I have an integer field in a table.
I want to read the first digit of this field, then up to that digit read next digits.
For example consider this field: 355560
I read the first digit (3)
Then read 3 digits after 3 : (555)
How would I write my select query?
SELECT SUBSTR (355560, 2, SUBSTR (355560, 1, 1))
FROM DUAL;
select substr('355560', 2, substr('355560', 0, 1)) from dual
try this one
SELECT SUBSTR(355560,1,1) FROM DUAL
OUTPUT : 3
SELECT SUBSTR(355560,2,3) FROM DUAL
OUTPUT : 555
As I understand your question:
select concat(substr(35550,1,1),substr(35550,2,3)) from dual
Related
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
135 ;1111776698 ;AB555678765
I have the above string and what I am looking for is to retrieve all the digits before the first occurrence of ;.
But the number of characters before the first occurrence of ; varies i.e. it may be a 4 digit number or 3 digit number.
I have played with regex_instr and instr, but I unable to figure this out.
The query should return all the digits before the first occurrence of ;
This answer assumes that you are using Oracle database. I don't know of way to do this using REGEX_INSTR alone, but we can do with REGEXP_REPLACE using capture groups:
SELECT REGEXP_REPLACE('135 ;1111776698 ;AB555678765', '^\s*(\d{3,4})\s*;.*', '\1')
FROM dual;
Demo
Here is the regex pattern being used:
^\s*(\d{3,4})\s*;.*
This allows, from the start of the string, any amount of leading whitespace, followed by a 3 or 4 digit number, followed again by any amount of whitespace, then a semicolon. The .* at the end of the pattern just consumes whatever remains in your string. Note (\d{3,4}), which captures the 3-4 digit number, which is then available in the replacement as \1.
Using INSTR,SUBTSR and TRIM should work ( based on your comment that there are "just white spaces and digits" )
select TRIM(SUBSTR(s,1, INSTR(s,';')-1)) FROM t;
Demo
The following using regexp_substr() should work:
SELECT s, REGEXP_SUBSTR(s, '^[^;]*')
Make sure you try all possible values in that first position, even those you don't expect and make sure they are handled as you want them to be. Always expect the unexpected! This regex matches the first subgroup of zero or more optional digits (allows a NULL to be returned) when followed by an optional space then a semi-colon, or the end of the line. You may need to tighten (or loosen) up the matching rules for your situation, just make sure to test even for incorrect values, especially if the input comes from user-entered data.
with tbl(id, str) as (
select 1, '135 ;1111776698 ;AB555678765' from dual union all
select 2, ' 135 ;1111776698 ;AB555678765' from dual union all
select 3, '135;1111776698 ;AB555678765' from dual union all
select 4, ';1111776698 ;AB555678765' from dual union all
select 5, ';135 ;1111776698 ;AB555678765' from dual union all
select 6, ';;1111776698 ;AB555678765' from dual union all
select 7, 'xx135 ;1111776698 ;AB555678765' from dual union all
select 8, '135;1111776698 ;AB555678765' from dual union all
select 9, '135xx;1111776698 ;AB555678765' from dual
)
select id, regexp_substr(str, '(\d*?)( ?;|$)', 1, 1, NULL, 1) element_1
from tbl
order by id;
ID ELEMENT_1
---------- ------------------------------
1 135
2 135
3 135
4
5
6
7 135
8 135
9
9 rows selected.
To get the desired result, you should use REGEX_SUBSTR as it will substring your desired data from the string you give. Here is the example of the Query.
Solution to your example data:
SELECT REGEXP_SUBSTR('135 ;1111776698 ;AB555678765','[^;]+',1,1) FROM DUAL;
So what it does, Regex splits the string on the basis of ; separator. You needed the first occurrence so I gave arguments as 1,1.
So if you need the second string 1111776698 as your output you can give an argument as 1,2.
The syntax for Regexp_substr is as following:
REGEXP_SUBSTR( string, pattern [, start_position [, nth_appearance [, match_parameter [, sub_expression ] ] ] ] )
Here is the link for more examples:
https://www.techonthenet.com/oracle/functions/regexp_substr.php
Let me know if this works for you. Best luck.
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 working with an Oracle DB and I'm trying to find and extract ALL occurrences in a string matching a specific pattern...
It's supposed to be 3 letters, 3 numbers and then maybe a letter or not
I tried this:
SELECT REGEXP_SUBSTR(my_column, '[A-Za-z]{3}(\d)(\d)(\d)') AS values
FROM my_table
but it only returns the first occurrence.
Using
REGEXP_SUBSTR(my_column, '[A-Za-z]{3}(\d)(\d)(\d)', 0, 0, 'i')
doesn't work either
Does anybody have any ideas?
Edit:
I'm trying to extract it from PLSQL files. So its pretty much like SQL queries like
select *
from abc123
where some_value = 'some_value'
Try this query to break ABC123CDE456FGHI789 squence
with mine as (select 'ABC123CDE456FGH789' hello from dual)
select regexp_substr(hello, '[A-Za-z]{3}(\d){3}', 1, level) STR from mine
connect by regexp_substr(hello, '[A-Za-z]{3}(\d){3}', 1, level) is not null
Output
ABC123
CDE456
GHI789
For get specific postion then you want to use
select regexp_substr('ABC123CDE456FGH789', '[A-Za-z]{3}(\d){3}', 1, i) STR from dual
change i value as per position like
select regexp_substr('ABC123CDE456FGH789', '[A-Za-z]{3}(\d){3}', 1, 1) STR from dual
Output :
ABC123
Try to get number in 1.2.3 (suppose it is a domain of a country)
SELECT str,level,REGEXP_SUBSTR(str, '[0-9]', 1, LEVEL) AS substr
FROM (
SELECT country_domain str from country where regexp_like(country_domain, '[0-9]')
)
CONNECT BY LEVEL <= REGEXP_count(str, '[0-9]');
Output
STR LEVEL SUBSTR
1.2.3 1 1
1.2.3 2 2
1.2.3 3 3
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