How to remove the carriage space using SQL query. I tried the replace (..) idea but its not working at all ? The duplicate idea is not working
I guess you should go with below code -
REPLACE(REPLACE( col_name, CHR(10)), CHR(13))
This will replace the spaces and carriage both and might solve your problem.
I generally use the test1 option:
select replace(replace('teste
return test', chr (13), ''), chr (10), ' ') test1,
translate('teste
return test', chr(10) || chr(13) || chr(09), ' ') test2
from dual;
Related
I need to remove accentes, spaces and special characters in Oracle SQL.
(I thought about using pipe and replace)
My code so far:
SELECT NR_CONHEC || ' - ' || RAZAO_SOCIAL_TRANSP FROM geq_gl_ctms_frete
For example:
7590 - J. T. TRANSPORTES LTDA. - ME
Needed output:
7590-JTTRANSPORTESLTDA-ME
You can try
select regexp_replace('7590 - J. T. TRANSPORTES LTDA. - ME','[[:space:]]|\.')
as "Result String"
from dual;
Result String
-------------------------
7590-JTTRANSPORTESLTDA-ME
For your case, replace with the following query :
SELECT regexp_replace( NR_CONHEC || ' - ' || RAZAO_SOCIAL_TRANSP,'[[:space:]]|\.')
FROM geq_gl_ctms_frete;
and if you want to remove more special chars. and accentes other than dot, add whichever you want after |\. such as |\%|\&|\è..., then those characters %, & or è are removed from the string.
this will work:
SELECT NR_CONHEC||'-'|| reexp_replace(regexp_replace(RAZAO_SOCIAL_TRANSP
RAZAO_SOCIAL_TRANSP,'.',''),' ','') FROM geq_gl_ctms_frete;
How about this?
SQL> with test (col) as
2 (select '7590 - J. T. TRANSPORTES LTDA. - ME' from dual)
3 select regexp_replace(col, '[^([:alpha:][:digit:]-)]') result
4 from test;
RESULT
-------------------------
7590-JTTRANSPORTESLTDA-ME
SQL>
I.e. replace anything that isn't a letter, a number or a hyphen with nothing.
try:
select regexp_replace (NR_CONHEC|| ' - ' || RAZAO_SOCIAL_TRANSP, '[^-0-9a-zA-Z]') as rez from geq_gl_ctms_frete ;
Currently in address column in test table,i have data in following format,
12th street
Test avenue
Test_City
but in the output,i would require it in following format,
12th street Test avenue Test_City.
Could any one please tell me the query to use to display it in the required manner.
You can try this:
select regexp_replace(your_address,'[[:space:]]',' ') from your_tab;
Just strip chr(13) and chr(10) from the string:
declare
teststring varchar2 (32767) := ' This is the value
that I chose';
begin
dbms_output.put_line (teststring);
dbms_output.put_line (replace (replace (teststring, chr (13), ''), chr (10), ' '));
end;
The result:
This is the value
that I chose
This is the value that I chose
Two spaces since I put in two returns in the text.
Try this:
translate(' example ', chr(10) || chr(13) || chr(09), ' ')
The above will replace all line breaks (chr(10)), all tabs (chr(09)) and all carriage returns (chr(13)) with a space (' ').
Just replace ' example ' with your field name on wish you want to have the characters removed.
Another solution would be:
select trim(chr(13) FROM trim(chr(10) FROM your_column)) from your_table
I have an issue with a file that's created from data (in format of bytes) returned from the database.
the problem is that there is a few newlines in the file.
i would like to know if there is a way to write a where clause to check if some record has a newline character?
Using the CHR function to look for the ASCII value:
select *
from your_table
where instr(your_text_col, chr(10)) > 0;
If you want to search for carriage returns, that would be chr(13).
You can write something like:
SELECT id
FROM table_name
WHERE field_name LIKE '%'||CHR(10)||'%'
;
(|| is the concatenation operator; CHR(10) is the tenth ASCII character, i.e. a newline.)
Depending on the platform, a newline will generally either be a CHR(10) (Unix) or a CHR(13) followed by a CHR(10) (Windows). There are other options for other more esoteric platforms, but 99.999% of the time, it will be one of these two.
You can search the data in a column looking for one or both characters
SELECT instr( column_name, CHR(10) ) position_of_first_lf,
instr( column_name, CHR(13) || CHR(10) ) position_of_first_cr_lf
FROM table_name
WHERE instr( column_name, CHR(10) ) > 0
This worked the best for me.
select * from label_master where regexp_like (text, chr(10));
One column of my query output contains char 13 ( new line character). To replace it with nothing I am trying to use below function but it is giving me below error ORA-00936: missing expression
select
replace(AUDITOR_COMMENTS,char(13),'')
from csa_sli_all.T_CONV_QUOTE
When I put char(13) in quote 'char(13)' error goes but it will not do as desired.
I think I cannot include char(13) in quotes .
I am using Oracle Database 10g Release 10.2.0.1.0 - 64bit Production
The function isn't char it's chr try calling:
select
replace(AUDITOR_COMMENTS,chr(13),'')
from csa_sli_all.T_CONV_QUOTE
try chr(13) instead of char(13) and see if it works
Try this :
REPLACE(col_name, CHR(13) + CHR(10), '')
or
REPLACE(REPLACE( col_name, CHR(10) ), CHR(13) )
replace(your_data, chr(13), '')
try this as #sebastian said
select
replace(AUDITOR_COMMENTS,chr(13),'')
from csa_sli_all.T_CONV_QUOTE
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);