I need to fetch distinct rows for a particular column from a table having over 200 columns. How can I achieve this?
I used
Select distinct col1 , * from table;
but it failed.
Please suggest an elegant way to achieve this.
Regards,
Tarun
The solution if you want one row for each distinct value in a particular column is:
SELECT col1, MAX(col2), MAX(col3), MAX(col4), ...
FROM mytable
GROUP BY col1
I chose MAX() arbitrarily. It could also be MIN() or some other aggregate function. The point is if you use GROUP BY to make sure to get one row per value in col1, then all the other columns must be inside aggregate functions.
There is no way to write MAX(*) to get that treatment for all columns. Sorry, you will have to write out all the column names (at least those that you need in this query, which might not be all 200).
We can generate a sequence of ROW_NUMBER() for every COL1 and then select the first entry of every sequence.
SELECT * FROM
(
SELECT E.* , ROW_NUMBER() OVER( PARTITION BY COL1 ORDER BY 1) AS ID
FROM YOURTABLE E
) MYDATA
WHERE MYDATA.ID = 1
A working example in fiddle
Related
I don't understand why I can't use this in my code :
SELECT MAX(SMTHNG), COUNT(MAX(SMTHNG))
FROM SomeTable;
Searched for an answer but didn't find it in documentation about these aggregate functions.
Also I get an SQL-compiler error "Invalid column name "SMTHNG"".
You want to know what the maximum SMTHNG in the table is with:
SELECT MAX(SMTHNG) FROM SomeTable;
This is an aggregation without GROUP BY and hence results in one single row containing the maximum SMTHNG.
Now you also want to know how often this SMTHNG occurs and you add COUNT(MAX(SMTHNG)). This, however, does not work, because you can not aggregate an aggregate directly.
This doesn't work either:
SELECT ANY_VALUE(max_smthng), COUNT(*)
FROM (SELECT MAX(smthng) AS max_smthng FROM sometable) t;
because the sub query only contains one row, so it's too late to count.
So, either use a sub query and select from the table again:
SELECT ANY_VALUE(smthng), COUNT(*)
FROM sometable
WHERE smthng = (SELECT MAX(smthng) FROM sometable);
Or count per SMTHNG before looking for the maximum. Here is how to get the counts:
SELECT smthng, COUNT(*)
FROM sometable
GROUP BY smthng;
And the easiest way to get the maximum from this result is:
SELECT TOP(1) smthng, COUNT(*)
FROM sometable
GROUP BY smthng
ORDER BY COUNT(*) DESC;
First of all, please read my comment.
Depending on what you're trying to achieve, the statement have to be changed.
If you want to count the highest values in SMTHNG field, you may try this:
SELECT T1.SMTHNG, COUNT(T1.SMTHNG)
FROM SomeTable T1 INNER JOIN
(
SELECT MAX(SMTHNG) AS A
FROM SomeTable
) T2 ON T1.SMTHNG = T2.A
GROUP BY T1.SMTHNG;
use cte like below or subquery
with cte as
(
select count(*) as cnt ,col from table_name
group by col
) select max(cnt) from cte
you can not use double aggregate function at a time on same column
I have a table with data where in Column A I have groups of repeating Data (one after another).
I want to select only first row of each group based on values in column A only (no other criteria). Mind you, I want all corresponding columns selected also for the mentioned new found row (I don't want to exclude them).
Can someone help me with a proper query.
Here is a sample:
SAMPLE
Thanks!
#standardSQL
SELECT row.*
FROM (
SELECT ARRAY_AGG(t LIMIT 1)[OFFSET(0)] row
FROM `project.dataset.table` t
GROUP BY columnA
)
you can try smth like this:
#standardSQL
SELECT
* EXCEPT(rn)
FROM (
SELECT
*,
ROW_NUMBER() OVER(PARTITION BY columnA ORDER BY columnA) AS rn
FROM
your_dataset.your_table)
WHERE rn = 1
that will return:
Row columnA col2 ...
1 AC1001 Z_Creation
2 ACO112BISPIC QN
...
Add LIMIT 1 at the end of the query
something like
SELECT name, year FROM person_table ORDER BY year LIMIT 1
You can now use qualify for a more concise solution:
select
*
from
your_dataset.your_table
where true
qualify ROW_NUMBER() OVER(PARTITION BY columnA ORDER BY columnA) = 1
In BigQuery the physical sequence of rows is not significant. “BigQuery does not guarantee a stable ordering of rows in a table. Only the result of a query with an explicit ORDER BY clause has well-defined ordering.”[1].
First, you need to define which property will determine the first row of your group, then you can run Vasily Bronsky’s query by changing ORDER BY with that property. Which means either you should add another column to the table to store the order of the rows or select one from the columns you have.
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.
All these attempts fail:
SELECT DISTINCT(row1,row2), * FROM foo
SELECT *, DISTINCT (row1,row2) FROM foo
SELECT *, DISTINCT row1,row2 FROM foo
I want to fetch all rows but not duplicate rows on row1 & row2
PS. I don't want to use GROUP By.
Thanks!
You can only use DISTINCT on the entire row, not part of it.
PS. I don't want to use GROUP By.
That's a shame, because GROUP BY would have been the correct way to do this.
I need, if possible, a t-sql query that, returning the values from an arbitrary table, also returns a incremental integer column with value = 1 for the first row, 2 for the second, and so on.
This column does not actually resides in any table, and must be strictly incremental, because the ORDER BY clause could sort the rows of the table and I want the incremental row in perfect shape always.
The solution must run on SQL Server 2000
For SQL 2005 and up
SELECT ROW_NUMBER() OVER( ORDER BY SomeColumn ) AS 'rownumber',*
FROM YourTable
for 2000 you need to do something like this
SELECT IDENTITY(INT, 1,1) AS Rank ,VALUE
INTO #Ranks FROM YourTable WHERE 1=0
INSERT INTO #Ranks
SELECT SomeColumn FROM YourTable
ORDER BY SomeColumn
SELECT * FROM #Ranks
Order By Ranks
see also here Row Number
You can start with a custom number and increment from there, for example you want to add a cheque number for each payment you can do:
select #StartChequeNumber = 3446;
SELECT
((ROW_NUMBER() OVER(ORDER BY AnyColumn)) + #StartChequeNumber ) AS 'ChequeNumber'
,* FROM YourTable
will give the correct cheque number for each row.
Try ROW_NUMBER()
http://msdn.microsoft.com/en-us/library/ms186734.aspx
Example:
SELECT
col1,
col2,
ROW_NUMBER() OVER (ORDER BY col1) AS rownum
FROM tbl
It is ugly and performs badly, but technically this works on any table with at least one unique field AND works in SQL 2000.
SELECT (SELECT COUNT(*) FROM myTable T1 WHERE T1.UniqueField<=T2.UniqueField) as RowNum, T2.OtherField
FROM myTable T2
ORDER By T2.UniqueField
Note: If you use this approach and add a WHERE clause to the outer SELECT, you have to added it to the inner SELECT also if you want the numbers to be continuous.