I have a Emptbl in which I have EmpType Column.
In EmpType I have following data for example :
E0123
M0123
E1245
E4578
M1245
E0478
M4789
E4762
Now I want to get only those emp data which have same EmpType for example below data:
E0123
M0123
E1245
M1245
And want to show this data as group by as 0123 and 1245
So how to get above data? I use UNION but it does not get valida data.
Thanks
Try this:
select substring(emptype, 2, len(emptype))
from emptbl
group by substring(emptype, 2, len(emptype))
having count(*) > 1
The hard-coded 2 is based on your sample data. If instead you had an arbitrary number of letters before the numeric part, e.g. 'ABCDEFG0123', you could use patindex to get the starting index for your substring like so;
select substring(emptype, patindex('%[0-9]%',emptype), len(emptype)
Select A.*
From EmpTbl A
Inner Join
EmpTbl B
On SubString(A.EmpType, 2, 4) = SubString(B.EmpType, 2, 4) And
SubString(A.EmpType, 1, 1) <> SubString(B.EmpType , 1, 1)
;with CTE as (Select Name,SUBSTRING(Name,2,5) as Number, ROW_NUMBER()
OVER (PARTITION By SUBSTRING(Name,2,5) ORDER BY Name) AS Row
from #Temp)
Select Temp.Name
From CTE C
Cross Apply (Select Name FRom CTE T Where T.Number=C.Number) as Temp
Where C.Row>1
Here is the fiddle sample
select top 4 id from
(
select id,rn=row_number()over(partition by right(id,3) order by right(id,3)) from #t
)x
SEE IT LIVE
This is the smallest query that works:
select substr(a.emptype, 2) num
join emptbl a
join emptbl b on substr(a.emptype, 2) = substr(b.emptype, 2)
and a.emptype != b.emptype
Related
I have a table that contains comma-separated values in a column In Postgres.
ID PRODS
--------------------------------------
1 ,142,10,75,
2 ,142,87,63,
3 ,75,73,2,58,
4 ,142,2,
Now I want a query where I can give a comma-separated string and it will tell me the number of matches between the input string and the string present in the row.
For instance, for input value ',142,87,', I want the output like
ID PRODS No. of Match
------------------------------------------------------------------------
1 ,142,10,75, 1
2 ,142,87,63, 2
3 ,75,73,2,58, 0
4 ,142,2, 1
Try this:
SELECT
*,
ARRAY(
SELECT
*
FROM
unnest(string_to_array(trim(both ',' from prods), ','))
WHERE
unnest = ANY(string_to_array(',142,87,', ','))
)
FROM
prods_table;
Output is:
1 ,142,10,75, {142}
2 ,142,87,63, {142,87}
3 ,75,73,2,58, {}
4 ,142,2, {142}
Add the cardinality(anyarray) function to the last column to get just a number of matches.
And consider changing your database design.
Check This.
select T.*,
COALESCE(No_of_Match,'0')
from TT T Left join
(
select ID,count(ID) No_of_Match
from (
select ID,unnest(string_to_array(trim(t.prods, ','), ',')) A
from TT t)a
Where A in ('142','87')
group by ID
)B
On T.Id=b.id
Demo Here
OutPut
If you install the intarray extension, this gets quite easy:
select id, prods, cardinality(string_to_array(trim(prods, ','), ',')::int[] & array[142,87])
from bad_design;
Otherwise it's a bit more complicated:
select bd.id, bd.prods, m.matches
from bad_design bd
join lateral (
select bd.id, count(v.p) as matches
from unnest(string_to_array(trim(bd.prods, ','), ',')) as l(p)
left join (
values ('142'),('87') --<< these are your input values
) v(p) on l.p = v.p
group by bd.id
) m on m.id = bd.id
order by bd.id;
Online example: http://rextester.com/ZIYS97736
But you should really fix your data model.
with data as
(
select *,
unnest(string_to_array(trim(both ',' from prods), ',') ) as v
from myTable
),
counts as
(
select id, count(t) as c from data
left join
( select unnest(string_to_array(',142,87,', ',') ) as t) tmp on tmp.t = data.v
group by id
order by id
)
select t1.id, t1.prods, t2.c as "No. of Match"
from myTable t1
inner join counts t2 on t1.id = t2.id;
I'm not new in sql and t-sql, but at past I've never used recursive query - all problems were solved with WHILE or CURSOR. I just got 1 question - how to organaze recursion query for following problem: I want to manipulate with last row of data in certain partition. Can't understand how to stop my recursion at last level of partition.
CREATE TABLE #temp
(i int
, s int
, v int);
INSERT INTO #temp
SELECT 1, 1, 10
UNION
SELECT 1, 2, 20
UNION
SELECT 2, 1, 5
UNION
SELECT 2, 2, 5
UNION
SELECT 2, 3, 2
WITH CTE AS
(
SELECT i
, s
, v
FROM #temp
WHERE s=1
UNION ALL
SELECT t.i
, t.s
, t.v + cte.v as new_v
FROM #temp t
INNER JOIN cte
ON (cte.i=t.i)
WHERE t.s>1
)
SELECT *
FROM cte
OPTION(MAXRECURSION 0)
I want to get 5 rows as result:
result
I know that it could be solved with OUTER APPLY, JOINS, WHILE or CURSOR methods. Could you please share any features for my to understand how to get same result with recurcive cte query? SUM function there is just for example - for that problem recurcive query is best way cause I will use many scalar functions in big CASE which will use value from last row in partition and value of current row partition.
Thanks.
Sorry for my bad english level.
Will it be correctly if I'll try same problem with following example? I guess that need to correctly say in which order way recursive query gonna do any data manipulating. So below code which will help you understand what did I want to solve:
CREATE TABLE #temp
(i_key int
, step int
, step_h int
, value int);
INSERT INTO #temp
SELECT 1, 1, NULL, 20
UNION
SELECT 1, 2, 1, 20
UNION
SELECT 2, 1, NULL, 10
UNION
SELECT 2, 2, 1, 10
UNION
SELECT 2, 3, 2, 5
WITH CTE AS
(
SELECT i_key
, step
, value
FROM #temp
WHERE step=1
--AND i_key=2
UNION ALL
SELECT t.i_key
, t.step
, CASE
WHEN cte.value - t.value <=0 THEN 0
ELSE cte.value - t.value
END as value
FROM #temp t
INNER JOIN cte
ON (cte.i_key=t.i_key
AND cte.step=t.step_h)
--WHERE t.step>1
)
SELECT *
FROM CTE
OPTION(MAXRECURSION 0)
Is parent-child structure always need for solving this problems?
So i guess it could be done with another join (without column of parent-child).
AND cte.step=t.step-1
For your particular example, recursion is unnecessary. All you need is SQL Server 2012 or later version:
select t.*,
sum(t.v) over(partition by t.i order by t.s) as [RT]
from #temp t
order by t.i, t.s;
If you need to access previos / next row, there are lag() / lead() ranking functions that were introduced in the same aforementioned version of SQL Server.
EDIT: Ah, I see. You simply want to know how to write recursive CTEs properly. Here is a (seemingly) correct code for your second example:
with cte as (
select t.i_key, t.step, t.value
from #temp t
where t.step_h is null
union all
select c.i_key, t.step, case
when c.value < t.value then 0
else c.value - t.value
end as [Value]
from #temp t
inner join cte c on c.step = t.step_h
and c.i_key = t.i_key
)
select *
from cte c
order by c.i_key, c.step;
In the end, it stops by itself when an iteration does not produce any new rows.
I need to identify duplicate sets of data and give those sets who's data is similar a group id.
id threshold cost
-- ---------- ----------
1 0 9
1 100 7
1 500 6
2 0 9
2 100 7
2 500 6
I have thousands of these sets, most are the same with different id's. I need find all the like sets that have the same thresholds and cost amounts and give them a group id. I'm just not sure where to begin. Is the best way to iterate and insert each set into a table and then each iterate through each set in the table to find what already exists?
This is one of those cases where you can try to do something with relational operators. Or, you can just say: "let's put all the information in a string and use that as the group id". SQL Server seems to discourage this approach, but it is possible. So, let's characterize the groups using:
select d.id,
(select cast(threshold as varchar(8000)) + '-' + cast(cost as varchar(8000)) + ';'
from data d2
where d2.id = d.id
for xml path ('')
order by threshold
) as groupname
from data d
group by d.id;
Oh, I think that solves your problem. The groupname can serve as the group id. If you want a numeric id (which is probably a good idea, use dense_rank():
select d.id, dense_rank() over (order by groupname) as groupid
from (select d.id,
(select cast(threshold as varchar(8000)) + '-' + cast(cost as varchar(8000)) + ';'
from data d2
where d2.id = d.id
for xml path ('')
order by threshold
) as groupname
from data d
group by d.id
) d;
Here's the solution to my interpretation of the question:
IF OBJECT_ID('tempdb..#tempGrouping') IS NOT NULL DROP Table #tempGrouping;
;
WITH BaseTable AS
(
SELECT 1 id, 0 as threshold, 9 as cost
UNION SELECT 1, 100, 7
UNION SELECT 1, 500, 6
UNION SELECT 2, 0, 9
UNION SELECT 2, 100, 7
UNION SELECT 2, 500, 6
UNION SELECT 3, 1, 9
UNION SELECT 3, 100, 7
UNION SELECT 3, 500, 6
)
, BaseCTE AS
(
SELECT
id
--,dense_rank() over (order by threshold, cost ) as GroupId
,
(
SELECT CAST(TblGrouping.threshold AS varchar(8000)) + '/' + CAST(TblGrouping.cost AS varchar(8000)) + ';'
FROM BaseTable AS TblGrouping
WHERE TblGrouping.id = BaseTable.id
ORDER BY TblGrouping.threshold, TblGrouping.cost
FOR XML PATH ('')
) AS MultiGroup
FROM BaseTable
GROUP BY id
)
,
CTE AS
(
SELECT
*
,DENSE_RANK() OVER (ORDER BY MultiGroup) AS GroupId
FROM BaseCTE
)
SELECT *
INTO #tempGrouping
FROM CTE
-- SELECT * FROM #tempGrouping;
UPDATE BaseTable
SET BaseTable.GroupId = #tempGrouping.GroupId
FROM BaseTable
INNER JOIN #tempGrouping
ON BaseTable.Id = #tempGrouping.Id
IF OBJECT_ID('tempdb..#tempGrouping') IS NOT NULL DROP Table #tempGrouping;
Where BaseTable is your table, and and you don't need the CTE "BaseTable", because you have a data table.
You may need to take extra-precautions if your threshold and cost fields can be NULL.
Suppose I have a list of values, such as 1, 2, 3, 4, 5 and a table where some of those values exist in some column. Here is an example:
id name
1 Alice
3 Cindy
5 Elmore
6 Felix
I want to create a SELECT statement that will include all of the values from my list as well as the information from those rows that match the values, i.e., perform a LEFT OUTER JOIN between my list and the table, so the result would be like follows:
id name
1 Alice
2 (null)
3 Cindy
4 (null)
5 Elmore
How do I do that without creating a temp table or using multiple UNION operators?
If in Microsoft SQL Server 2008 or later, then you can use Table Value Constructor
Select v.valueId, m.name
From (values (1), (2), (3), (4), (5)) v(valueId)
left Join otherTable m
on m.id = v.valueId
Postgres also has this construction VALUES Lists:
SELECT * FROM (VALUES (1, 'one'), (2, 'two'), (3, 'three')) AS t (num,letter)
Also note the possible Common Table Expression syntax which can be handy to make joins:
WITH my_values(num, str) AS (
VALUES (1, 'one'), (2, 'two'), (3, 'three')
)
SELECT num, txt FROM my_values
With Oracle it's possible, though heavier From ASK TOM:
with id_list as (
select 10 id from dual union all
select 20 id from dual union all
select 25 id from dual union all
select 70 id from dual union all
select 90 id from dual
)
select * from id_list;
the following solution for oracle is adopted from this source. the basic idea is to exploit oracle's hierarchical queries. you have to specify a maximum length of the list (100 in the sample query below).
select d.lstid
, t.name
from (
select substr(
csv
, instr(csv,',',1,lev) + 1
, instr(csv,',',1,lev+1 )-instr(csv,',',1,lev)-1
) lstid
from (select ','||'1,2,3,4,5'||',' csv from dual)
, (select level lev from dual connect by level <= 100)
where lev <= length(csv)-length(replace(csv,','))-1
) d
left join test t on ( d.lstid = t.id )
;
check out this sql fiddle to see it work.
Bit late on this, but for Oracle you could do something like this to get a table of values:
SELECT rownum + 5 /*start*/ - 1 as myval
FROM dual
CONNECT BY LEVEL <= 100 /*end*/ - 5 /*start*/ + 1
... And then join that to your table:
SELECT *
FROM
(SELECT rownum + 1 /*start*/ - 1 myval
FROM dual
CONNECT BY LEVEL <= 5 /*end*/ - 1 /*start*/ + 1) mypseudotable
left outer join myothertable
on mypseudotable.myval = myothertable.correspondingval
Assuming myTable is the name of your table, following code should work.
;with x as
(
select top (select max(id) from [myTable]) number from [master]..spt_values
),
y as
(select row_number() over (order by x.number) as id
from x)
select y.id, t.name
from y left join myTable as t
on y.id = t.id;
Caution: This is SQL Server implementation.
fiddle
For getting sequential numbers as required for part of output (This method eliminates values to type for n numbers):
declare #site as int
set #site = 1
while #site<=200
begin
insert into ##table
values (#site)
set #site=#site+1
end
Final output[post above step]:
select * from ##table
select v.id,m.name from ##table as v
left outer join [source_table] m
on m.id=v.id
Suppose your table that has values 1,2,3,4,5 is named list_of_values, and suppose the table that contain some values but has the name column as some_values, you can do:
SELECT B.id,A.name
FROM [list_of_values] AS B
LEFT JOIN [some_values] AS A
ON B.ID = A.ID
T-SQL query for finding first missing sequence string (prefix+no)
Sequence can have a prefix + a continuing no.
ex sequence will be
ID
-------
AUTO_500
AUTO_501
AUTO_502
AUTO_504
AUTO_505
AUTO_506
AUTO_507
AUTO_508
So above the missing sequence is AUTO_503 or if there is no missing sequence then it must return next sequence.
Also starting no is to specified ex. 500 in this case and prefix can be null i.e. no prefix only numbers as sequence.
You could LEFT JOIN the id numbers on shifted(+1) values to find gaps in sequential order:
SELECT
MIN(a.offsetnum) AS first_missing_num
FROM
(
SELECT 500 AS offsetnum
UNION
SELECT CAST(REPLACE(id, 'AUTO_', '') AS INT) + 1
FROM tbl
) a
LEFT JOIN
(SELECT CAST(REPLACE(id, 'AUTO_', '') AS INT) AS idnum FROM tbl) b ON a.offsetnum = b.idnum
WHERE
a.offsetnum >= 500 AND b.idnum IS NULL
SQLFiddle Demo
Using a recursive CTE to dynamically generate the sequence between the min and max of the ID Numbers maybe over complicated things a bit but it seems to work -
LIVE ON FIDDLE
CREATE TABLE tbl (
id VARCHAR(55)
);
INSERT INTO tbl VALUES
('AUTO_500'),
('AUTO_501'),
('AUTO_502'),
('AUTO_504'),
('AUTO_505'),
('AUTO_506'),
('AUTO_507'),
('AUTO_508'),
('509');
;WITH
data_cte(id)AS
(SELECT [id] = CAST(REPLACE(id, 'AUTO_', '') AS INT) FROM tbl)
,maxmin_cte(minId, maxId)AS
(SELECT [minId] = min(id),[maxId] = max(id) FROM data_cte)
,recursive_cte(n) AS
(
SELECT [minId] n from maxmin_cte
UNION ALL
SELECT (1 + n) n FROM recursive_cte WHERE n < (SELECT [maxId] from maxmin_cte)
)
SELECT x.n
FROM
recursive_cte x
LEFT OUTER JOIN data_cte y ON
x.n = y.id
WHERE y.id IS NULL
Check this solution.Here you just need to add identity column.
CREATE TABLE tbl (
id VARCHAR(55),
idn int identity(0,1)
);
INSERT INTO tbl VALUES
('AUTO_500'),
('AUTO_501'),
('AUTO_502'),
('AUTO_504'),
('AUTO_505'),
('AUTO_506'),
('AUTO_507'),
('AUTO_508'),
('509');
SELECT min(idn+500) FROM tbl where 'AUTO_'+cast((idn+500) as varchar)<>id
try this:
with cte as(
select cast(REPLACE(id,'AUTO_','') as int)-500+1 [diff],ROW_NUMBER()
over(order by cast(REPLACE(id,'AUTO_','') as int)) [rnk] from tbl)
select top 1 'AUTO_'+cast(500+rnk as varchar(50)) [ID] from cte
where [diff]=[rnk]
order by rnk desc
SQL FIddle Demo
Had a similar situation, where we have R_Cds that were like this R01005
;with Active_R_CD (R_CD)
As
(
Select Distinct Cast(Replace(R_CD,'R', ' ') as Int)
from table
where stat = 1)
select Arc.R_CD + 1 as 'Gaps in R Code'
from Active_R_CD as Arc
left outer join Active_R_CD as r on ARC.R_CD + 1 = R.R_CD
where R.R_CD is null
order by 1