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Like if I have a string "123456,852369,7852159,1596357"
The out put looking for "1234,8523,7852,1596"
Requirement is....we want to collect 4 char after every ',' separator
like split, substring and again concat
select
REGEXP_REPLACE('MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MDCB,MDCB,MDCB,MDCB,MDCB,MDCB', '([^,]+)(,\1)+', '\1')
from dual;
we want to collect 4 char after every ',' separator
Here is an approach using regexp_replace:
select regexp_replace(
'123456,852369,7852159,1596357',
'([^,]{4})[^,]*(,|$)',
'\1\2'
)
from dual
Regexp breakdown:
([^,]{4}) 4 characters others than "," (capture that group as \1)
[^,]* 0 to n characters other than "," (no capture)
(,|$) either character "," or the end of string (capture this as \2)
The function replaces each match with capture 1 (the 4 characters we want) followed by capture 2 (the separator, if there is one).
Demo:
RESULT
1234,8523,7852,1596
One option might be to split the string, extract 4 characters and aggregate them back:
SQL> with test (col) as
2 (select '123456,852369,7852159,1596357' from dual)
3 select listagg(regexp_substr(col, '[^,]{4}', 1, level), ',')
4 within group (order by level) result
5 from test
6 connect by level <= regexp_count(col, ',') + 1;
RESULT
--------------------------------------------------------------------------------
1234,8523,7852,1596
SQL>
With REGEX_REPLACE:
select regexp_replace(the_string, '(^|,)([^,]{4})[^,]*', '\1\2')
from mytable;
This looks for
the beginning of the string or the comma
then four characters that are not a comma
then any number of trailing characters that are not a comma
And only keeps
the beginning or the comma
the four characters that follow
Demo: https://dbfiddle.uk/efUFvKyO
spark.sql("select case when length(medicare)>0 then lpad(med,13,'0') else '' end as med from input").show(false)
In the above query I'm able to trim the characters to a specified length but what is the procedure to get the characters to a specified length with spaces before that.
sample input: 1234(10 spaces plus 1234)
sample output: 1234(9 spaces plus 1234)
Whatever the input is it should trim to 13 characters.
sample input: 1234567890123(10 spaces plus 1234567890123)
sample output:1234567890123(13 characters)
Please help me out on this...Thanks in advance
You can take the right-most 13 spaces and pad that to 13:
select lpad(right(medicare, 13), 13, ' ')
The right() function is just to handle the case where medicare might have more than 13 characters.
EDIT:
Based on the comment:
select lpad(left(ltrim(medicare), 13), 13, ' ')
How do I use an ORACLE REGEX function to remove all leading and trailing line break characters and spaces?
For example, assume I have the following string where refers to actual invisible carriage return line feed characters. Here's the input:
"
SELECT *
FROM
TABLE
"
And here's the desire output:
"SELECT *
FROM
TABLE"
This would do it if regex_replace() is a requirement:
select regexp_replace('
SELECT *
FROM
TABLE
', '^\s*|\s*$', '') as hello
from dual
See https://www.techonthenet.com/oracle/functions/regexp_replace.php for documentation.
A single regexp_replace is sufficient, eg.
select regexp_replace('
select frut
from prut
','^[[:space:]]*(.*[^[:space:]])[[:space:]]*$','\1',1,1,'mn') from dual;
results in
select frut
from prut
Given the following sample statement, I simply can't get rid of the leading and trailing white spaces no matter if using a combination of RTRIM() and LTRIM() or REGEXP_REPLACE():
select
test_column
,length(test_column) len
,regexp_replace(test_column, '(^[[:blank:]]+)|([[:blank:]]+$)','') rxp
,length(regexp_replace(test_column, '(^[[:blank:]]+)|([[:blank:]]+$)','')) len_rxp --22 characters expected, but is 26
,rtrim(ltrim(test_column)) rltrim
,length(rtrim(ltrim(test_column))) len_rltrim --22 characters expected, but is 26
from(
select ' ABCDEF Hijklmnopqr S32 ' test_column --22 characters without and 29 including whitespaces
from dual);
What's the matter?
You can use the following:
select regexp_replace(test_column, '^(\t|\s)*(.*)(\t|\s)*$', '\2')
from (
select ' ABCDEF Hijklmnopqr S32 ' test_column
from dual
);
This should divide your string in 3 parts (leading, meaningful text, ending) and return only the second one, thus cutting away the trailing and ending sequences of spaces and tabs
I need to trim New Line (Chr(13) and Chr(10) and Tab space from the beginning and end of a String) in an Oracle query. I learnt that there is no easy way to trim multiple characters in Oracle. "trim" function trims only single character. It would be a performance degradation if i call trim function recursivelly in a loop using a function. I heard regexp_replace can match the whitespaces and remove them.
Can you guide of a reliable way to use regexp_replace to trim multiple tabspaces or new lines or combinations of them in beginning and end of a String. If there is any other way, Please guide me.
If you have Oracle 10g, REGEXP_REPLACE is pretty flexible.
Using the following string as a test:
chr(9) || 'Q qwer' || chr(9) || chr(10) ||
chr(13) || 'qwerqwer qwerty' || chr(9) ||
chr(10) || chr(13)
The [[:space:]] will remove all whitespace, and the ([[:cntrl:]])|(^\t) regexp will remove non-printing characters and tabs.
select
tester,
regexp_replace(tester, '(^[[:space:]]+)|([[:space:]]+$)',null)
regexp_tester_1,
regexp_replace(tester, '(^[[:cntrl:]^\t]+)|([[:cntrl:]^\t]+$)',null)
regexp_tester_2
from
(
select
chr(9) || 'Q qwer' || chr(9) || chr(10) ||
chr(13) || 'qwerqwer qwerty' || chr(9) ||
chr(10) || chr(13) tester
from
dual
)
Returning:
REGEXP_TESTER_1: "Qqwerqwerqwerqwerty"
REGEXP_TESTER_2: "Q qwerqwerqwer qwerty"
Hope this is of some use.
This how I would implement it:
REGEXP_REPLACE(text,'(^[[:space:]]*|[[:space:]]*$)')
How about the quick and dirty translate function?
This will remove all occurrences of each character in string1:
SELECT translate(
translate(
translate(string1, CHR(10), '')
, CHR(13), '')
, CHR(09), '') as massaged
FROM BLAH;
Regexp_replace is an option, but you may see a performance hit depending on how complex your expression is.
You could use both LTRIM and RTRIM.
select rtrim(ltrim('abcdab','ab'),'ab') from dual;
If you want to trim CHR(13) only when it comes with a CHR(10) it gets more complicated. Firstly, translated the combined string to a single character. Then LTRIM/RTRIM that character, then replace the single character back to the combined string.
select replace(rtrim(ltrim(replace('abccccabcccaab','ab','#'),'#'),'#'),'#','ab') from dual;
TRANSLATE (column_name, 'd'||CHR(10)||CHR(13), 'd')
The 'd' is a dummy character, because translate does not work if the 3rd parameter is null.
For what version of Oracle? 10g+ supports regexes - see this thread on the OTN Discussion forum for how to use REGEXP_REPLACE to change non-printable characters into ''.
I know this is not a strict answer for this question, but I've been working in several scenarios where you need to transform text data following these rules:
No spaces or ctrl chars at the beginning of the string
No spaces or ctrl chars at the end of the string
Multiple ocurrencies of spaces or ctrl chars will be replaced to a single space
Code below follow the rules detailed above:
WITH test_view AS (
SELECT CHR(9) || 'Q qwer' || CHR(9) || CHR(10) ||
CHR(13) || ' qwerqwer qwerty ' || CHR(9) ||
CHR(10) || CHR(13) str
FROM DUAL
) SELECT
str original
,TRIM(REGEXP_REPLACE(str, '([[:space:]]{2,}|[[:cntrl:]])', ' ')) fixed
FROM test_view;
ORIGINAL FIXED
---------------------- ----------------------
Q qwer Q qwer qwerqwer qwerty
qwerqwer qwerty
1 row selected.
If at all anyone is looking to convert data in 1 variable that lies in 2 or 3 different lines like below
'Data1
Data2'
And you want to display data as 'Data1 Data2' then use below
select TRANSLATE ('Data1
Data2', ''||CHR(10), ' ') from dual;
it took me hrs to get the right output. Thanks to me I just saved you 1 or 2 hrs :)
In cases where the Oracle solution seems overly convoluted, I create a java class with static methods and then install it as a package in Oracle. This might not be as performant, but you will eventually find other cases (date conversion to milliseconds for example) where you will find the java fallback helpful.
Below code can be used to Remove New Line and Table Space in text column
Select replace(replace(TEXT,char(10),''),char(13),'')
Try the code below.
It will work if you enter multiple lines in a single column.
create table products (prod_id number , prod_desc varchar2(50));
insert into products values(1,'test first
test second
test third');
select replace(replace(prod_desc,chr(10),' '),chr(13),' ') from products where prod_id=2;
Output :test first test second test third
TRIM(BOTH chr(13)||chr(10)||' ' FROM str)
Instead of using regexp_replace multiple time use (\s) as given below;
SELECT regexp_replace('TEXT','(\s)','')
FROM dual;
Fowloing code remove newline from both side of string:
select ltrim(rtrim('asbda'||CHR(10)||CHR(13) ,''||CHR(10)||CHR(13)),''||CHR(10)||CHR(13)) from dual
but in most cases this one is just enought :
select rtrim('asbda'||CHR(10)||CHR(13) ,''||CHR(10)||CHR(13))) from dual
UPDATE My_Table
SET Mycolumn1 =
TRIM (
TRANSLATE (Mycolumn1,
CHR (10) || CHR (11) || CHR (13),
' '))
WHERE ( INSTR (Mucolumn1, CHR (13)) > 0
OR INSTR (Mucolumn1, CHR (10)) > 0
OR INSTR (Mucolumn1, CHR (11)) > 0);