I am trying to find count and distinct of multiple values but its not worikng in db2
select count(distinct col1, col2) from table
it throws syntax error that count has multiple columns.
any way to achieve this
column 1 column 2 date
1 a 2022-12-01
1 a 2022-12-01
2 a 2022-11-30
2 b 2022-11-30
1 b 2022-12-01
i want output
column1 column2 date count
1 a 2022-12-01 2
2 a 2022-11-30 1
2 b 2022-11-30 1
1 a 2022-12-01 1
The following query returns exactly what you want.
WITH MYTAB (column1, column2, date) AS
(
VALUES
(1, 'a', '2022-12-01')
, (1, 'a', '2022-12-01')
, (2, 'a', '2022-11-30')
, (2, 'b', '2022-11-30')
, (1, 'b', '2022-12-01')
)
SELECT
column1
, column2
, date
, COUNT (*) AS CNT
FROM MYTAB
GROUP BY
column1
, column2
, date
COLUMN1
COLUMN2
DATE
CNT
1
a
2022-12-01
2
1
b
2022-12-01
1
2
a
2022-11-30
1
2
b
2022-11-30
1
fiddle
Not exactly sure of what you are looking for...
but
select count(distinct col1), count(distinct col2) from table
or
select count(distinct col1 CONCAT col2) from table
Are how I would interpret "distinct count of multiple values" in a table..
Related
I have a table of datestamped events that I need to bundle into 7-day groups, starting with the earliest occurrence of each event_id.
The final output should return each bundle's start and end date and 'value' column of the most recent event from each bundle.
There is no predetermined start date, and the '7-day' windows are arbitrary, not 'week of the year'.
I've tried a ton of examples from other posts but none quite fit my needs or use things I'm not sure how to refactor for BigQuery
Sample Data;
Event_Id
Event_Date
Value
1
2022-01-01
010203
1
2022-01-02
040506
1
2022-01-03
070809
1
2022-01-20
101112
1
2022-01-23
131415
2
2022-01-02
161718
2
2022-01-08
192021
3
2022-02-12
212223
Expected output;
Event_Id
Start_Date
End_Date
Value
1
2022-01-01
2022-01-03
070809
1
2022-01-20
2022-01-23
131415
2
2022-01-02
2022-01-08
192021
3
2022-02-12
2022-02-12
212223
You might consider below.
CREATE TEMP FUNCTION cumsumbin(a ARRAY<INT64>) RETURNS INT64
LANGUAGE js AS """
bin = 0;
a.reduce((c, v) => {
if (c + Number(v) > 6) { bin += 1; return 0; }
else return c += Number(v);
}, 0);
return bin;
""";
WITH sample_data AS (
select 1 event_id, DATE '2022-01-01' event_date, '010203' value union all
select 1 event_id, '2022-01-02' event_date, '040506' value union all
select 1 event_id, '2022-01-03' event_date, '070809' value union all
select 1 event_id, '2022-01-20' event_date, '101112' value union all
select 1 event_id, '2022-01-23' event_date, '131415' value union all
select 2 event_id, '2022-01-02' event_date, '161718' value union all
select 2 event_id, '2022-01-08' event_date, '192021' value union all
select 3 event_id, '2022-02-12' event_date, '212223' value
),
binning AS (
SELECT *, cumsumbin(ARRAY_AGG(diff) OVER w1) bin
FROM (
SELECT *, DATE_DIFF(event_date, LAG(event_date) OVER w0, DAY) AS diff
FROM sample_data
WINDOW w0 AS (PARTITION BY event_id ORDER BY event_date)
) WINDOW w1 AS (PARTITION BY event_id ORDER BY event_date)
)
SELECT event_id,
MIN(event_date) start_date,
ARRAY_AGG(
STRUCT(event_date AS end_date, value) ORDER BY event_date DESC LIMIT 1
)[OFFSET(0)].*
FROM binning GROUP BY event_id, bin;
I have a need to select all rows from a table (main table) and join to another table (child table). In the results set, I want to include one column from the child table, that is only the first row / line number with a column value in a specified list. If there is no match for the specified list, it should be (null)
Desired Result:
ORDER_NO
ORDER_DATE
ORDER CUST
ORDER_VALUE
ITEM
1
02/14/2022
12345
$1,000.00
APPLES
2
02/13/2022
67890
$5,000.00
(null)
3
02/12/2022
45678
$100.00
PEARS
Example:
Main Table: Order Table
Order Number (Handle)
Order Date,
Order Customer,
Order Value
ORDER_NO
ORDER_DATE
ORDER CUST
ORDER_VALUE
1
02/14/2022
12345
$1,000.00
2
02/13/2022
67890
$5,000.00
3
02/12/2022
45678
$100.00
Child Table: Order Details Tbl
Order Number (Handle)
Line Number = Order Line No
Ordered Item,
Ordered Qty
ORDER_NO
LINE_NO
ITEM
1
10
APPLES
1
20
ORANGES
1
30
LETTUCE
2
10
BROCCOLI
2
20
CAULIFLOWER
2
30
LETTUCE
3
10
KALE
3
20
RADISHES
3
30
PEARS
In this example, the returned column is essentially the first line of the order that is a fruit, not a vegetable. And if the order includes no matching fruit, null is returned.
What my code is thus far:
SELECT
MAIN.ORDER_NO,
MAIN.ORDER_DATE,
MAIN.ORDER_CUST,
MAIN.ORDER_VALUE,
B.ITEM
FROM
MAIN
LEFT JOIN
(
SELECT
CHILD.ORDER_NO,
CHILD.LINE_NO,
CHILD.ITEM
FROM
CHILD
WHERE
CHILD.ORDER_NO||'_'||LINE_NO IN
(
SELECT
CHILD.ORDER_NO||'_'||MIN(LINE_NO) AS ORDER_LINE_NO
FROM
CHILD
WHERE
CHILD.ITEM IN ('APPLES','ORANGES','PEACHES','PEARS','GRAPES')
GROUP BY
CHILD.ORDER_NO
)
) B ON MAIN.ORDER_NO = B.ORDER_NO
'''
This code is of course not working as desired, as table 'B' is including all results from CHILD.
From Oracle 12, you can use:
SELECT o.*,
d.item
FROM orders o
LEFT OUTER JOIN LATERAL(
SELECT *
FROM order_details d
WHERE o.order_no = d.order_no
AND item IN ('APPLES','ORANGES','PEACHES','PEARS','GRAPES')
ORDER BY line_no ASC
FETCH FIRST ROW ONLY
) d
ON (1 = 1)
In earlier versions you can use:
SELECT o.*,
d.item
FROM orders o
LEFT OUTER JOIN(
SELECT d.*,
ROW_NUMBER() OVER (PARTITION BY order_no ORDER BY line_no ASC)
AS rn
FROM order_details d
WHERE item IN ('APPLES','ORANGES','PEACHES','PEARS','GRAPES')
) d
ON (o.order_no = d.order_no AND rn = 1)
Which, for the sample data:
CREATE TABLE orders (ORDER_NO, ORDER_DATE, ORDER_CUST, ORDER_VALUE) AS
SELECT 1, DATE '2022-02-14', 12345, 1000.00 FROM DUAL UNION ALL
SELECT 2, DATE '2022-02-13', 67890, 5000.00 FROM DUAL UNION ALL
SELECT 3, DATE '2022-02-12', 45678, 100.00 FROM DUAL;
CREATE TABLE Order_Details (ORDER_NO, LINE_NO, ITEM) AS
SELECT 1, 10, 'APPLES' FROM DUAL UNION ALL
SELECT 1, 20, 'ORANGES' FROM DUAL UNION ALL
SELECT 1, 30, 'LETTUCE' FROM DUAL UNION ALL
SELECT 2, 10, 'BROCCOLI' FROM DUAL UNION ALL
SELECT 2, 20, 'CAULIFLOWER' FROM DUAL UNION ALL
SELECT 2, 30, 'LETTUCE' FROM DUAL UNION ALL
SELECT 3, 10, 'KALE' FROM DUAL UNION ALL
SELECT 3, 20, 'RADISHES' FROM DUAL UNION ALL
SELECT 3, 30, 'PEARS' FROM DUAL;
Both output:
ORDER_NO
ORDER_DATE
ORDER_CUST
ORDER_VALUE
ITEM
1
2022-02-14 00:00:00
12345
1000
APPLES
2
2022-02-13 00:00:00
67890
5000
null
3
2022-02-12 00:00:00
45678
100
PEARS
db<>fiddle here
I have a postgres table test_table that looks like this:
date | test_hour
------------+-----------
2000-01-01 | 1
2000-01-01 | 2
2000-01-01 | 3
2000-01-02 | 1
2000-01-02 | 2
2000-01-02 | 3
2000-01-02 | 4
2000-01-03 | 1
2000-01-03 | 2
I need to select all the dates which don't have test_hour = 1, 2, and 3, so it should return
date
------------
2000-01-03
Here is what I have tried:
SELECT date FROM test_table WHERE test_hour NOT IN (SELECT generate_series(1,3));
But that only returns dates that have extra hours beyond 1, 2, 3
You can use aggregation and conditional HAVING clauses, like so:
SELECT mydate
FROM mytable
GROUP BY mydate
HAVING
MAX(CASE WHEN test_hour = 1 THEN 1 END) != 1
OR MAX(CASE WHEN test_hour = 2 THEN 1 END) != 1
OR MAX(CASE WHEN test_hour = 3 THEN 1 END) != 1
Another possibility would be to join it against the series (or another subquery containing the hours) and do a [distinct] count on the hours aggregatet per date:
select date from tst
inner join (select generate_series(1,3) "hour") hours on hours.hour = tst.hour
group by tst.date
having count(distinct tst.hour) < 3;
or
select date from tst
where hour in (select generate_series(1,3))
group by date
having count(distinct tst.hour) < 3;
[You don't need the distinct if date/hour combinations in Your table are unique]
A solution using set difference, giving you exactly the rows that are missing:
(SELECT DISTINCT
date, all_hour
FROM test_table
CROSS JOIN generate_series(1,3) all_hour)
EXCEPT
(TABLE test_table)
And a solution using an array aggregate and the array contains operator:
SELECT date
FROM test_table
GROUP BY date
HAVING NOT array_agg(test_hour) #> ARRAY(SELECT generate_series(1,3))
(online demos)
Bit stuck on a problem. Trying to find the difference between two dates in postgreSQL.
I have a table emp with many employees in it:
emp_id, date
1, 31-10-2017
1, 08-08-2017
1, 02-06-2017
I want it to look like this:
emp_id, max_date, penultimate_date, difference
1, 31-10-2017, 08-08-2017, 84 days
Obviously you can use max(date) and group by the emp_id, however how do you retrieve the penultimate date. I have used a few functions like:
order by date desc limit 1 offset 1
I have also tried to put these in sub queries but that hasn,t worked as there are many employee numbers and I need one row for each employee.
Can anyone help???
Thanks,
pp84
as kindly suggested by #Haleemur Ali, order by date desc limit 1 offset 1 would not work with several emp_id:
t=# with d(emp_id, date)as (values(1, '31-10-2017'::date),(1, '08-08-2017'),(1, '02-06-2017' ),(2,'2016-01-01'),(2,'2016-02-02'),(2,'2016-03-03'))
select distinct emp_id
, max(date) over (partition by emp_id) max_date
, nth_value(date,2) over (partition by emp_id) penultimate_date
, max(date) over (partition by emp_id) - nth_value(date,2) over (partition by emp_id) diff
from d
;
emp_id | max_date | penultimate_date | diff
--------+------------+------------------+------
2 | 2016-03-03 | 2016-02-02 | 30
1 | 2017-10-31 | 2017-08-08 | 84
(2 rows)
Time: 0.756 ms
WITH emps (emp_id, date) AS (
VALUES (1, '2017-10-31'::DATE)
, (1, '2017-08-08'::DATE)
, (1, '2017-08-08'::DATE)
)
SELECT DISTINCT ON (emp_id)
emp_id
, "date" max_date
, LEAD("date") OVER w penultimate_date
, "date" - LEAD("date") OVER w difference
FROM emps
WINDOW w AS (PARTITION BY emp_id)
ORDER BY emp_id, date DESC
When ordered in descending order, the LEAD("date") w will give the value of the date value from the next row.
The DISTINCT ON limits the resultset to 1 row (the first row encountered) per emp_id.
With our ordering this first row must contain the greatest date, and the LEAD(...) over w therefore returns the penultimate date. This gives us the following result:
emp_id | max_date | penultimate_date | difference
--------+------------+------------------+------------
1 | 2017-10-31 | 2017-08-08 | 84
(1 row)
Think of a table like below:
unique_id
a_column
b_column
a_int
b_int
date_created
Let's say data is like:
-unique_id -a_column -b_column -a_int -b_int -date_created
1z23 abc 444 0 1 27.12.2016 18:03:00
2c31 abc 444 0 0 26.12.2016 13:40:00
2e22 qwe 333 0 1 28.12.2016 15:45:00
1b11 qwe 333 1 1 27.12.2016 19:00:00
3a33 rte 333 0 1 15.11.2016 11:00:00
4d44 rte 333 0 1 27.09.2016 18:00:00
6e66 irt 333 0 1 22.12.2016 13:00:00
7q77 aaa 555 1 0 27.12.2016 18:00:00
I want to get the unique_id s where b_int is 1, b_column is 333 and considering a_column, a_int column must always be 0, if there are any records with a_int = 1 even if there are records with a_int = 0 these records must not be shown in the result. Desired result is: " 3a33 , 6e66 " when grouped by a_column and ordered by date_created and got top1 for each unique a_column.
I tried lots of "with ties" and "over(partition by" samples, searched questions, but couldn't manage to do it. This is what I could do:
select unique_id
from the_table
where b_column = '333'
and b_int = 1
and a_column in (select a_column
from the_table
where b_column = '333'
and b_int = 1
group by a_column
having sum(a_int) = 0)
order by date_created desc;
This query returns the result like this " 3a33 ,4d44, 6e66 ". But I don't want "4d44".
You were on the right track with the partitions and window functions. This solution uses ROW_NUMBER to assign a value to the a_column so we can see where there is more than 1. The 1 is the most recent date_created. Then you select from the result set where the row_counter is 1.
;WITH CTE
AS (
SELECT unique_id
, a_column
, ROW_NUMBER() OVER (
PARTITION BY a_column ORDER BY date_created DESC
) AS row_counter --This assigns a 1 to the most recent date_created and partitions by a_column
FROM #test
WHERE a_column IN (
SELECT a_column
FROM #test
WHERE b_column = '333'
AND b_int = 1
GROUP BY a_column
HAVING MAX(a_int) < 1
)
)
SELECT unique_ID
FROM cte
WHERE row_counter = 1