in my table one column contains data as below
BMS/430301420-XN/0
I need to use substr function in oracle and output to be taken as
430301420-XN
the one I used is as below
substr(buy_id,5),substr(substr(buy_id,5),instr(buy_id,'/',2))
but it is not working please help me
If you know the format of the string and you always want to start on the fifth character and remove the last two, then:
select substr(str, 5, -2)
If you just want the part between the slashes, then use regexp_substr():
select replace(regexp_substr(str, '/.*/'), '/', '')
Easiest way is a Regular Expression, find the string between the slashes but don't include them in the result:
regexp_substr(buy_id, '(?<=/).*(?=/)')
With a combination of SUBSTR and INSTR:
SQL> WITH DATA AS(
2 SELECT 'BMS/430301420-XN/0' str FROM dual UNION ALL
3 SELECT 'BMSABC/430301420-XN/0' str FROM dual UNION ALL
4 SELECT 'BMS/430301420-XN/012345' str FROM dual
5 )
6 SELECT str,
7 SUBSTR(str, instr(str, '/', 1, 1)+1, instr(str, '/', 1, 2)
8 -instr(str, '/', 1, 1)-1) new_str
9 FROM DATA;
STR NEW_STR
----------------------- -----------------------
BMS/430301420-XN/0 430301420-XN
BMSABC/430301420-XN/0 430301420-XN
BMS/430301420-XN/012345 430301420-XN
SQL>
The above uses the logic to find the substring between the first and second occurrence of the /.
This will also Work :D
select Column_Name as OLD , substr(''||to_char(Column_Name)||'',instr
(''||to_char(Column_Name)||'','/',1)+1,(instr(''||to_char(Column_Name)
||'','/',1,2)-instr(''||to_char(Column_Name)||'','/',1,1)-1)) as NEW from Table_Name;
Same Use Of substr and instr
my answer is :
select
substr('BMS/430301420-XN/0',
(instr('BMS/430301420-XN/0','/') +1),
(instr('BMS/430301420-XN/0','/',(instr('BMS/430301420-XN/0','/')+1))-instr('BMS/430301420-XN/0','/')-1 ))
from dual
you can see this sample :
http://www.sqlfiddle.com/#!4/9eecb7/863/0
Related
I need to replace all characters with nothing before the . character and also replace all [ and ] with nothing.
Please see examples below:
from
to
[PINWHEEL_ASSET].[MX5530]
MX5530
[PINWHEEL_TRADE].[AR5403]
AR5403
The parts before and after the . dot are variables.
with
sample_data (my_string) as (
select '[PINWHEEL_ASSET].[MX5530]' from dual
)
select rtrim(substr(my_string, instr(my_string, '.') + 2), ']') as second_part
from sample_data
;
SECOND_PART
-----------
MX5530
This assumes that the input string looks exactly like this: [first].[second], where "first" and "second" are (possibly empty) strings that do not contain periods or closing brackets.
Yet another option is to use regular expressions (see line #6).
Sample data:
SQL> with test (col) as
2 (select '[PINWHEEL_ASSET].[MX5530]' from dual union all
3 select '[PINWHEEL_TRADE].[AR5403]' from dual
4 )
Query begins here:
5 select col,
6 regexp_substr(col, '\w+', 1, 2) result
7 from test;
COL RESULT
------------------------- --------------------
[PINWHEEL_ASSET].[MX5530] MX5530
[PINWHEEL_TRADE].[AR5403] AR5403
SQL>
I have a table with a "Link" attribute. It has the following meaning:
INC102
INC1020
INC10200
I want to get the following result:
INC102
INC1020
INC10200
I need to leave the INC and the numbers after it without .
Tell me which command will help here? Since I understand that "Substr" will not work here.
I am use SQL Developer - Oracle
If you just want to extract "INC" with the following digits, use regexp_substr():
select regexp_substr(link, 'INC[0-9]+')
Here is a db<>fiddle.
Since I understand that "Substr" will not work here.
Says who?
SQL> with test (col) as
2 (select 'INC102' from dual union all
3 select 'INC1020' from dual union all
4 select 'INC10200' from dual
5 )
6 select
7 substr(col,
8 instr(col, '>') + 1,
9 instr(col, '<', instr(col, '>')) - instr(col, '>') - 1
10 ) result
11 from test;
RESULT
------------------------------------------------------------------------------------------------------------------------
INC102
INC1020
INC10200
SQL>
What does it do?
lines #1 - 5: sample data
line #8: starting point of the SUBSTR function is one character after the first > sign
line #9: length (used as the 3rd parameter of the SUBSTR) is position of the first < that follows the first > minus position of the first >
And that's it ... why wouldn't it work?
You could treat [<>] as the word delimiter and take the second word:
with test (col) as
( select 'INC102' from dual union all
select 'INC1020' from dual union all
select 'INC10200' from dual
)
select regexp_substr(col,'[^<>]+', 1, 2)
from test;
REGEXP_SUBSTR(COL,'[^<>]+',1,2)
-------------------------------
INC102
INC1020
INC10200
I have a string like "1490/2334/5166400411000434" from which I need to derive value after second slash. I tried below logic
select REGEXP_SUBSTR('1490/2334/5166400411000434','[^/]+',1,3) from dual;
it is working fine. But when i dont have value between first and second slash it is returining blank.
For example my string is "1490//5166400411000434" and am trying
select REGEXP_SUBSTR('1490//5166400411000434','[^/]+',1,3) from dual;
it is returning blank. Please suggest me what i am missing.
If I understand well, you may need
regexp_substr(t, '(([^/]*/){2})([^/]*)', 1, 1, 'i', 3)
This handles the first 2 parts like 'xxx/' and then checks for a sequence of non / characters; the parameter 3 is used to get the 3rd matching subexpression, which is what you want.
For example:
with test(t) as (
select '1490/2334/5166400411000434' from dual union all
select '1490//5166400411000434' from dual union all
select '1490//5166400411000434/ramesh/3344' from dual
)
select t, regexp_substr(t, '(([^/]*/){2})([^/]*)', 1, 1, 'i', 3) as substr
from test
gives:
T SUBSTR
---------------------------------- ----------------------------------
1490/2334/5166400411000434 5166400411000434
1490//5166400411000434 5166400411000434
1490//5166400411000434/ramesh/3344 5166400411000434
You can REVERSE() your string and take the value before the first slash. And then reverse again to obtain the desired output.
select reverse(regexp_substr(reverse('1490//5166400411000434'), '[^/]+', 1, 1)) from dual;
It can also be done with basic substring and instr function:
select reverse(SUBSTR(reverse('1490//5166400411000434'), 0, INSTR(reverse('1490//5166400411000434'), '/')-1)) from dual;
Use other options in REGEXP_SUBSTR to match a pattren
select REGEXP_SUBSTR('1490//5166400411000434','(/\d*)/(\d+)',1,1,'x',2) from dual
Basically it is finding the pattren of two / including digits starting from 1 with 1 appearance and ignoring whitespaces ('x') then outputting 2nd subexpression that is in second expression within ()
... pattern,1,1,'x',subexp2)
I'm struggling with a query in Oracle SQL, wanting to get some timings out of some text stored in an Oracle db.
Table :
kde_test (myString varchar(50))
Table contents (3 records):
'task1 - 6m'
'task2 - 66m'
'task3 - 666m'
I would like to get only the interesting part of the string, being the timings, so I would like to get only '6', '66' & '666' as results.
Searched this forum a bit, and got up with this query eventually, but it seems I do not completely get it, as the results it returns are :
6m
66m
666m
select
CASE
WHEN myString like 'task1%' THEN substr(myString,9,INSTR(myString,'m',1,1)-1)
WHEN myString like 'task2%' THEN substr(myString,9,INSTR(myString,'m',1,1)-1)
WHEN myString like 'task3%' THEN substr(myString,9,INSTR(myString,'m',1,1)-1)
END
from kde_test
where myString like 'task%'
EDIT :
Since some solutions (thanks already for quick response) take into account the specific values (eg. all 3 records ending on '6m'), maybe it best to take into account the values could be :
Table contents (3 records):
'task1 - 6m'
'task2 - 58m'
'task3 - 123m'
you can use this way too
select regexp_replace(SUBSTR('task3 - 666m' ,
INSTR('task3 - 666m', '-',1, 1)+1, length('task3 - 666m')), '[A-Za-z]')
from dual
result :666
Use SUBSTR and INSTR and make it dynamic.
SUBSTR(str,
instr(str, ' - ', 1, 1) +3,
instr(str, 'm', 1, 1) -
instr(str, ' - ', 1, 1) -3
)
For example,
SQL> WITH DATA AS(
2 SELECT 'task1 - 6m' str FROM dual UNION ALL
3 SELECT 'task2 - 66m' str FROM dual UNION ALL
4 SELECT 'task3 - 666m' str FROM dual UNION ALL
5 SELECT 'task4 - 58m' str FROM dual UNION ALL
6 SELECT 'task5 - 123m' str FROM dual
7 )
8 SELECT str,
9 SUBSTR(str, instr(str, ' - ', 1, 1) +3,
10 instr(str, 'm', 1, 1) - instr(str, ' - ', 1, 1) -3) new_st
11 FROM DATA;
STR NEW_STR
------------ ------------
task1 - 6m 6
task2 - 66m 66
task3 - 666m 666
task4 - 58m 58
task5 - 123m 123
SQL>
You can use the regex_substr function. \d+ means one or more digits, and $ anchors the end of the string.
select regexp_substr(str, '\d+m$')
from mytable
Example at SQL Fiddle.
In order to correct your current query, you should change the following string - "INSTR(myString,'m',1,1)-1" to "INSTR(myString,'m',1,1)-9".
However, the other answers provided above seem like a more elegant solution to your problem.
I did feel the need to publish this just to clarify what wasn't working well in current query - in INSTR function returns the position of the m letter, and then used as the length of the string to print. What my fix does is telling the query to print everything from the 9th character until the position of the m letter, which results in the task time required.
I have tried to divide this into two parts
First pick the string after -
regexp_substr ('task1 - 1234m', '[^ _ ]+',1, 3) --results 1234m
Fetch the number part of the string fetched from output of first
regexp_substr(regexp_substr ('task1 - 1234m', '[^ _ ]+',1, 3),'[[:digit:]]*')
--output 1234
So,the final query is
SELECT regexp_substr(regexp_substr (mystring, '[^ _ ]+',1, 3),'[[:digit:]]*')
FROM kde_test;
Use This:-
select substr(replace(myString,'m',''),9) output
from kde_test
where myString like 'task%'
I have a string and I would like to split that string by delimiter at a certain position.
For example, my String is F/P/O and the result I am looking for is:
Therefore, I would like to separate the string by the furthest delimiter.
Note: some of my strings are F/O also for which my SQL below works fine and returns desired result.
The SQL I wrote is as follows:
SELECT Substr('F/P/O', 1, Instr('F/P/O', '/') - 1) part1,
Substr('F/P/O', Instr('F/P/O', '/') + 1) part2
FROM dual
and the result is:
Why is this happening and how can I fix it?
Therefore, I would like to separate the string by the furthest delimiter.
I know this is an old question, but this is a simple requirement for which SUBSTR and INSTR would suffice. REGEXP are still slower and CPU intensive operations than the old subtsr and instr functions.
SQL> WITH DATA AS
2 ( SELECT 'F/P/O' str FROM dual
3 )
4 SELECT SUBSTR(str, 1, Instr(str, '/', -1, 1) -1) part1,
5 SUBSTR(str, Instr(str, '/', -1, 1) +1) part2
6 FROM DATA
7 /
PART1 PART2
----- -----
F/P O
As you said you want the furthest delimiter, it would mean the first delimiter from the reverse.
You approach was fine, but you were missing the start_position in INSTR. If the start_position is negative, the INSTR function counts back start_position number of characters from the end of string and then searches towards the beginning of string.
You want to use regexp_substr() for this. This should work for your example:
select regexp_substr(val, '[^/]+/[^/]+', 1, 1) as part1,
regexp_substr(val, '[^/]+$', 1, 1) as part2
from (select 'F/P/O' as val from dual) t
Here, by the way, is the SQL Fiddle.
Oops. I missed the part of the question where it says the last delimiter. For that, we can use regex_replace() for the first part:
select regexp_replace(val, '/[^/]+$', '', 1, 1) as part1,
regexp_substr(val, '[^/]+$', 1, 1) as part2
from (select 'F/P/O' as val from dual) t
And here is this corresponding SQL Fiddle.