SQL select all columns for two distinct columns - sql

The question might be not new but i need help from you .I have sql db table and i want to select all columns for the rows which have distinct values of two columns
.For instance i have a table named 'information' as below.
What i want is to select all rows with distinct values of 'ctc_card_no' and 'tarehe' .Can anyone please help me as i have struggled to get the results
I need results to be like

I think you want to use not exists:
select t.*
from t
where not exists (select 1
from t t2
where t2.ctc_card_no = t.ctc_card_no and t2.tarehe = t.tarehe and
t2.id <> t.id
);
This returns all rows that appear only once in the table, which is how I interpret your question.
EDIT:
If you just want the distinct pairs, you can use group by:
select min(id), ctc_card_no, tarehe
from t
group by ctc_card_no, tarehe;

Maybe something like this?
select *
from information
where (ctc_card_no, tarehe) in
(select ctc_card_no, tarehe
from information
group by ctc_card_no, tarehe
having count(*) = 1
);

Related

SQL : Find Numbers of Rows in a Table according to Criteria

I want to get numbers of rows in a table according to certain criteria.
Please see the below table:-
Herein I want to get numbers of rows according to Column StationTo.
I want to get numbers of rows of each StationTo entries.
You could group by the StationTo and use the aggregate count(*) function:
SELECT StationTo, COUNT(*)
FROM mytable
GROUP BY StationTo
EDIT:
If you just want the number of rows for a single StationTo, you could use a where clause:
SELECT COUNT(*)
FROM mytable
WHERE StationTo = 'P11004400000'
Hi have you master table for stationTo records?
select s.stationto, count(data.*) from stationtomaster
left join data on data.stationto=stationtomaster.stationto
group by s.stationto
Select StationTo,Date, count(*) from table group by StationTo, Datemeaning all the stationTo having rows display their count.
or select count(distinct StationTo) from table or Select count(*) from table where stationTo='yourvalue'

How to count unique rows in Oracle

I have an oracle database table with a lot of columns. I'd like to count the number of fully unique rows. The only thing I could find is:
SELECT COUNT(DISTINCT col_name) FROM table;
This however would require me listing all the columns and I haven't been able to come up with syntax that will do that for me. I'm guessing the reason for that is that this query would be very low performance? Is there a recommended way of doing this?
How about
SELECT COUNT(*)
FROM (SELECT DISTINCT * FROM Table)
It depends on what you are trying to accomplish.
To get a count of the distinct rows by specific column, so that you know what data exists, and how many of that distinct data there are:
SELECT DISTINCT
A_CODE, COUNT(*)
FROM MY_ARCHV
GROUP BY A_CODE
--This informs me there are 93 unique codes, and how many of each of those codes there are.
Another method
--How to count how many of a type value exists in an oracle table:
select A_CDE, --the value you need to count
count(*) as numInstances --how many of each value
from A_ARCH -- the table where it resides
group by A_CDE -- sorting method
Either way, you get something that looks like this:
A_CODE Count(*)
1603 32
1600 2
1605 14
I think you want a count of all distinct rows from a table like this
select count(1) as c
from (
select distinct *
from tbl
) distinct_tbl;
SELECT DISTINCT **col_name**, count(*) FROM **table_name** group by **col_name**

Find duplicated rows that are not exactly same

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.

sql query - filtering duplicate values to create report

I am trying to list all the duplicate records in a table. This table does not have a Primary Key and has been specifically created only for creating a report to list out duplicates. It comprises of both unique and duplicate values.
The query I have so far is:
SELECT [OfficeCD]
,[NewID]
,[Year]
,[Type]
FROM [Test].[dbo].[Duplicates]
GROUP BY [OfficeCD]
,[NewID]
,[Year]
,[Type]
HAVING COUNT(*) > 1
This works right and gives me all the duplicates - that is the number of times it occurs.
But I want to display all the values in my report of all the columns. How can I do that without querying for each record separately?
For example:
Each table has 10 fields and [NewID] is the field which is occuring multiple times.I need to create a report with all the data in all the fields where newID has been duplicated.
Please help.
Thank you.
You need a subquery:
SELECT * FROM yourtable
WHERE NewID IN (
SELECT NewID FROM yourtable
GROUP BY OfficeCD,NewID,Year,Type
HAVING Count(*)>1
)
Additionally you might want to check your tags: You tagged mysql, but the Syntax lets me think you mean sql-server
Try this:
SELECT * FROM [Duplicates] WHERE NewID IN
(
SELECT [NewID] FROM [Duplicates] GROUP BY [NewID] HAVING COUNT(*) > 1
)
select d.*
from Duplicates d
inner join (
select NewID
from Duplicates
group by NewID
having COUNT(*) > 1
) dd on d.NewID = dd.NewID

SELECT *, COUNT(*) in SQLite

If i perform a standard query in SQLite:
SELECT * FROM my_table
I get all records in my table as expected. If i perform following query:
SELECT *, 1 FROM my_table
I get all records as expected with rightmost column holding '1' in all records. But if i perform the query:
SELECT *, COUNT(*) FROM my_table
I get only ONE row (with rightmost column is a correct count).
Why is such results? I'm not very good in SQL, maybe such behavior is expected? It seems very strange and unlogical to me :(.
SELECT *, COUNT(*) FROM my_table is not what you want, and it's not really valid SQL, you have to group by all the columns that's not an aggregate.
You'd want something like
SELECT somecolumn,someothercolumn, COUNT(*)
FROM my_table
GROUP BY somecolumn,someothercolumn
If you want to count the number of records in your table, simply run:
SELECT COUNT(*) FROM your_table;
count(*) is an aggregate function. Aggregate functions need to be grouped for a meaningful results. You can read: count columns group by
If what you want is the total number of records in the table appended to each row you can do something like
SELECT *
FROM my_table
CROSS JOIN (SELECT COUNT(*) AS COUNT_OF_RECS_IN_MY_TABLE
FROM MY_TABLE)