I have the following data in a table, this is a single column shown from a table that has multiple columns, but only data from this column needs to be pulled into two column output using a query:
+----------------+--+
| DataText | |
| 1 DEC20 DDD | |
| 1 JUL20 DDD | |
| 1 JAN21 DDD | |
| 1 JUN20 DDD500 | |
| 1 JUN20 DDD500 | |
| 1 JUN20DDDD500 | |
| 1 JUN20DDDD500 | |
| 1 JUL20 DDD800 | |
| 1 JUL20 DDD800 | |
| 1 JUL20DDDD800 | |
| 1 JUL20DDDD400 | |
| 1 JUL20DDDD400 | |
+----------------+--+
Required result: distinct values based on the first 13 characters of the data, split into two columns based on "long data", and "short data", BUT only giving the first 13 characters in output for both columns:
+-------------+-------------+
| ShortData | LongData |
| 1 DEC20 DDD | 1 JUN20 DDD |
| 1 JUL20 DDD | 1 JUN20DDDD |
| 1 JAN21 DDD | 1 JUL20 DDD |
| | 1 JUL20DDDD |
+-------------+-------------+
Something like:
Select
(Select DISTINCT LEFT(DataText,13)
From myTable)
Where LEN(DataText)=13) As ShortData
,
(Select DISTINCT LEFT(DataText,13)
From myTable)
Where LEN(DataText)>13) As LongData
I would also like to query/"scan" the table only once if possible. I can't get any of the SO examples modified to make such a query work.
This is quite ugly, but doable. As a starter, you need a column that defines the order of the rows - I assumed that you have such a column, and that is called id.
Then you can select the distinct texts, put them in separate groups depending on their length, and finally pivot:
select
max(case when grp = 0 then dataText end) shortData,
max(case when grp = 1 then dataText end) longData
from (
select
dataText,
grp,
row_number() over(partition by grp order by id) rn
from (
select
id,
case when len(dataText) <= 13 then 0 else 1 end grp,
substring(dataText, 1, 13) dataText
from (select min(id) id, dataText from mytable group by dataText) t
) t
) t
group by rn
If you are content with ordering the records by the string column itself, it is a bit simpler (and, for your sample data, it produces the same results):
select
max(case when grp = 0 then dataText end) shortData,
max(case when grp = 1 then dataText end) longData
from (
select
dataText,
grp,
row_number() over(partition by grp order by dataText) rn
from (
select distinct
case when len(dataText) <= 13 then 0 else 1 end grp,
substring(dataText, 1, 13) dataText
from mytable
) t
) t
group by rn
Demo on DB Fiddle:
shortData | longData
:---------- | :------------
1 DEC20 DDD | 1 JUL20 DDD80
1 JAN21 DDD | 1 JUL20DDDD40
1 JUL20 DDD | 1 JUL20DDDD80
null | 1 JUN20 DDD50
null | 1 JUN20DDDD50
Related
I have a table:
| id | Number |Address
| -----| ------------|-----------
| 1 | 0 | NULL
| 1 | 1 | NULL
| 1 | 2 | 50
| 1 | 3 | NULL
| 2 | 0 | 10
| 3 | 1 | 30
| 3 | 2 | 20
| 3 | 3 | 20
| 4 | 0 | 75
| 4 | 1 | 22
| 4 | 2 | 30
| 5 | 0 | NULL
I need to get: the NUMBER of the last ADDRESS change for each ID.
I wrote this select:
select dh.id, dh.number from table dh where dh =
(select max(min(t.history)) from table t where t.id = dh.id group by t.address)
But this select not correctly handling the case when the address first changed, and then changed to the previous value. For example id=1: group by return:
| Number |
| -------- |
| NULL |
| 50 |
I have been thinking about this select for several days, and I will be happy to receive any help.
You can do this using row_number() -- twice:
select t.id, min(number)
from (select t.*,
row_number() over (partition by id order by number desc) as seqnum1,
row_number() over (partition by id, address order by number desc) as seqnum2
from t
) t
where seqnum1 = seqnum2
group by id;
What this does is enumerate the rows by number in descending order:
Once per id.
Once per id and address.
These values are the same only when the value is 1, which is the most recent address in the data. Then aggregation pulls back the earliest row in this group.
I answered my question myself, if anyone needs it, my solution:
select * from table dh1 where dh1.number = (
select max(x.number)
from (
select
dh2.id, dh2.number, dh2.address, lag(dh2.address) over(order by dh2.number asc) as prev
from table dh2 where dh1.id=dh2.id
) x
where NVL(x.address, 0) <> NVL(x.prev, 0)
);
fellow stackers
I have a data set like so:
+---------+------+--------+
| user_id | date | metric |
+---------+------+--------+
| 1 | 1 | 1 |
| 1 | 2 | 1 |
| 1 | 3 | 1 |
| 2 | 1 | 1 |
| 2 | 2 | 1 |
| 2 | 3 | 0 |
| 2 | 4 | 1 |
+---------+------+--------+
I am looking to flag those customers who has 3 consecutive "1"s in the metric column. I have a solution as below.
select distinct user_id
from (
select user_id
,metric +
ifnull( lag(metric, 1) OVER (PARTITION BY user_id ORDER BY date), 0 ) +
ifnull( lag(metric, 2) OVER (PARTITION BY user_id ORDER BY date), 0 )
as consecutive_3
from df
) b
where consecutive_3 = 3
While it works it is not scalable. As one can imagine what the above query would look like if I were looking for a consecutive 50.
May I ask if there is a scalable solution? Any cloud SQL will do. Thank you.
If you only want such users, you can use a sum(). Assuming that metric is only 0 or 1:
select user_id,
(case when max(metric_3) = 3 then 1 else 0 end) as flag_3
from (select df.*,
sum(metric) over (partition by user_id
order by date
rows between 2 preceding and current row
) as metric_3
from df
) df
group by user_id;
By using a windowing clause, you can easily expand to as many adjacent 1s as you like.
Consider the following dataset:
+---------------------+
| ID | NAME | VALUE |
+---------------------+
| 1 | a | 0.2 |
| 1 | b | 8 |
| 1 | c | 3.5 |
| 1 | d | 2.2 |
| 2 | b | 4 |
| 2 | c | 0.5 |
| 2 | d | 6 |
| 3 | a | 2 |
| 3 | b | 4 |
| 3 | c | 3.6 |
| 3 | d | 0.2 |
+---------------------+
I'm tying to develop a sql select statement that returns the top or distinct ID where NAME 'a' and 'b' both exist and both of the corresponding VALUE's are >= '1'. Thus, the desired output would be:
+---------------------+
| ID | NAME | VALUE |
+---------------------+
| 3 | a | 2 |
+----+-------+--------+
Appreciate any assistance anyone can provide.
You can try to use MIN window function and some condition to make it.
SELECT * FROM (
SELECT *,
MIN(CASE WHEN NAME = 'a' THEN [value] end) OVER(PARTITION BY ID) aVal,
MIN(CASE WHEN NAME = 'b' THEN [value] end) OVER(PARTITION BY ID) bVal
FROM T
) t1
WHERE aVal >1 and bVal >1 and aVal = [Value]
sqlfiddle
This seems like a group by and having query:
select id
from t
where name in ('a', 'b')
having count(*) = 2 and
min(value) >= 1;
No subqueries or joins are necessary.
The where clause filters the data to only look at the "a" and "b" records. The count(*) = 2 checks that both exist. If you can have duplicates, then use count(distinct name) = 2.
Then, you want the minimum value to be 1, so that is the final condition.
I am not sure why your desired results have the "a" row, but if you really want it, you can change the select to:
select id, 'a' as name,
max(case when name = 'a' then value end) as value
you can use in and sub-query
select top 1 * from t
where t.id in
(
select id from t
where name in ('a','b')
group by id
having sum(case when value>1 then 1 else 0)>=2
)
order by id
I have a table in the form of :
| ID | COURSE | PASS |
---------------------------
| 1 | 1 | 1 |
| 1 | 2 | 1 |
| 1 | 3 | 1 |
| 1 | 4 | 0 |
| 1 | 5 | 0 |
and I want row in the form:
| ID | FAILED | PASSED |
---------------------------
| 1 | 4,5 | 1,2,3 |
the only i figured is something like this:
select NVL(passed.id, failed.id), passed.test, failed.test from
(select id, listagg(course, ',') within group (order by course) test from table1 where pass = 1 group by id ) passed
full outer join
(select id, listagg(course, ',') within group (order by course) test from table1 where pass = 0 group by id ) failed
on passed.id = failed.id
is there a way to do it in a single query ?
Try
select id,
listagg(case when pass = 1 then course end, ',') within group (order by course) passed,
listagg(case when pass = 0 then course end, ',') within group (order by course) failed
from table1
group by id
Here is a sqlfiddle demo
I am doing a row_number() over some data.
My problem is that I do not wnat to take into account data that is 0.
Here's my data sample:
+-----------+-----+
| Alex | 1 |
| Liza | 2 |
| Harry | 0 |
| Marge | 24 |
| Bla | 0 |
| Something | 234 |
+-----------+-----+
Here's what I want:
+-----------+--------+------------+
| name | number | row_number |
+-----------+--------+------------+
| Harry | 0 | 0 |
| Bla | 0 | 0 |
| Something | 234 | 1 |
| Marge | 24 | 2 |
| Liza | 2 | 3 |
| Alex | 1 | 4 |
+-----------+--------+------------+
as you can see the third column is the row_number()
this is what I have so far:
select name, number, row_number() over (partition by name order by name,number)
from myTable
How do I get the query above to return 0 for all 0's in the source data and not count the 0's at all towards the row_number sequence?
Try this SQL Fiddle:
SELECT name, number, 0 as [row_number]
FROM myTable
WHERE number = 0
UNION ALL
SELECT name, number, row_number() OVER (ORDER BY number DESC)
FROM myTable
WHERE number <> 0
Try this:
select name, number,
(case when number = 0 then 0
else row_number() over (partition by iszero order by number desc)
end) as row_number
from (select t.*, (case when number = 0 then 0 else 1 end) as iszero
from myTable t
) t
Since you are ordering by numbers descending, the following works if no numbers are negative:
select name, number,
(case when number = 0 then 0
else row_number() over (order by number desc)
end) as row_number
from myTable t