Is there a way to convert the first letter uppercase in Oracle SQl without using the Initcap Function?
I have the problem, that I must work with the DISTINCT keyword in SQL clause and the Initcap function doesn´t work.
Heres is my SQL example:
select distinct p.nr, initcap(p.firstname), initcap(p.lastname), ill.describtion
from patient p left join illness ill
on p.id = ill.id
where p.deleted = 0
order by p.lastname, p.firstname;
I get this error message: ORA-01791: not a SELECTed expression
When SELECT DISTINCT, you can't ORDER BY columns that aren't selected. Use column aliases instead, as:
select distinct p.nr, initcap(p.firstname) fname, initcap(p.lastname) lname, ill.describtion
from patient p left join illness ill
on p.id = ill.id
where p.deleted = 0
order by lname, fname
this would do it, but i think you need to post your query as there may be a better solution
select upper(substr(<column>,1,1)) || substr(<column>,2,9999) from dual
To change string to String, you can use this:
SELECT
regexp_replace ('string', '[a-z]', upper (substr ('string', 1, 1)), 1, 1, 'i')
FROM dual;
This assumes that the first letter is the one you want to convert. It your input text starts with a number, such as 2 strings then it won't change it to 2 Strings.
You can also use the column number instead of the name or alias:
select distinct p.nr, initcap(p.firstname), initcap(p.lastname), ill.describtion
from patient p left join illness ill
on p.id = ill.id
where p.deleted = 0
order by 3, 2;
WITH inData AS
(
SELECT 'word1, wORD2, word3, woRD4, worD5, word6' str FROM dual
),
inRows as
(
SELECT 1 as tId, LEVEL as rId, trim(regexp_substr(str, '([A-Za-z0-9])+', 1, LEVEL)) as str
FROM inData
CONNECT BY instr(str, ',', 1, LEVEL - 1) > 0
)
SELECT tId, LISTAGG( upper(substr(str, 1, 1)) || substr(str, 2) , '') WITHIN GROUP (ORDER BY rId) AS camelCase
FROM inRows
GROUP BY tId;
Related
I have the following problem: Show all rows in table where column first_name contains at least 2 vowels (a, e, i, o, u), and the number of occurences of each vowel is the same.
Valid example: Alexander, "e" appears 2 times, "a" appears 2 times. That is coreect.
Invalid example: Jonathan, it has 2 vowels (a, o), but "o" appears once, and "a" appears twice, the number of occurences is not equal.
I've solved this problem by calculating each vowel, and then verify every case (A E, A I, A O etc. Shortly, each combination of 2, 3, 4, 5). With that solution, I have a very long WHERE. Is there any shorter way and more elegant and simple?
This is how I solved it in TSQL in MS SQL Server 2019.
I know its not exactly what you wanted. Just an interesting thing to try. Thanks for that.
DROP TABLE IF EXISTS #Samples
SELECT n.Name
INTO #Samples
FROM
(
SELECT 'Ravi' AS Name
UNION
SELECT 'Tim'
UNION
SELECT 'Timothe'
UNION
SELECT 'Ian'
UNION
SELECT 'Lijoo'
UNION
SELECT 'John'
UNION
SELECT 'Jami'
) AS n
SELECT g.Name,
IIF(MAX (g.Repeat) = MIN (g.Repeat) AND SUM (g.Appearance) >= 2, 'Valid', 'Invalid') AS Validity
FROM
(
SELECT v.value,
s.Name,
SUM (LEN (s.Name) - LEN (REPLACE (s.Name, v.value, ''))) AS Repeat,
SUM (IIF(s.Name LIKE '%' + v.value + '%', 1, 0)) AS Appearance
FROM STRING_SPLIT('a,e,i,o,u', ',') AS v
CROSS APPLY #Samples AS s
GROUP BY v.value,
s.Name
) AS g
WHERE g.Repeat > 0
GROUP BY g.Name
Output
we can replace STRING_SPLIT with a temp table for supporting lower versions
DROP TABLE IF EXISTS #Vowels
SELECT C.Vowel
INTO #Vowels
FROM
(
SELECT 'a' AS Vowel
UNION
SELECT 'e'
UNION
SELECT 'i'
UNION
SELECT 'o'
UNION
SELECT 'u'
) AS C
SELECT g.Name,
IIF(MAX (g.Repeat) = MIN (g.Repeat) AND SUM (g.Appearance) >= 2, 'Valid', 'Invalid') AS Validity
FROM
(
SELECT v.Vowel,
s.Name,
SUM (LEN (s.Name) - LEN (REPLACE (s.Name, v.Vowel, ''))) AS Repeat,
SUM (IIF(s.Name LIKE '%' + v.Vowel + '%', 1, 0)) AS Appearance
FROM #Vowels AS v
CROSS APPLY #Samples AS s
GROUP BY v.Vowel,
s.Name
) AS g
WHERE g.Repeat > 0
GROUP BY g.Name
From Oracle 12, you can use:
SELECT name
FROM table_name
CROSS JOIN LATERAL(
SELECT 1
FROM (
-- Step 2: Count the frequency of each vowel
SELECT letter,
COUNT(*) As frequency
FROM (
-- Step 1: Find all the vowels
SELECT REGEXP_SUBSTR(LOWER(name), '[aeiou]', 1, LEVEL) AS letter
FROM DUAL
CONNECT BY LEVEL <= REGEXP_COUNT(LOWER(name), '[aeiou]')
)
GROUP BY letter
)
-- Step 3: Filter out names where the number of vowels are
-- not equal or the vowels do not occur at least twice
-- and there are not at least 2 different vowels.
HAVING MIN(frequency) >= 2
AND MIN(frequency) = MAX(frequency)
AND COUNT(*) >= 2
);
Which, for the sample data:
CREATE TABLE table_name (name) AS
SELECT 'Alexander' FROM DUAL UNION ALL
SELECT 'Johnaton' FROM DUAL UNION ALL
SELECT 'Anna' FROM DUAL;
Outputs:
NAME
Alexander
db<>fiddle here
Hello people here again with another oracle SQL question.
Im having some problems spliting values from a column to another one.
So there it goes.. im having this query :
SELECT MONEDA ,
LISTAGG (MONTO , ';') WITHIN GROUP (ORDER BY MONTO) MONTO,
REGEXP_SUBSTR(MONTO, '[^;]+', 1, 1) col_one,
REGEXP_SUBSTR(MONTO, '[^;]+', 1, 2) col_two
FROM (SELECT SUM(ZMT.AMOUNT) AS MONTO,
ZMT.T_TYPE AS tipo,
JSON_VALUE(MSG, '$.glAccount.currency.code') AS moneda
FROM Z_MAMBU_TRANSACTIONS ZMT JOIN POSTING_ONLINE0182 PO ON PO.RESP_REFERENCE0182 = ZMT.TRANSACTIONID
WHERE TO_CHAR(ZMT.CREATIONDATE, 'YYYY-MM-DD') = '2021-04-20' AND
PO.POSTING_RESPCODE0182 = 0 AND
(JSON_VALUE(MSG, '$.type') = 'DEBIT') OR (JSON_VALUE(MSG, '$.type') = 'CREDIT')
GROUP BY T_TYPE, JSON_VALUE(MSG, '$.glAccount.currency.code')
ORDER BY T_TYPE)
GROUP BY MONEDA
the result is the next:
https://i.stack.imgur.com/QMgYr.png
What i need to do is SPLIT the "MONTO" values with the ";" as separator to other 2 columns (col_one and col_two). As you can see in the result he is spliting me only the second value not the first.
After that i need to make the substract from the values that i split.
This is an exaple of what i need :
MONEDA MONTO COL_ONE COL_TWOV
COL 174579148065,39;175491229711,9 174579148065,39 175491229711,9
DOL 30300300300;30300300300 30300300300 30300300300
THANK YOU GUYS!
I agree with Tim - substr + instr do the job just nicely. If you, for some reason, want to try regular expressions, see if this helps (sample data in lines #1 - 4; query begins at line #5):
SQL> with result (moneda, monto) as
2 (select 'COL', '174579148065,39;175491229711,9' from dual union all
3 select 'DOL', '30300300300;30300300300' from dual
4 )
5 select moneda,
6 regexp_substr(monto, '\d+(,\d+)?', 1, 1) col_one,
7 regexp_substr(monto, '\d+(,\d+)?', 1, 2) col_two
8 from result;
MONEDA COL_ONE COL_TWO
---------- -------------------- --------------------
COL 174579148065,39 175491229711,9
DOL 30300300300 30300300300
SQL>
I would just use the base string functions here and avoid regex altogether. Going by your sample data given at the very end of your question:
SELECT
MONEDA,
MONTO,
SUBSTR(MONTO, 1, INSTR(MONTO, ';') - 1) AS COL_ONE,
SUBSTR(MONTO, INSTR(MONTO, ';') + 1, LENGTH(MONTO) - INSTR(MONTO, ';')) AS COL_TWO
FROM yourTable;
Demo
Given two comma separated (un-ordered) lists of numbers, I want to extract only the differences between them (using regexp probably).
e.g.:
select
'1010484,1025781,1051394,1069679' as list_1,
'1005923,1010484,1025781,1034010,1044261,1048311,1051394' as list_2
What I wish for is a result such as:
l1_additional_data: 1069679
l2_additional_data: 1005923,1034010,1044261,1048311
How can this be done?
I'm using Vertica, BTW - That means that no hierarchic ("connect by") queries could be used here.
Thanks in advance!
There's a relevant post that will be helpful - Splitting string into multiple rows in Oracle
I don't know vertica but based on oracle You could go with:
with list1 as
(
select
regexp_substr(list_1 ,'[^,]+', 1, level) as list_1_rows
from (
select
'1010484,1025781,1051394,1069679' as list_1
from dual)
connect by
regexp_substr(list_1 ,'[^,]+', 1, level) is not null),
list2 as (select
regexp_substr(list_2 ,'[^,]+', 1, level) as list_2_rows
from (
select
'1005923,1010484,1025781,1034010,1044261,1048311,1051394' as list_2
from dual)
connect by regexp_substr(list_2 ,'[^,]+', 1, level) is not null)
select * from list1
full outer join list2
on list1.list_1_rows = list2.list_2_rows
where list_1_rows is null or list_2_rows is null
OK, Here's my solution - But it's not very efficient, and it probably won't scale (in terms of performance):
WITH lists AS
(SELECT'1010484,1025781,1051394,1069679' AS list_1, '1005923,1010484,1025781,1034010,1044261,1048311,1051394' AS list_2 )
, numbers AS
(SELECT row_number() over() i
FROM system_columns limit 100)
SELECT group_concat(parsed_code_1) list_1_additions,
group_concat(parsed_code_2) list_2_additions
FROM(SELECT parsed_code_1
FROM(SELECT split_part(list_1, ',', i) parsed_code_1
FROM lists
CROSS JOIN numbers
WHERE i <= regexp_count(list_1, ',')+1) l
WHERE parsed_code_1 IS NOT NULL) a
FULL OUTER JOIN
(SELECT parsed_code_2
FROM(SELECT split_part(list_2, ',', i) parsed_code_2
FROM lists
CROSS JOIN numbers
WHERE i <= regexp_count(list_2, ',')+1) l
WHERE parsed_code_2 IS NOT NULL) b ON(parsed_code_1 = parsed_code_2)
WHERE parsed_code_1 IS NULL OR parsed_code_2 IS NULL
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
I need to check if a partial name matches full name. For example:
Partial_Name | Full_Name
--------------------------------------
John,Smith | Smith William John
Eglid,Timothy | Timothy M Eglid
I have no clue how to approach this type of matching.
Another thing is that name and last name may come in the wrong order, making it harder.
I could do something like this, but this only works if names are in the same order and 100% match
decode(LOWER(REGEXP_REPLACE(Partial_Name,'[^a-zA-Z'']','')), LOWER(REGEXP_REPLACE(Full_Name,'[^a-zA-Z'']','')), 'Same', 'Different')
you could use this pattern on the text provided - works for most engines
([^ ,]+),([^ ,]+)(?=.*\b\1\b)(?=.*\b\2\b)
Demo
WITH
/*
tab AS
(
SELECT 'Smith William John' Full_Name, 'John,Smith' Partial_Name FROM dual
UNION ALL SELECT 'Timothy M Eglid', 'Eglid,timothy' FROM dual
UNION ALL SELECT 'Tim M Egli', 'Egli,Tim,M2' FROM dual
UNION ALL SELECT 'Timot M Eg', 'Eg' FROM dual
),
*/
tmp AS (
SELECT Full_Name, Partial_Name,
trim(CASE WHEN instr(Partial_Name, ',') = 0 THEN Partial_Name
ELSE regexp_substr(Partial_Name, '[^,]+', 1, lvl+1)
END) token
FROM tab t CROSS JOIN (SELECT lvl FROM (SELECT LEVEL-1 lvl FROM dual
CONNECT BY LEVEL <= (SELECT MAX(LENGTH(Partial_Name) - LENGTH(REPLACE(Partial_Name, ',')))+1 FROM tab)))
WHERE LENGTH(Partial_Name) - LENGTH(REPLACE(Partial_Name, ',')) >= lvl
)
SELECT Full_Name, Partial_Name
FROM tmp
GROUP BY Full_Name, Partial_Name
HAVING count(DISTINCT token)
= count(DISTINCT CASE WHEN REGEXP_LIKE(Full_Name, token, 'i')
THEN token ELSE NULL END);
In the tmp each partial_name is splitted on tokens (separated by comma)
The resulting query retrieves only those rows which full_name matches all the corresponding tokens.
This query works with the dynamic number of commas in partial_name. If there can be only zero or one commas then the query will be much easier:
SELECT * FROM tab
WHERE instr(Partial_Name, ',') > 0
AND REGEXP_LIKE(full_name, substr(Partial_Name, 1, instr(Partial_Name, ',')-1), 'ix')
AND REGEXP_LIKE(full_name, substr(Partial_Name,instr(Partial_Name, ',')+1), 'ix')
OR instr(Partial_Name, ',') = 0
AND REGEXP_LIKE(full_name, Partial_Name, 'ix');
This is what I ended up doing... Not sure if this is the best approach.
I split partials by comma and check if first name present in full name and last name present in full name. If both are present then match.
CASE
WHEN
instr(trim(lower(Full_Name)),
trim(lower(REGEXP_SUBSTR(Partial_Name, '[^,]+', 1, 1)))) > 0
AND
instr(trim(lower(Full_Name)),
trim(lower(REGEXP_SUBSTR(Partial_Name, '[^,]+', 1, 2)))) > 0
THEN 'Y'
ELSE 'N'
END AS MATCHING_NAMES