT EXISTS query.
One option you can try is to add a count of rows per group using the same partioning critera and then filter accordingly. Not entirely sure about your query without seeing it in context and with sample data - there's no aggregation so why use group by?
However can you try something along these lines
select * from (
select ID,EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE,
Row_Number() over(partition by EXTERNAL_TRANSACTION_ID order by ID) as SEQNUM,
Count(*) over(partition by EXTERNAL_TRANSACTION_ID) Qty
from AC_POS_TRANSACTION_TRK
where [RESULT] ='Success'
)x
where SEQNUM in (1,2) and Qty>1
This should do the job.
With Qry As (
-- Your original query goes here
),
Select Qry.*
From Qry
Where Exists (
Select *
From Qry Qry1
Where Qry1.EXTERNAL_TRANSACTION_ID = Qry.EXTERNAL_TRANSACTION_ID
And Qry1.SEQNUM = 1
)
And Exists (
Select *
From Qry Qry2
Where Qry2.EXTERNAL_TRANSACTION_ID = Qry.EXTERNAL_TRANSACTION_ID
And Qry2.SEQNUM = 2
)
BTW, your original query looks problematic to me, specifically I think that instead of a GROUP BY columns those columns should be in the PARTITION BY clause of the OVER statement, but without knowing more about the table structures and what you're trying to achieve, I could not say for sure.
I am looking for a distinct list of the CUSTOMER_NAME field from my table. Normally I would simply do
SELECT
distinct
[CUSTOMER_NAME]
FROM [iData3].[dbo].[N241650]
or
SELECT
[CUSTOMER_NAME]
FROM [iData3].[dbo].[N241650]
Group by [CUSTOMER_NAME]
But I am limited in my query. Due to software restrictions, the query can only be of the form
SELECT * from
[iData3].[dbo].[N241650]
where ...
How do I get a distinct list of customer names given these restrictions? I essentially need to cram everything into the WHERE clause. I'm thinking possibly WHERE EXISTS or NOT EXISTS but I haven't used those conditions before so I'm not certain if they'd be useful.
This is not possible because... is acceptable if the disappointing answer.
You can use row_number() function :
SELECT TOP (1) WITH TIES [CUSTOMER_NAME]
FROM [iData3].[dbo].[N241650]
ORDER BY ROW_NUMBER() OVER (PARTITION BY CUSTOMER_NAME ORDER BY ?)
? indicates something identity or primary/unique column which you have.
You can group by that column to achieve the same result.
select CUSTOMER_NAME
from ...
group by CUSTOMER_NAME
order by CUSTOMER_NAME;
Another alternative is to use a stored procedure.
If you can't escape from the *, then you can't GROUP BY and if you just have a WHERE then you will need a key (unique set of columns) to be able to filter correctly or else you can't differentiate dupicates (and end up selecting more than 1 row with the same customer name).
It's a bit convoluted, but try with this. It will get you 1 row per each CUSTOMER_NAME.
SELECT
*
from
[iData3].[dbo].[N241650]
where
[N241650].KeyColumn IN
(
SELECT
Z.KeyColumn
FROM
(
SELECT
X.KeyColumn,
Ranking = ROW_NUMBER() OVER (PARTITION BY X.CUSTOMER_NAME ORDER BY X.KeyColumn ASC)
FROM
[iData3].[dbo].[N241650] AS X
WHERE
X.KeyColumn IS NOT NULL
) AS Z
WHERE
Z.Ranking = 1
)
The ORDER BY inside the OVER will determine which row you get for each CUSTOMER_NAME.
If you have multiple columns for your key, then you will have to switch the IN for an EXISTS against multiple columns (you can't do a multiple column IN in SQL Server).
SELECT
*
from
[iData3].[dbo].[N241650]
where
EXISTS (
SELECT
'key columns match'
FROM (
SELECT
X.KeyColumn1,
X.KeyColumn2,
Ranking = ROW_NUMBER() OVER (PARTITION BY X.CUSTOMER_NAME ORDER BY X.KeyColumn1 ASC)
FROM
[iData3].[dbo].[N241650] AS X
) AS Z
WHERE
Z.Ranking = 1 AND
[N241650].KeyColumn1 = Z.KeyColumn1 AND
[N241650].KeyColumn2 = Z.KeyColumn2
)
You need something unique in each row. If you have that, you can use:
SELECT CUSTOMER_NAME
FROM [iData3].[dbo].[N241650]
WHERE pk = (SELECT MIN(n2.pk)
FROM [iData3].[dbo].[N241650] n2
WHERE n2.CUSTOMER_NAME = N241650.N241650
);
pk is the unique column.
To describe my query problem, the following data is helpful:
A single table contains the columns ID (int), VAL (varchar) and ORD (int)
The values of VAL may change over time by which older items identified by ID won't get updated but appended. The last valid item for ID is identified by the highest ORD value (increases over time).
T0, T1 and T2 are points in time where data got entered.
How do I get in an efficient manner to the Result set?
A solution must not involve materialized views etc. but should be expressible in a single SQL-query. Using Postgresql 9.3.
The correct way to select groupwise maximum in postgres is using DISTINCT ON
SELECT DISTINCT ON (id) sysid, id, val, ord
FROM my_table
ORDER BY id,ord DESC;
Fiddle
You want all records for which no newer record exists:
select *
from mytable
where not exists
(
select *
from mytable newer
where newer.id = mytable.id
and newer.ord > mytable.ord
)
order by id;
You can do the same with row numbers. Give the latest entry per ID the number 1 and keep these:
select sysid, id, val, ord
from
(
select
sysid, id, val, ord,
row_number() over (partition by id order by ord desc) as rn
from mytable
)
where rn = 1
order by id;
Left join the table (A) against itself (B) on the condition that B is more recent than A. Pick only the rows where B does not exist (i.e. A is the most recent row).
SELECT last_value.*
FROM my_table AS last_value
LEFT JOIN my_table
ON my_table.id = last_value.id
AND my_table.ord > last_value.ord
WHERE my_table.id IS NULL;
SQL Fiddle
Can i select all rows that have same column value (for example SSN field) but display them all separably. ?
I've searched for this answer but they all have "count(*) and group by" section that demands the rows to be exactly same.
Try This:
SELECT A, B FROM MyTable
WHERE A IN
(
SELECT A FROM MyTable GROUP BY A HAVING COUNT(*)>1
)
I have done with SQL server. But hope this is what you need
Here is another approach, which only references the table once, using an analytic function instead of a subquery to get the duplicate counts It might be faster; it also might not, depending on the particular data.
SELECT * FROM (
SELECT col1, col2, col3, ssn, COUNT(*) OVER (PARTITION BY ssn) ssn_dup_count
)
WHERE ssn_dup_count > 1
ORDER BY ssn_dup_count DESC
SELECT
*
FROM
MyTable
WHERE
EXISTS
(
SELECT
NULL
FROM
MyTable MT
WHERE
MyTable.SameColumnName = MT.SameColumnName
AND MyTable.DifferentColumnName <> MT.DifferentColumnName)
This will fetch the required data and show them in order so that we can see the grouped data together.
SELECT * FROM TABLENAME
WHERE SSN IN
(
SELECT SSN FROM TABLENAMEGROUP BY SSN HAVING COUNT(SSN)>1
)
ORDER BY SSN
Here SSN is the column names fro which similar value check is done.