finding the duplicate column depending on the other column - sql

I have 3 columns as
Column1 column2 column3
GU1 1 a
GU1 2 a
GU1 3 a
GU2 4 b
GU3 5 c
GU4 6 a
GU4 7 b
I would like to filter out the column1 where the values are having multiple column3 values
In this example, I want to pull GU4 where it is having both a & b in the column3.

Use distinct in your count
select column1
from your_table
group by column1
having count(distinct column3) = 1

Related

Divide Column in Half in BigQuery

using BigQuery, I would like to be able to divide one column, column1, into two separate columns, column2, and column3 with 50% of all records in column1 in column2 and 50% of all records in column1 in column 3. Ex column1 has 8 records of the number 2. I'd like to create a column2 with 4 records of the number 2 and column3 with 4 records of the number 2.
Is there a query to write this in BigQuery?
Column1
2
2
2
2
2
2
2
2
Column2
2
2
2
2
Column3
2
2
2
2
try:
SELECT
Column1 AS Column2
FROM `my-project.my-dataset.my-table`
WHERE 1=1
QUALIFY ROW_NUMBER() OVER (ORDER BY Column1) <= (
SELECT COUNT(*)/2
FROM `my-project.my-dataset.my-table`
);
SELECT
Column1 AS Column3
FROM `my-project.my-dataset.my-table`
WHERE 1=1
QUALIFY ROW_NUMBER() OVER (ORDER BY Column1) > (
SELECT COUNT(*)/2
FROM `my-project.my-dataset.my-table`
);
This will give you 2 results: One for each Column2 and Column3 with the first and second half of the data respectively order by Column1 (to use analytical functions you always have to specify an ORDER BY inside an OVER clause)
For random order try:
CREATE TEMP TABLE a AS (
SELECT Column1 as Column2
FROM `my-project.my-dataset.my-table`
WHERE 1=1
QUALIFY
ROW_NUMBER() OVER (ORDER BY RAND()) <= (SELECT COUNT(*)/2 FROM `my-project.my-dataset.my-table`)
);
SELECT Column1 as Column3
FROM `my-project.my-dataset.my-table`
WHERE Column1 NOT IN (SELECT * FROM a);
SELECT * FROM a
In this case you'll get 3 results: first one is the temporary table creation and the other 2 are the columns 2 and 3.

SQL sum values in columns for each row

I have the following table:
column1 | column2 | column3
1 3 4
5 7 6
how do I sum the values of say, column 2 and 3, to return the sum?
The expected result is:
res
7
13
You can do maths within a select statement, so the following will work:
SELECT column2 + column3 AS res FROM table
This works in postgresql.
select sum(col2+col3) from (
select col1, col2,col3,row_number() over() as rows from column_sum ) as foo
group by rows order by rows;

Show 2 columns in 2 different rows - SQL

I need to separate two columns into two rows in sql
I have this:
Column1 Column2 Column3
Car 2 5
Boat 4
Truck 6
And I want this:
Column1 Column2
Car 2
Car 5
Boat 4
Truck 6
How can I do this in SQL?
This operation is unpivoting. I would recommend apply:
select t.column1, v.col
from t cross apply
(values (col1), (col2)) v(col)
where v.col is not null;
This should do it
select Column1, Column2
from tbl where Column2 is not null and Column2 <> ''
union
select Column1, Column3
from tbl where Column3 is not null and Column3 <> ''

MS SQL count AS to new column

Good day
I have problem with my table and counting
TABLE1
COLUMN1 COLUMN2
3 jjd
5 jd
3 jjd
4 kg
5 jd
48 gjh
446 djj
… …
I need
TABLE1
COLUMN1 COLUMN2 COLUMN3
3 jj 2
5 jd 2
4 kg 1
48 gjh 1
446 djj 1
... ... …
Iam doing but not working well.
SELECT * , COUNT(Column1) as column3 FROM TABLE1
Thanks for help withh my counting
Use a GROUP BY and ORDER BY with DESC to put them in COUNT total order.
SELECT COLUMN1, COLUMN2, COUNT(Column1) AS COLUMN3
FROM Table1
GROUP BY COLUMN1, COLUMN2
ORDER BY COUNT(Column1) DESC
Output
COLUMN1 COLUMN2 COLUMN3
5 jd 2
3 jjd 2
4 kg 1
446 djj 1
48 gjh 1
SQL Fiddle: http://sqlfiddle.com/#!6/89f49/4/0
Try by using group by
SELECT COLUMN1 ,
COLUMN2 ,
COUNT(Column1) As COLUMN3 FROM cte_TABLE1
Group by COLUMN1 ,COLUMN2
Order by COLUMN1
By using Window function
SELECT DISTINCT COLUMN1 ,
COLUMN2 ,
COUNT(Column1)OVER(Partition by COLUMN1,COLUMN2 ORder by COLUMN1 ) As COLUMN3 FROM cte_TABLE1
Result
COLUMN1 COLUMN2 column3
-----------------------
3 jjd 2
4 kg 1
5 jd 2
48 gjh 1
446 djj 1
Using OVER we can achieve it easily
SELECT COLUMN1 ,
COLUMN2 ,
COUNT(Column1)OVER(Partition by COLUMN1,COLUMN2 ORder by COLUMN1 ) As COLUMN3 FROM cte_TABLE1
Group By COLUMN1,COLUMN2

Take values of different columns in the same 1 column table

I am working on Ms Sql Server 2008 R2.
now i want one column which has results of 3 different columns of same table..
Let me explain with Figure.
Table: mainTable
Id Column1 Column2 Column3
-------------------------------------
1 urla urlb urlc
2 urld urle urlf
3 urlg urlh urli
Now i want one column
Table Name: ResultTable
Id ColumnResult
-------------
1 urla
2 urlb
3 urlc
4 urld
5 urle
6 urlf
7 urlg
8 urlh
9 urli
Thanks in Advance.
Sahil Patel
You can use UNION ALL:
SELECT
Id = ROW_NUMBER() OVER(ORDER BY ColumnResult),
Column1
FROM (
SELECT Id, Column1 AS ColumnResult FROM mainTable UNION ALL
SELECT Id, Column2 FROM mainTable UNION ALL
SELECT Id, Column3 FROM mainTable
)t