i want to find the minimum missing number of a column named (s_no) and the table named (test_table) in oracle and I write the following code..
select
min_s_no-1+level missing_number
from (
select min(s_no) min_s_no, max(s_no) max_s_no
from test_table
) connect by level <= max_s_no-min_s_no+1
minus
select s_no from test_table
;
it gives me all the missing number as a result. But I want to select the minimum
number. Can any one help me please.
thanks in advance.
Using analytical function LEAD you can get the number from the next row in ascending order. Comparing of this value with with the original number increased by 1 you get the missing values (if two numbers do not match).
To get the first missing value in ascending order is the same selecting the MIN value:
select
num,
lead(num) over (order by num) num_lead,
case when num + 1 != lead(num) over (order by num) then num + 1 end as missing_num
from test_data
order by num;
NUM NUM_LEAD MISSING_NUM
---------- ---------- -----------
4 5
5 6
6 9 7
9 10
10 13 11
13
-- first missing number = MIN missing number
select min(missing_num)
from (
select
case when num + 1 != lead(num) over (order by num) then num + 1 end as missing_num
from test_data
);
MIN(MISSING_NUM)
----------------
7
ADDENDUM
A good practice in writing SQL is to consider edge cases - here a table that contains a complete interval without holes. The first missing value will be the successor of the last number.
select nvl(min(missing_num),max(num)+1) first_missing_value
from (
select
num,
case when num + 1 != lead(num) over (order by num) then num + 1 end as missing_num
from test_data
);
A complete table return no MISSING_NUM, so the original query return NULL. Using the NVL the expected result is provided.
The best way to find the gaps is to use analytic functiions lead or lag. An example with lag:
with test_data as (
select 1 num from dual union all
select 4 from dual union all
select 6 from dual union all
select 8 from dual union all
select 3 from dual union all
select 9 from dual union all
select 0 from dual
)
select min(gap) min_gap
from (
select num, lag(num) over (order by num)+1 gap
from test_data
)
where num != gap
;
MIN_GAP
------------------
2
More about how to find the gaps here
In Oracle 12.1 and above, MATCH_RECOGNIZE can do quick work of this kind of problems:
Edited. Initially I was picking the "next number" where a gap exists (in the example, the value 9). But that is not what the OP wants, he wants the first missing number (7 in this case). I edited to change the measures clause, to find the first missing number as requested. End Edit
with test_data (num) as (
select 4 from dual union all
select 5 from dual union all
select 6 from dual union all
select 9 from dual union all
select 10 from dual union all
select 13 from dual
)
-- end of test data; when you use the SQL query below,
-- replace test_data and num with your actual table and column names.
select result as num
from test_data
match_recognize (
order by num
measures last(b.num) + 1 as result
pattern ( ^ a b* c )
define b as num = prev(num) + 1,
c as num > prev(num) + 1
)
;
NUM
---
7
I have a table with two columns:
OLD_REVISIONS |NEW_REVISIONS
-----------------------------------
1,25,26,24 |1,26,24,25
1,56,55,54 |1,55,54
1 |1
1,2 |1
1,96,95,94 |1,96,94,95
1 |1
1 |1
1 |1
1 |1
1,2 |1,2
1 |1
1 |1
1 |1
1 |1
For each row there will be a list of revisions for a document (comma separated)
The comma separated list might be the same in both columns but the order/sort might be different - e.g.
2,1 |1,2
I would like to find all the instances where the highest revision in the OLD_REVISIONS column is lower than than the highest revision in NEW_REVISIONS
The following would fit that criteria
OLD_REVISIONS |NEW_REVISIONS
-----------------------------------
1,2 |1
1,56,55,54 |1,55,54
I tried a solution using the MINUS option (joining the table to itself) but it returns differences even for when the list is the same but in the wrong order
I tried the function GREATEST (i.e where greatest(new_Revisions) < greatest(old_revisions)) but i am not sure why greatest(OLD_REVISIONS) always just returns the comma separated value. It does not return the max value. I suspect it is comparing strings because the columns are VARCHAR.
Also, MAX function expects a single number.
Is there another way i can achieve the above? I am looking for a pure SQL option so i can print out the results (or a PL/SQL option that can print out the results)
Edit
Apologies for not mentioning this but for the NEW_REVISIONS i do actually have the data in a table where each revision is in a separate row:
"DOCNUMBER" "REVISIONNUMBER"
67 1
67 24
67 25
67 26
75 1
75 54
75 55
75 56
78 1
79 1
79 2
83 1
83 96
83 94
Just to give some content, a few weeks ago i suspected that there are revisions disappearing.
To investigate this, i decided to take a count of all revisions for all documents and take a snapshot to compare later to see if revisions are indeed missing.
The snapshot that i took contained the following columns:
docnumber, count, revisions
The revisions were stored in a comma separated list using the listagg function.
The trouble i have now is the on live table, new revisions have been added so when i compare the main table and the snapshot using a MINUS i get a difference because
of the new revisions in the main table.
Even though in the actual table the revisions are individual rows, in the snapshot table i dont have the individual rows.
I am thinking the only way to recreate the snapshot in the same format and compare them find out if maximum revision in the main table is lower than the max revision in the snapshot table (hence why im trying to find out how to find out the max in a comma separated string)
Enjoy.
select xmlcast(xmlquery(('max((' || OLD_REVISIONS || '))') RETURNING CONTENT) as int) as OLD_REVISIONS_max
,xmlcast(xmlquery(('max((' || NEW_REVISIONS || '))') RETURNING CONTENT) as int) as NEW_REVISIONS_max
from t
;
Assuming your base table has an id column (versions of what?) - here is a solution based on splitting the rows.
Edit: If you like this solution, check out vkp's solution, which is better than mine. I explain why his solution is better in a Comment to his Answer.
with
t ( id, old_revisions, new_revisions ) as (
select 101, '1,25,26,24', '1,26,24,25' from dual union all
select 102, '1,56,55,54', '1,55,54' from dual union all
select 103, '1' , '1' from dual union all
select 104, '1,2' , '1' from dual union all
select 105, '1,96,95,94', '1,96,94,95' from dual union all
select 106, '1' , '1' from dual union all
select 107, '1' , '1' from dual union all
select 108, '1' , '1' from dual union all
select 109, '1' , '1' from dual union all
select 110, '1,2' , '1,2' from dual union all
select 111, '1' , '1' from dual union all
select 112, '1' , '1' from dual union all
select 113, '1' , '1' from dual union all
select 114, '1' , '1' from dual
)
-- END of TEST DATA; the actual solution (SQL query) begins below.
select id, old_revisions, new_revisions
from (
select id, old_revisions, new_revisions, 'old' as flag,
to_number(regexp_substr(old_revisions, '\d+', 1, level)) as rev_no
from t
connect by level <= regexp_count(old_revisions, ',') + 1
and prior id = id
and prior sys_guid() is not null
union all
select id, old_revisions, new_revisions, 'new' as flag,
to_number(regexp_substr(new_revisions, '\d+', 1, level)) as rev_no
from t
connect by level <= regexp_count(new_revisions, ',') + 1
and prior id = id
and prior sys_guid() is not null
)
group by id, old_revisions, new_revisions
having max(case when flag = 'old' then rev_no end) !=
max(case when flag = 'new' then rev_no end)
order by id -- ORDER BY is optional
;
ID OLD_REVISION NEW_REVISION
--- ------------ ------------
102 1,56,55,54 1,55,54
104 1,2 1
You can compare every value by putting together the revisions in the same order using listagg function.
SELECT listagg(o,',') WITHIN GROUP (ORDER BY o) old_revisions,
listagg(n,',') WITHIN GROUP (ORDER BY n) new_revisions
FROM (
SELECT DISTINCT rowid r,
regexp_substr(old_revisions, '[^,]+', 1, LEVEL) o,
regexp_substr(new_revisions, '[^,]+', 1, LEVEL) n
FROM table
WHERE regexp_substr(old_revisions, '[^,]+', 1, LEVEL) IS NOT NULL
CONNECT BY LEVEL<=(SELECT greatest(MAX(regexp_count(old_revisions,',')),MAX(regexp_count(new_revisions,',')))+1 c FROM table)
)
GROUP BY r
HAVING listagg(o,',') WITHIN GROUP (ORDER BY o)<>listagg(n,',') WITHIN GROUP (ORDER BY n);
This could be a way:
select
OLD_REVISIONS,
NEW_REVISIONS
from
REVISIONS t,
table(cast(multiset(
select level
from dual
connect by level <= length (regexp_replace(t.OLD_REVISIONS, '[^,]+')) + 1
) as sys.OdciNumberList
)
) levels_old,
table(cast(multiset(
select level
from dual
connect by level <= length (regexp_replace(t.NEW_REVISIONS, '[^,]+')) + 1
)as sys.OdciNumberList
)
) levels_new
group by t.ROWID,
OLD_REVISIONS,
NEW_REVISIONS
having max(to_number(trim(regexp_substr(t.OLD_REVISIONS, '[^,]+', 1, levels_old.column_value)))) >
max(to_number(trim(regexp_substr(t.new_REVISIONS, '[^,]+', 1, levels_new.column_value))))
This uses a double string split to pick the values from every field, and then simply finds the rows where the max values among the two collections match your requirement.
You should edit this by adding some unique key in the GROUP BYclause, or a rowid if you don't have any unique key on your table.
One way to do is to split the columns on comma separation using regexp_substr and checking if the max and min values are different.
Sample Demo
with rownums as (select t.*,row_number() over(order by old_revisions) rn from t)
select old_revisions,new_revisions
from rownums
where rn in (select rn
from rownums
group by rn
connect by regexp_substr(old_revisions, '[^,]+', 1, level) is not null
or regexp_substr(new_revisions, '[^,]+', 1, level) is not null
having max(cast(regexp_substr(old_revisions,'[^,]+', 1, level) as int))
<> max(cast(regexp_substr(new_revisions,'[^,]+', 1, level) as int))
)
Comments say normalise data. I agree but also I understand it may be not possible. I would try something like query below:
select greatest(val1, val2), t1.r from (
select max(val) val1, r from (
select regexp_substr(v1,'[^,]+', 1, level) val, rowid r from tab1
connect by regexp_substr(v1, '[^,]+', 1, level) is not null
) group by r) t1
inner join (
select max(val) val2, r from (
select regexp_substr(v2,'[^,]+', 1, level) val, rowid r from tab1
connect by regexp_substr(v2, '[^,]+', 1, level) is not null
) group by r) t2
on (t1.r = t2.r);
Tested on:
create table tab1 (v1 varchar2(100), v2 varchar2(100));
insert into tab1 values ('1,3,5','1,4,7');
insert into tab1 values ('1,3,5','1,2,9');
insert into tab1 values ('1,3,5','1,3,5');
insert into tab1 values ('1,3,5','1,4');
and seems to work fine. I left rowid for reference. I guess you have some id in table.
After your edit I would change query to:
select greatest(val1, val2), t1.r from (
select max(val) val1, r from (
select regexp_substr(v1,'[^,]+', 1, level) val, DOCNUMBER r from tab1
connect by regexp_substr(v1, '[^,]+', 1, level) is not null
) group by DOCNUMBER) t1
inner join (
select max(DOCNUMBER) val2, DOCNUMBER r from NEW_REVISIONS) t2
on (t1.r = t2.r);
You may write a PL/SQL function parsing the string and returning the maximal number
select max_num( '1,26,24,25') max_num from dual;
MAX_NUM
----------
26
The query ist than very simple:
select OLD_REVISIONS NEW_REVISIONS
from revs
where max_num(OLD_REVISIONS) < max_num(NEW_REVISIONS);
A prototyp function without validation and error handling
create or replace function max_num(str_in VARCHAR2) return NUMBER as
i number;
x varchar2(1);
n number := 0;
max_n number := 0;
pow number := 0;
begin
for i in 0.. length(str_in)-1 loop
x := substr(str_in,length(str_in)-i,1);
if x = ',' then
-- check max number
if n > max_n then
max_n := n;
end if;
-- reset
n := 0;
pow := 0;
else
n := n + to_number(x)*power(10,pow);
pow := pow +1;
end if;
end loop;
return(max_n);
end;
/
I have a table EmployeeTable.
If I want only that records where employeename have character of 1 to 5
will be palindrome and there also condition like total character is more then 10 then 4 to 8 if character less then 7 then 2 to 5 and if character less then 5 then all char will be checked and there that are palindrome then only display.
Examples :- neen will be display
neetan not selected
kiratitamara will be selected
I try this something on string function like FOR first case like name less then 5 character long
SELECT SUBSTRING(EmployeeName,1,5),* from EmaployeeTable where
REVERSE (SUBSTRING(EmployeeName,1,5))=SUBSTRING(EmployeeName,1,5)
I want to do that without string functions,
Can anyone help me on this?
You need at least SUBSTRING(), I have a solution like this:
(In SQL Server)
DECLARE #txt varchar(max) = 'abcba'
;WITH CTE (cNo, cChar) AS (
SELECT 1, SUBSTRING(#txt, 1, 1)
UNION ALL
SELECT cNo + 1, SUBSTRING(#txt, cNo + 1, 1)
FROM CTE
WHERE SUBSTRING(#txt, cNo + 1, 1) <> ''
)
SELECT COUNT(*)
FROM (
SELECT *, ROW_NUMBER() OVER (ORDER BY cNo DESC) as cRevNo
FROM CTE t1 CROSS JOIN
(SELECT Max(cNo) AS strLength FROM CTE) t2) dt
WHERE
dt.cNo <= dt.strLength / 2
AND
dt.cChar <> (SELECT dti.cChar FROM CTE dti WHERE dti.cNo = cRevNo)
The result will shows the count of differences and 0 means no differences.
Note :
Current solution is Non-Case-Sensitive for change it to a Case-Sensitive you need to check the strings in a case-sensitive collation like Latin1_General_BIN
You can use this solution as a SVF or something like that.
I dont realy understand why you dont want to use string functions in your query, but here is one solution. Compute everything beforehand:
Add Column:
ALTER TABLE EmployeeTable
ADD SubString AS
SUBSTRING(EmployeeName,
(
CASE WHEN LEN(EmployeeName)>10
THEN 4
WHEN LEN(EmployeeName)>7
THEN 2
ELSE 1 END
)
,
(
CASE WHEN LEN(EmployeeName)>10
THEN 8
WHEN LEN(EmployeeName)>7
THEN 5
ELSE 5 END
)
PERSISTED
GO
ALTER TABLE EmployeeTable
ADD Palindrome AS
REVERSE(SUBSTRING(EmployeeName,
(
CASE WHEN LEN(EmployeeName)>10
THEN 4
WHEN LEN(EmployeeName)>7
THEN 2
ELSE 1 END
)
,
(
CASE WHEN LEN(EmployeeName)>10
THEN 8
WHEN LEN(EmployeeName)>7
THEN 5
ELSE 5 END
)) PERSISTED
GO
Then your query will looks like:
SELECT * from EmaployeeTable
where Palindrome = SubString
BUT!
This is not a good idea. Please tell us, why you dont want to use string functios.
You could do it building a list of palindrome words using a recursive query that generates palindrome words till a length o n characters and then selects employees with the name matching a palindrome word. This may be a really inefficient way, but it does the trick
This is a sample query for Oracle, PostgreSQL should support this feature as well with little differences on syntax. I don't know about other RDBMS.
with EmployeeTable AS (
SELECT 'ADA' AS employeename
FROM DUAL
UNION ALL
SELECT 'IDA' AS employeename
FROM DUAL
UNION ALL
SELECT 'JACK' AS employeename
FROM DUAL
), letters as (
select chr(ascii('A') + rownum - 1) as letter
from dual
connect by ascii('A') + rownum - 1 <= ascii('Z')
), palindromes(word, len ) as (
SELECT WORD, LEN
FROM (
select CAST(NULL AS VARCHAR2(100)) as word, 0 as len
from DUAL
union all
select letter as word, 1 as len
from letters
)
union all
select l.letter||p.word||l.letter AS WORD, len + 1 AS LEN
from palindromes p
cross join letters l
where len <= 4
)
SEARCH BREADTH FIRST BY word SET order1
CYCLE word SET is_cycle TO 'Y' DEFAULT 'N'
select *
from EmployeeTable
WHERE employeename IN (
SELECT WORD
FROM palindromes
)
DECLARE #cPalindrome VARCHAR(100) = 'SUBI NO ONIBUS'
SET #cPalindrome = REPLACE(#cPalindrome, ' ', '')
;WITH tPalindromo (iNo) AS (
SELECT 1
WHERE SUBSTRING(#cPalindrome, 1, 1) = SUBSTRING(#cPalindrome, LEN(#cPalindrome), 1)
UNION ALL
SELECT iNo + 1
FROM tPalindromo
WHERE SUBSTRING(#cPalindrome, iNo + 1, 1) = SUBSTRING(#cPalindrome, LEN(#cPalindrome) - iNo, 1)
AND LEN(#cPalindrome) > iNo
)
SELECT IIF(MAX(iNo) = LEN(#cPalindrome), 'PALINDROME', 'NOT PALINDROME')
FROM tPalindromo
With a condition like no two adjacent characters ( from a to z) should be in same case;
I need to change helloworld to HeLlOwOrLd , and used a query like :
SELECT listagg(jumping_char,'') WITHIN GROUP(ORDER BY rn) jumped_word
FROM
(SELECT rn,
CASE
WHEN mod(rn, 2) = 1
THEN upper(split_word)
ELSE lower(split_word)
END jumping_char
FROM
(SELECT regexp_substr('helloworld','.',LEVEL)split_word,
ROWNUM rn
FROM dual
CONNECT BY LEVEL <= LENGTH('helloworld')
)
);
Now I got a string like hello2world should becomes HeLlO2wOrLd.
Any simple ,different queries are appreciated and thanks in advance.
If I understand you correctly, you want to "skip" over non-characters in the input. You can achieve that by using regexp_count() with an offset of rn (instead of simply using rn) in your old solution:
SELECT listagg(jumping_char,'') WITHIN GROUP(ORDER BY rn) jumped_word
FROM
(SELECT rn,
CASE
when mod (regexp_count('hello2world', '[a-zA-Z]', rn), 2) = 1
THEN upper(split_word)
ELSE lower(split_word)
END jumping_char
FROM
(SELECT regexp_substr('hello2world','.',LEVEL)split_word,
ROWNUM rn
FROM dual
CONNECT BY LEVEL <= LENGTH('hello2world')
)
);
UPDATE:
Here's an alternative solution using the MODEL clause, just for completeness' sake:
with t as
(select 'hello2world' txt from dual)
select listagg(case
when mod(v2.char_cnt, 2) = 1
then upper(v2.txt)
else lower(v2.txt)
end,
'') within group(order by v2.rn)
from (
select
v1.txt,
rownum as rn,
sum(case
when regexp_like(txt, '[a-zA-Z]')
then 1
else 0
end) over (partition by 1 order by rownum) as char_cnt
from (
SELECT TXT
FROM T
MODEL
RETURN UPDATED ROWS
PARTITION BY(ROWNUM RN)
DIMENSION BY (0 POSITION)
MEASURES (TXT ,length(txt) NB_MOT)
RULES
(TXT[FOR POSITION FROM 1 TO NB_MOT[0] INCREMENT 1] =
substr(txt[0], CV(POSITION), 1) )
) v1
) v2