I want remove specific value from comma separated sting using oracle.
Sample Input -
col
1,2,3,4,5
Suppose i want to remove 3 from the string.
Sample Output -
col
1,2,4,5
Please suggest how i can do this using oracle query.
Thanks.
Here is a solution that uses only standard string functions (rather than regular expressions) - which should result in faster execution in most cases; it removes 3 only when it is the first character followed by comma, the last character preceded by comma, or preceded and followed by comma, and it removes the comma that precedes it in the middle case and it removes the comma that follows it in the first and third case.
It is able to remove two 3's in a row (which some of the other solutions offered are not able to do) while leaving in place consecutive commas (which presumably stand in for NULL) and do not disturb numbers like 38 or 123.
The strategy is to first double up every comma (replace , with ,,) and append and prepend a comma (to the beginning and the end of the string). Then remove every occurrence of ,3,. From what is left, replace every ,, back with a single , and finally remove the leading and trailing ,.
with
test_data ( str ) as (
select '1,2,3,4,5' from dual union all
select '1,2,3,3,4,4,5' from dual union all
select '12,34,5' from dual union all
select '1,,,3,3,3,4' from dual
)
select str,
trim(both ',' from
replace( replace(',' || replace(str, ',', ',,') || ',', ',3,'), ',,', ',')
) as new_str
from test_data
;
STR NEW_STR
------------- ----------
1,2,3,4,5 1,2,4,5
1,2,3,3,4,4,5 1,2,4,4,5
12,34,5 12,34,5
1,,,3,3,3,4 1,,,4
4 rows selected.
Note As pointed out by MT0 (see Comments below), this will trim too much if the original string begins or ends with commas. To cover that case, instead of wrapping everything within trim(both ',' from ...) I should wrap the rest within a subquery, and use something like substr(new_str, 2, length(new_str) - 2) in the outer query.
Here is one method:
select trim(both ',' from replace(',' || '1,2,3,4,5' || ',', ',' || '3' || ',', ','))
That said, storing comma-delimited strings is a really, really bad idea. There is almost no reason to do such a thing. Oracle supports JSON, XML, and nested tables -- all of which are better alternatives.
The need to remove an element suggests a poor data design.
You can convert the list rows using an XMLTABLE, filter to remove the unwanted rows and then re-aggregate them:
SELECT LISTAGG( x.value.getStringVal(), ',' ) WITHIN GROUP ( ORDER BY idx )
FROM XMLTABLE(
( '1,2,3,4,5' )
COLUMNS value XMLTYPE PATH '.',
idx FOR ORDINALITY
) x
WHERE x.value.getStringVal() != 3;
For a simple filter this is probably not worth it and you should use something like (based on #mathguy's solution):
SELECT SUBSTR( new_list, 2, LENGTH( new_list ) - 2 ) AS new_list
FROM (
SELECT REPLACE(
REPLACE(
',' || REPLACE( :list, ',', ',,' ) || ',',
',' || :value_to_replace || ','
),
',,',
','
) AS new_list
FROM DUAL
)
However, if the filtering is more complicated then it might be worth converting the list to rows, filtering and re-aggregating.
I do not knwo how to do this in Oracle, but with SQL-Server I'd use a trick:
convert the list to XML by replacing the comma with tags
use XQuery to filter the data
reconcatenate
This is SQL Server syntax but might point you the direction:
declare #s varchar(100)='1,2,2,3,3,4';
declare #exclude int=3;
WITH Casted AS
(
SELECT CAST('<x>' + REPLACE(#s,',','</x><x>') + '</x>' AS XML) AS TheXml
)
SELECT x.value('.','int')
FROM Casted
CROSS APPLY TheXml.nodes('/x[text()!=sql:variable("#exclude")]') AS A(x)
UPDATE
I just found this answer which seems to show pretty well how to start...
I agree with Gordon regarding the fact that storing comma delimited data in a column is a really bad idea.
I just preceed the csv with a ',', then use the replace function followed by a left trim function to clean-up the preceeding ','.
SCOTT#tst>VAR b_number varchar2(5);
SCOTT#tst>EXEC :b_number:= '3';
PL/SQL procedure successfully completed.
SCOTT#tst>WITH srce AS (
2 SELECT
3 ',' || '3,1,2,3,3,4,5,3' col
4 FROM
5 dual
6 ) SELECT
7 ltrim(replace(col,',' ||:b_number),',') col
8 FROM
9 srce;
COL
1,2,4,5
Related
I am working with a field called codes that is a delimited list of values, separated by commas. Within each item there is a title ending in a colon and then a code number following the colon. I want a list of only the code numbers after each colon.
Example Value:
name-form-na-stage0:3278648990379886572,rules-na-unwanted-sdfle2:6886328308933282817,us-disdg-order-stage1:1273671130817907765
Desired Output:
3278648990379886572,6886328308933282817,1273671130817907765
The title does always start with a letter and the end with a colon so I can see how REGEXP_REPLACE might work to replace any string between starting with a letter and ending with a colon with '' might work but I am not good at REGEXP_REPLACE patterns. Chat GPT is down fml.
Side note, if anyone knows of a good guide for understanding pattern notation for regular expressions it would be much appreciated!
I tried this and it is not working REGEXP_REPLACE(REPLACE(REPLACE(codes,':', ' '), ',', ' ') ,' [^0-9]+ ', ' ')
This solution assumes a few things:
No colons anywhere else except immediately before the numbers
No number at the very start
At a high level, this query finds how many colons there are, splits the entire string into that many parts, and then only keeps the number up to the comma immediately after the number, and then aggregates the numbers into a comma-delimited list.
Assuming a table like this:
create temp table tbl_string (id int, strval varchar(1000));
insert into tbl_string
values
(1, 'name-form-na-stage0:3278648990379886572,rules-na-unwanted-sdfle2:6886328308933282817,us-disdg-order-stage1:1273671130817907765');
with recursive cte_num_of_delims AS (
select max(regexp_count(strval, ':')) AS num_of_delims
from tbl_string
), cte_nums(nums) AS (
select 1 as nums
union all
select nums + 1
from cte_nums
where nums <= (select num_of_delims from cte_num_of_delims)
), cte_strings_nums_combined as (
select id,
strval,
nums as index
from cte_nums
cross join tbl_string
), prefinal as (
select *,
split_part(strval, ':', index) as parsed_vals
from cte_strings_nums_combined
where parsed_vals != ''
and index != 1
), final as (
select *,
case
when charindex(',', parsed_vals) = 0
then parsed_vals
else left(parsed_vals, charindex(',', parsed_vals) - 1)
end as final_vals
from prefinal
)
select listagg(final_vals, ',')
from final
I have Values which are stored like 1,2,3,4,5 in the database.
I want it back like '1','2','3','4','5'.
I am trying with string_agg(format('''%s''', ticker_information.user_groups), ',')
but giving me result '1,2,3,4,5'
Any solution ? Or let me know If I am doing wrong.
Thanks
Try this if you just want a string back with the quotes
WITH sample AS (
SELECT '1,2,3,4,5'::text as test
)
SELECT
'''' || array_to_string(
string_to_array(test, ','),
''','''
) || ''''
FROM sample
You can create an array from your csv string using unnest, wrap the elements with quote_literal() and then aggregate them again. You can achieve this with a subquery ..
SELECT array_to_string(array_agg(i),',') FROM
(SELECT quote_literal(unnest(string_to_array(user_groups,',')))
FROM ticker_information) j (i);
array_to_string
---------------------
'1','2','3','4','5'
Or with a LATERAL :
SELECT array_to_string(array_agg(quote_literal(j.i)),',')
FROM ticker_information,
LATERAL unnest(string_to_array(user_groups,',')) j (i);
array_to_string
---------------------
'1','2','3','4','5'
Another option would be with regular expressions.. but it could get nasty if the elements of your csv contain commas.
Demo: db<>fiddle
I have a string as follows: first, last (123456) the expected result should be 123456. Could someone help me in which direction should I proceed using Oracle?
It will depend on the actual pattern you care about (I assume "first" and "last" aren't literal hard-coded strings), but you will probably want to use regexp_substr.
For example, this matches anything between two brackets (which will work for your example), but you might need more sophisticated criteria if your actual examples have multiple brackets or something.
SELECT regexp_substr(COLUMN_NAME, '\(([^\)]*)\)', 1, 1, 'i', 1)
FROM TABLE_NAME
Your question is ambiguous and needs clarification. Based on your comment it appears you want to select the six digits after the left bracket. You can use the Oracle instr function to find the position of a character in a string, and then feed that into the substr to select your text.
select substr(mycol, instr(mycol, '(') + 1, 6) from mytable
Or if there are a varying number of digits between the brackets:
select substr(mycol, instr(mycol, '(') + 1, instr(mycol, ')') - instr(mycol, '(') - 1) from mytable
Find the last ( and get the sub-string after without the trailing ) and convert that to a number:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE test ( str ) AS
SELECT 'first, last (123456)' FROM DUAL UNION ALL
SELECT 'john, doe (jr) (987654321)' FROM DUAL;
Query 1:
SELECT TO_NUMBER(
TRIM(
TRAILING ')' FROM
SUBSTR(
str,
INSTR( str, '(', -1 ) + 1
)
)
) AS value
FROM test
Results:
| VALUE |
|-----------|
| 123456 |
| 987654321 |
I have a column that contains data like:
AB-123 XYZ
ABCD-456 AAA
BCD-789 BBB
ZZZ-963
Y-85
and this is what i need from those string:
123
456
789
963
85
I need the characters from the left after the dash('-') character, then ends before the space character is read.
Thank You guys.
Note: Original tag on this question was Oracle and this answer is based on that tag. Now that, tag is updated to SqlServer, this answer is no longer valid, if somebody looking for Oracle solution, this may help.
Use regular expression to arrive at sub string.
select trim(substr(regexp_substr('ABCD-456 AAA','-[0-9]+ '),2)) from dual
'-[0-9]+ ' will grab any string pattern which starts with dash has one or more digits and ends with a ' ' and returns number with dash
substr will remove '-' from above output
trim will remove any trailing ' '
Check This.
Using Substring and PatIndex.
select
SUBSTRING(colnm, PATINDEX('%[0-9]%',colnm),
PATINDEX('%[^0-9]%',ltrim(RIGHT(colnm,LEN(colnm)-CHARINDEX('-',colnm)))))
from
(
select 'AB-123 XYZ' colnm union
select 'ABCD-456 AAA' union
select 'BCD-789 BBB' union
select 'ZX- 23 BBB'
)a
OutPut :
Try this
http://rextester.com/YTBPQD69134
CREATE TABLE Table1 ([col] varchar(12));
INSERT INTO Table1
([col])
VALUES
('AB-123 XYZ'),
('ABCD-456 AAA'),
('BCD-789 BBB');
select substring
(col,
charindex('-',col,1)+1,
charindex(' ',col,1)-charindex('-',col,1)
) from table1;
Assume all values has '-' and followed by a space ' '. Below solution will not tolerant to exception case:
SELECT
*,
SUBSTRING(Value, StartingIndex, Length) AS Result
FROM
-- You can replace this section of code with your table name
(VALUES
('AB-123 XYZ'),
('ABCD-456 AAA'),
('BCD-789 BBB')
) t(Value)
-- Use APPLY instead of sub-query is for debugging,
-- you can view the actual parameters in the select
CROSS APPLY
(
SELECT
-- Get the first index of character '-'
CHARINDEX('-', Value) + 1 AS StartingIndex,
-- Get the first index of character ' ', then calculate the length
CHARINDEX(' ', Value) - CHARINDEX('-', Value) - 1 AS Length
) b
I have a column in my oracle db as character and data stored in this is like here
30.170527093355002,72.615875338654 and
30.805165,71.82474
Now I want to get the separated by comma whole string. I mean I want to get the part of string before comma and also part after comma separately. Please any one tell me is there any built in function to do this that I can separate my while string by comma regardless of comma position where it exist.I have already tried floor function and substr but all in vain please help me to use any built in function or user defined function to full fill my requirements.
select
substr( COLNAME, 1, instr( COLNAME, ',') - 1 ) as p_1 ,
substr( COLNAME, instr( COLNAME, ',', - 1 ) + 1 ) as p_2
from YOURTABLE