So, i have a lot of strings like the ones below in my database:
product1:1stparty:single_aduls:android:
product2:3rdparty:married_adults:ios:
product3:3rdparty:other_adults:android:
I need a regex to get only the text after the product name and before the device category. So, in the first line I'd get 1stparty:single_aduls, in the second 3rdparty:married_adults and in the third 3rdparty:other_adults. I'm stuck and can't find a way to solve that. Could anyone help me please?
As a regular expression, you can use:
select regexp_extract('product1:1stparty:single_aduls:android:', '^[^:]*:(.*):[^:]*:$')
This returns every after the first colon and before the penultimate colon.
We can try using REGEXP_REPLACE here:
SELECT REGEXP_REPLACE(val, r"^.*?:|:[^:]+:$", "") AS output
FROM yourTable;
This approach removes either the leading ...: or trailing :...: from the column, leaving behind the content you want. Here is a demo showing that the regex replacement is working:
Demo
You can also use standard split function and access result array element by index, which is quite clear to read and understand.
with a as (
select split('product1:1stparty:single_aduls:android:', ':') as splitted
)
select splitted[ordinal(2)] || ':' || splitted[ordinal (3)] as subs
from a
Consider below example
with your_table as (
select 'product1:1stparty:single_aduls:android:' txt union all
select 'product2:3rdparty:married_adults:ios:' union all
select 'product3:3rdparty:other_adults:android:'
)
select *,
(
select string_agg(part, ':' order by offset)
from unnest(split(txt, ':')) part with offset
where offset in (1, 2)
) result
from your_table
with output
I have the following code...
TRIM(LEADING '/' FROM ci.Long_description_1) as 'Description',
I am getting the error message
Incorrect syntax near '/'
Whats the best way of writing this?
Thanks
If you are using SQL Server 2017+, you may use TRIM() with a small trick (by default TRIM() removes the specified characters from the start and the end of the string):
SELECT LEFT(
TRIM('/' FROM Long_description_1 + '?'),
LEN(TRIM('/' FROM Long_description_1 + '?')) - 1
) AS Description
FROM (VALUES
(NULL),
('abcd'),
('1234/'),
('////abcd'),
('/folder/subfolder/x.yz')
) ci (Long_description_1)
Result:
Description
----------------------
abcd
1234/
abcd
folder/subfolder/x.yz
I think you want to remove '/' if it is the first character of Long_description_1 column value. TRIM function in SQL Server will not work the way you are expecting.
For this you can write your query like following.
SELECT CASE
WHEN CHARINDEX('/', ci.Long_description_1) = 1
THEN RIGHT(ci.Long_description_1, LEN(ci.Long_description_1) - 1)
ELSE ci.Long_description_1
END AS [Description]
FROM YouTable
I have a column in a table ident_nums that contains different types of ids. I need to remove special characters(e.g. [.,/#&$-]) from that column and replace them with space; however, if the special characters are found at the beginning of the string, I need to remove it without placing a space. I tried to do it in steps; first, I removed the special characters and replaced them with space (I used
REGEXP_REPLACE) then found the records that contain spaces at the beginning of the string and tried to use the TRIM function to remove the white space, but for some reason is not working that.
Here is what I have done
Select regexp_replace(id_num, '[:(),./#*&-]', ' ') from ident_nums
This part works for me, I remove all the unwanted characters from the column, however, if the string in the column starts with a character I don't want to have space in there, I would like to remove just the character, so I tried to use the built-in function TRIM.
update ident_nums
set id_num = TRIM(id_num)
I'm getting an error ORA-01407: can't update ident_nums.id_num to NULL
Any ideas what I am doing wrong here?
It does work if I add a where clause,
update ident_nums
set id_num = TRIM(id_num) where id = 123;
but I need to update all the rows with the white space at the beginning of the string.
Any suggestions are welcome.
Or if it can be done better.
The table has millions of records.
Thank you
Regexp can be slow sometimes so if you can do it by using built-in functions - consider it.
As #Abra suggested TRIM and TRANSLATE is a good choice, but maybe you would prefer LTRIM - removes only leading spaces from string (TRIM removes both - leading and trailing character ). If you want to remove "space" you can ommit defining the trim character parameter, space is default.
select
ltrim(translate('#kdjdj:', '[:(),./#*&-]', ' '))
from dual;
select
ltrim(translate(orginal_string, 'special_characters_to_remove', ' '))
from dual;
Combination of Oracle built-in functions TRANSLATE and TRIM worked for me.
select trim(' ' from translate('#$one,$2-zero...', '#$,-.',' ')) as RESULT
from DUAL
Refer to this dbfiddle
I think trim() is the key, but if you want to keep only alpha numerics, digits, and spaces, then:
select trim(' ' from regexp_replace(col, '[^a-zA-Z0-9 ]', ' ', 1, 0))
regexp_replace() makes it possible to specify only the characters you want to keep, which could be convenient.
Thanks, everyone, It this query worked for me
update update ident_nums
set id_num = LTRIM(REGEXP_REPLACE(id_num, '[:space:]+', ' ')
where REGEXP_LIKE(id_num, '^[ ?]')
this should work for you.
SELECT id_num, length(id_num) length_old, NEW_ID_NUM, length(NEW_ID_NUM) len_NEW_ID_NUM, ltrim(NEW_ID_NUM), length(ltrim(NEW_ID_NUM)) length_after_ltrim
FROM (
SELECT id_num, regexp_replace(id_num, '[:(),./#*&-#]', ' ') NEW_ID_NUM FROM
(
SELECT '1234$%45' as id_num from dual UNION
SELECT '#SHARMA' as id_num from dual UNION
SELECT 'JACK TEST' as id_num from dual UNION
SELECT 'XYZ#$' as id_num from dual UNION
SELECT '#ABCDE()' as id_num from dual -- THe 1st character is space
)
)
I am having issues understanding what's wrong with this regex: \?.*
select REGEXP_REPLACE(longstringcolumn, '\?.*', '') as newstring from tablename
My example string aka 'longstring' has '?' character, and I am trying to match everything trailing '?' (including '?' itself).
I have checked my regexpr in online tools and my regex seems to be working.
Edit
Thanks guys for being so quick,
Here's a sample string (it's a url):
http://example.com/one/two/three?lang=en®ion=CN
I am trying to strip off everything after '?'. So this part:
?lang=en®ion=CN
This is the error I am being returned: Failed to parse regular expression "?": no argument for repetition operator: ?
I am really leaning towards this being a simple escape character issue but I can't figure it out somehow.
#standardSQL
SELECT REGEXP_REPLACE(longstringcolumn, '\\?.*', '') AS newstring
FROM tablename
or
#standardSQL
SELECT REGEXP_REPLACE(longstringcolumn, r'\?.*', '') AS newstring
FROM tablename
example below
#standardSQL
WITH tablename AS (
SELECT 'is this a question?abc ' AS longstringcolumn UNION ALL
SELECT 'this is not a question' union all
SELECT'http://example.com/one/two/three?lang=en®ion=CN'
)
SELECT REGEXP_REPLACE(longstringcolumn, r'\?.*', '') AS newstring
FROM tablename
with result as (where ? and all trailing chars are removed)
Row newstring
1 is this a question
2 this is not a question
Hope this shows what was wrong with your original query
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