use LAG with expression in oracle - oracle

I have a column (status) in a table that contain numbers and values are 1, 2 or 4.
I would like, in a SQL query, add a calculated column (bitStatus) that will store the bitwise oerator OR for the status column of the current line and the column in the previous line.
like so :
| id | status| bitStatus|
|----|-------|----------|
| 1 | 1 | 1 |
| 2 | 2 | 3 |
| 3 | 4 | 7 |
| 4 | 1 | 7 |
So what I did is to use LAG function in oracle but I coudn't figure out how to do it as long as I want to create only on calculated column bitStatus
my query is like :
select id, status,
BITOR(LAG(bitStatus) OVER (ORDER BY 1), status)) AS bitStatus
But as you know, I can't use LAG(bitStatus) when calculating bitStatus.
So how could I make it the desired table.
Thanks in advance.

Would this help?
lines #1 - 6 represent sample data
the TEMP CTE is here to fetch LAG status value (to improve readability)
the final select does the BITOR operation as bitor(a, b) = a - bitand(a, b) + b
SQL> with test (id, status) as
2 (select 1, 1 from dual union all
3 select 2, 2 from dual union all
4 select 3, 1 from dual union all
5 select 4, 4 from dual
6 ),
7 temp as
8 (select id, status,
9 lag(status) over (order by id) lag_status
10 from test
11 )
12 select id,
13 status,
14 status - bitand(status, nvl(lag_status, status)) + nvl(lag_status, status) as bitstatus
15 from temp
16 order by id;
ID STATUS BITSTATUS
---------- ---------- ----------
1 1 1
2 2 3
3 1 3
4 4 5
SQL>

Related

SQL help to count number of locations for each item/branch

I'm a SQL rookie, and am having trouble wrapping my head around how to do the following. I have a table that contains item information by branch. Within a branch an item can be in multiple locations. The data I need to extract needs to include a column that provides the total number of locations (count) the item is associated with for a given branch.
Output would look something like this:
I'm guessing this is a sub query, but to be honest I'm not sure how to get started... order in which this is done (subquery group by first, then join, etc)
In purely logical terms:
SELECT
a.Branch,
a.Item,
a.Loc,
COUNT(a.Branch||a.Item) AS 'LocCount'
FROM BranchInventoryFile a
GROUP BY a.Branch,a.Item
You can tackle this by using Oracle's Count Analytical functions found here. Be sure to read up on WINDOW/Partitioning functions as this unlocks quite a bit of functionality in SQL.
SQL:
SELECT
a.BRANCH,
a.ITEM,
a.LOC,
COUNT(a.ITEM) OVER (PARTITION BY a.BRANCH, a.ITEM) AS LOC_COUNT
FROM
BRANCH a;
Result:
| BRANCH | ITEM | LOC | LOC_COUNT |
|--------|------|------|-----------|
| 100 | A | 1111 | 2 |
| 100 | A | 1112 | 2 |
| 200 | A | 2111 | 1 |
| 200 | B | 1212 | 2 |
| 200 | B | 1212 | 2 |
| 300 | A | 1222 | 1 |
SQL Fiddle:
Here
total number of locations (count) the item is associated with for a given branch
The way you described it, you should
remove location from query:
SQL> with branchinventoryfile (branch, item, location) as
2 (select 100, 'A', 1111 from dual union all
3 select 100, 'A', 1112 from dual union all
4 select 200, 'A', 2111 from dual
5 )
6 select branch,
7 item,
8 count(distinct location) cnt
9 from BranchInventoryFile
10 group by branch, item;
BRANCH I CNT
---------- - ----------
100 A 2
200 A 1
SQL>
if you leave location in select, you have to group by it (and get wrong result):
6 select branch,
7 item,
8 location,
9 count(distinct location) cnt
10 from BranchInventoryFile
11 group by branch, item, location;
BRANCH I LOCATION CNT
---------- - ---------- ----------
100 A 1111 1
200 A 2111 1
100 A 1112 1
SQL>
or include locations, but aggregate them, e.g.
6 select branch,
7 item,
8 listagg(location, ', ') within group (order by null) loc,
9 count(distinct location) cnt
10 from BranchInventoryFile
11 group by branch, item;
BRANCH I LOC CNT
---------- - -------------------- ----------
100 A 1111, 1112 2
200 A 2111 1
SQL>

how to loop through each row of every group (while doing "group by") in Oracle table

I have a table like this:
I want to group by the table base on "customer_id" column and calculate "Day-day[0]" column. "Day-day[0]" is "Day" field in every group and "day[0]" is first row of the day in the group. At the same time, I have to calculate total risk which is in following:
This is the table after grouping by:
This is total risk formula:
In fact, I have to loop through each row of every group to calculate total risk.
My sample table is like this:
CREATE TABLE risk_test
(id VARCHAR2 (32) NOT NULL PRIMARY KEY,
customer_id varchar2 (40BYTE),
risk number,
day VARCHAR2(50 BYTE))
insert into risk_test values(1,102,15,1);
insert into risk_test values(2,102,16,1);
insert into risk_test values(3,104,11,1);
insert into risk_test values(4,102,17,2);
insert into risk_test values(5,102,10,2);
insert into risk_test values(6,102,13,3);
insert into risk_test values(7,104,14,2);
insert into risk_test values(8,104,13,2);
insert into risk_test values(9,104,17,1);
insert into risk_test values(10,104,16,2);
The sample answer is like this:
Would you please guide me how I can do this scenario in Oracle database?
Any help is really appreciated.
Using the sample data that was provided, I believe this query should calculate the risks properly:
Query
SELECT o.*,
ROUND (
SUM (day_minus_day0 * risk) OVER (PARTITION BY customer_id)
/ SUM (day_minus_day0) OVER (PARTITION BY customer_id),
5) AS total_risk
FROM (SELECT rt.*, (rt.day - MIN (rt.day) OVER (PARTITION BY customer_id)) + 1 AS day_minus_day0
FROM risk_test rt) o
ORDER BY customer_id, TO_NUMBER (day), TO_NUMBER (id);
Result
ID CUSTOMER_ID RISK DAY DAY_MINUS_DAY0 TOTAL_RISK
_____ ______________ _______ ______ _________________ _____________
1 102 15 1 1 13.77778
2 102 16 1 1 13.77778
4 102 17 2 2 13.77778
5 102 10 2 2 13.77778
6 102 13 3 3 13.77778
3 104 11 1 1 14.25
9 104 17 1 1 14.25
7 104 14 2 2 14.25
8 104 13 2 2 14.25
10 104 16 2 2 14.25
Your total risk calculation just looks like a weighted average to me. That is, the average risk of the rows for each customer, weighted according to the day offset (day-day[0]), so that risks in later days count for more.
To compute that, you need a common table expression to 1st compute the day-weighted risk for each row. Then you can just compute the weighted average by dividing.
The query below illustrates the approach, with comments.
-- This first WITH clause is just sample data. In your database you would
-- get rid of this and replace all references to "input" with your actual
-- table name
with input ( customer_id, risk, day ) AS (
SELECT 1053, 100, 1 FROM DUAL UNION ALL
SELECT 1053, 100, 1 FROM DUAL UNION ALL
SELECT 1053, 100, 2 FROM DUAL UNION ALL
SELECT 1053, 100, 2 FROM DUAL UNION ALL
SELECT 1053, 100, 3 FROM DUAL UNION ALL
SELECT 1054, 200, 1 FROM DUAL UNION ALL
SELECT 1054, 200, 1 FROM DUAL UNION ALL
SELECT 1054, 200, 3 FROM DUAL UNION ALL
SELECT 1054, 200, 3 FROM DUAL UNION ALL
SELECT 1054, 200, 4 FROM DUAL
),
-- This CTE computes the day offset for each row and multiplies by the risk to
-- compute a day-weighted risk.
-- I added +1 to the day_offset, otherwise risks on the 1st day would not contribute
-- to the total risk, which I think is not what you intended(?)
weighted_input AS (
SELECT i.customer_id,
i.risk,
i.day,
i.day - min(i.day) over ( partition by i.customer_id ) + 1 day_offset,
( i.day - min(i.day) over ( partition by i.customer_id ) + 1 ) * i.risk day_weighted_risk
FROM input i )
-- This is the main SELECT clause that gets all the weighted risks and computes
-- the group total risk, which appears the same in every row in each group.
SELECT wi.*,
sum(wi.day_weighted_risk) over ( partition by wi.customer_id ) / sum(wi.day_offset) over ( partition by wi.customer_id ) total_risk
FROM weighted_input wi;
+-------------+------+-----+------------+-------------------+------------+
| CUSTOMER_ID | RISK | DAY | DAY_OFFSET | DAY_WEIGHTED_RISK | TOTAL_RISK |
+-------------+------+-----+------------+-------------------+------------+
| 1053 | 100 | 1 | 1 | 100 | 100 |
| 1053 | 100 | 1 | 1 | 100 | 100 |
| 1053 | 100 | 2 | 2 | 200 | 100 |
| 1053 | 100 | 2 | 2 | 200 | 100 |
| 1053 | 100 | 3 | 3 | 300 | 100 |
| 1054 | 200 | 1 | 1 | 200 | 200 |
| 1054 | 200 | 1 | 1 | 200 | 200 |
| 1054 | 200 | 3 | 3 | 600 | 200 |
| 1054 | 200 | 3 | 3 | 600 | 200 |
| 1054 | 200 | 4 | 4 | 800 | 200 |
+-------------+------+-----+------------+-------------------+------------+
For your database, having the actual table and not needing the input CTE, it would be:
WITH weighted_input AS (
-- This CTE computes the day offset for each row and multiplies by the risk to
-- compute a day-weighted risk.
-- I added +1 to the day_offset, otherwise risks on the 1st day would not contribute
-- to the total risk, which I think is not what you intended(?)
SELECT i.customer_id,
i.risk,
i.day,
i.day - min(i.day) over ( partition by i.customer_id ) + 1 day_offset,
( i.day - min(i.day) over ( partition by i.customer_id ) + 1 ) * i.risk day_weighted_risk
FROM my_table i )
-- This is the main SELECT clause that gets all the weighted risks and computes
-- the group total risk, which appears the same in every row in each group.
SELECT wi.*,
sum(wi.day_weighted_risk) over ( partition by wi.customer_id ) / sum(wi.day_offset) over ( partition by wi.customer_id ) total_risk
FROM weighted_input wi;

Oracle sum group by date range

I have the following table
+-----------+-------+-------+
| Date | Type | Value |
+-----------+-------+-------+
| 1/1/2013 | A | 1 |
| 1/2/2013 | A | 3 |
| 1/3/2013 | A | 5 |
| 1/4/2013 | A | 6 |
| 1/6/2013 | A | 8 |
| 1/7/2013 | A | 1 |
| 1/8/2013 | A | 2 |
+-----------+-------+-------+
I want to sum the value for the previous 3 dates for a certain day so i used this query.
ie: sel_date = 1/3/2013.
select type, sum(value)
from table_name
where date <= seldate
and date > seldate - 3
group by type
Now the problem is, I want to output a table with a given date range computing for the previous 3 days for each date.
ie: sel_date range 1/3/2013 - 1/8/2013
+-----------+-------+------------+
| Date | Type | Sum(Value) |
+-----------+-------+------------+
| 1/3/2013 | A | 9 | // 5 + 3 + 1
| 1/4/2013 | A | 14 | // 6 + 5 + 3
| 1/5/2013 | A | 11 | // 0 + 6 + 5
| 1/6/2013 | A | 14 | // 8 + 0 + 6
| 1/7/2013 | A | 9 | // 1 + 8 + 0
| 1/8/2013 | A | 11 | // 2 + 1 + 8
+-----------+-------+------------+
Is there a way to do this in a single query. I tried reading on partitioning but it is leading me no where.
Use range between in windowing clause:
select dt, type, value,
sum(value) over (order by dt range between 2 preceding and current row) as sv
from t
Test data and output:
create table t (dt date, type varchar2(1), value number(5));
insert into t values (date '2013-01-01', 'A', 1);
insert into t values (date '2013-01-02', 'A', 3);
insert into t values (date '2013-01-03', 'A', 5);
insert into t values (date '2013-01-04', 'A', 6);
insert into t values (date '2013-01-05', 'A', 8);
insert into t values (date '2013-01-06', 'A', 1);
insert into t values (date '2013-01-07', 'A', 2);
insert into t values (date '2013-01-12', 'A', 2);
DT TYPE VALUE SV
----------- ---- ------ ----------
2013-01-01 A 1 1
2013-01-02 A 3 4
2013-01-03 A 5 9
2013-01-04 A 6 14
2013-01-05 A 8 19
2013-01-06 A 1 15
2013-01-07 A 2 11
2013-01-12 A 2 2
You can try with something like this:
with test(Date_, Type, Value ) as
(
select to_date('01/01/2013', 'mm/dd/yyyy'), 'A', 1 from dual union all
select to_date('01/02/2013', 'mm/dd/yyyy'), 'A', 3 from dual union all
select to_date('01/03/2013', 'mm/dd/yyyy'), 'A', 5 from dual union all
select to_date('01/04/2013', 'mm/dd/yyyy'), 'A', 6 from dual union all
select to_date('01/05/2013', 'mm/dd/yyyy'), 'A', 8 from dual union all
select to_date('01/06/2013', 'mm/dd/yyyy'), 'A', 1 from dual union all
select to_date('01/07/2013', 'mm/dd/yyyy'), 'A', 2 from dual
)
select *
from (
select date_, type,
value + nvl(lag(value, 1) over (partition by type order by date_), 0)
+ nvl(lag(value, 2) over (partition by type order by date_), 0) as value
from test
)
where date_ between to_date('01/03/2013', 'mm/dd/yyyy') and to_date('01/07/2013', 'mm/dd/yyyy')
This sums, for each row, the values of the two preceding ones, based on date; the external query is simply used to apply the filter, given that applying it in the internal query would lead to a wrong sum.
The LAG is used to read values from the rows that precede the current row by 1 or 2 positions.
You can use this:
select date1 ,type,
(select sum(t1.value) sumvalue from table_name t1 where t1.date1 between (t2.date1 - 2) and t2.date1 )
from table_name t2
where date1 between startDate and endDate
select t.date, sum(t.value) OVER(ORDER BY t.date ROWS BETWEEN 2 PRECEDING AND 0 FOLLOWING) as Pre_3row_sum
from table_name t

Oracle Lead by group

Hello I've got a problem with lead and retrieving the next value from the next group.
I've got this table:
TableA
-----------------
ID | value
-----------------
1 | 2.5
1 | 1
1 | 4.5
2 | 7
2 | 2
3 | 3
4 | 1
4 | 7
Expected result:
------------------------------
ID | value | lead_id
------------------------------
1 | 2.5 | 2
1 | 1 | 2
1 | 4.5 | 2
2 | 7 | 3
2 | 2 | 3
3 | 3 | 4
4 | 1 | NULL
4 | 7 | NULL
My SQL:
select ID, value, lead(id) OVER (order by id) lead_id from TableA
Is it possible to get that result ?
You can do this by adding in a windowing clause into the first_value analytic function:
with tablea as (select 1 id, 2.5 value from dual union all
select 1 id, 1 value from dual union all
select 1 id, 4.5 value from dual union all
select 2 id, 7 value from dual union all
select 2 id, 2 value from dual union all
select 3 id, 3 value from dual union all
select 4 id, 1 value from dual union all
select 4 id, 7 value from dual)
select id,
value,
first_value(id) over (order by id
range between 1 following and unbounded following) lead_id
from tablea;
ID VALUE LEAD_ID
---------- ---------- ----------
1 2.5 2
1 1 2
1 4.5 2
2 7 3
2 2 3
3 3 4
4 1
I think this gives the right output:
WITH g AS
(SELECT ID, lead(ID) OVER (ORDER BY ID) lead_id
FROM (SELECT DISTINCT ID FROM TableA) )
SELECT ID, VALUE, lead_id
FROM TableA
JOIN g USING (ID)
ORDER BY 1;
SELECT tablea.*, b.nextid FROM tablea
INNER JOIN (SELECT id, LEAD (id) OVER (ORDER BY id) nextid
FROM ( SELECT DISTINCT id
FROM tablea
ORDER BY id)) b
ON tablea.id = b.id
This should work.

Oracle "partition" a table at each new value

I have an Oracle table I need to "partition" :I use the terme loosely, I just need to detect groups and would like to display the group through a SELECT. Here's an example that might serve as a sample data (the four columns):
ID | Ref | Rank | Partition_group (only available for the 1st member)
1 | 1 | 1 | 1_A
2 | 1 | 2 | (null)
3 | 1 | 3 | 1_B
4 | 2 | 1 | (null)
5 | 2 | 2 | 2_A
...
It is sorted (the sort key would be the 'Ref' and a creation date). What I would need here, is to extract three groups:
IDs 1 and 2
ID 3
ID 5
What happens with ID 4 is not really important: it may be in its own group, or with the ID 5.
Two IDS should be in the same group if they have the same 'Ref' and if there hasn't been any 'Partition_group' change. In other words, at each change of 'Ref' or (logical or) 'Partition_group', I need to detect a new group. For instance, we could return something like that:
ID | Ref | Rank | Partition_group | Group
1 | 1 | 1 | 1_A | 1_A
2 | 1 | 2 | (null) | 1_A
3 | 1 | 3 | 1_B | 1_B
4 | 2 | 1 | (null) | (null) (or 2_A)
5 | 2 | 2 | 2_A | 2_A
...
I thought about writing a function or something, but it appears I don't have the rights to do so (yeah...) so I have to use plain Oracle SQL (11g).
I've been looking at CONNECT BY and OVER (analytical functions) but they don't seem to do the trick.
Has anyone been faced to such a problem? How would you resolve it?
Thanks in advance.
Assuming the input data is the first four columns, how about something like:
with sample_data as (select 1 id, 1 ref, 1 rank, '1_A' ptn_group from dual union all
select 2 id, 1 ref, 2 rank, null ptn_group from dual union all
select 3 id, 1 ref, 3 rank, '1_B' ptn_group from dual union all
select 4 id, 2 ref, 1 rank, null ptn_group from dual union all
select 5 id, 2 ref, 2 rank, '2_A' ptn_group from dual)
select id,
ref,
rank,
ptn_group,
last_value(ptn_group ignore nulls) over (partition by ref order by rank, id) grp1,
case when last_value(ptn_group ignore nulls) over (partition by ref order by rank, id) is null then
first_value(ptn_group ignore nulls) over (partition by ref order by rank, id rows between current row and unbounded following)
else last_value(ptn_group ignore nulls) over (partition by ref order by rank, id)
end grp2
from sample_data;
ID REF RANK PTN_GROUP GRP1 GRP2
---------- ---------- ---------- --------- ---- ----
1 1 1 1_A 1_A 1_A
2 1 2 1_A 1_A
3 1 3 1_B 1_B 1_B
4 2 1 2_A
5 2 2 2_A 2_A 2_A
I've given you two options to generate the grp, based on how you want to deal with rows where the first rows of the ptn_group are null - leave them null or pick up the first non-null value in the group.

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