Let's say I have three rows with value as
1 121/2808B|:6081
2 OD308B|:6081_1:
3 008312100001200|:6081_1
I want to display value only until B but want to exclude everything after B. So as you can see in above data:
from 121/2808B|:6081 I want only 121/2808B
from OD308B|:6081_1: only OD308B
from 008312100001200|:6081_1 only 008312100001200.
Thanks for the Help.
Try this: regexp_substr('<Your_string>','[^B]+')
SELECT
REGEXP_SUBSTR('121/2808B|:6081', '[^B]+')
FROM
DUAL;
REGEXP_S
--------
121/2808
SELECT
REGEXP_SUBSTR('OD308B|:6081_1:', '[^B]+')
FROM
DUAL;
REGEX
-----
OD308
SELECT
REGEXP_SUBSTR('008312100001200.', '[^B]+')
FROM
DUAL;
REGEXP_SUBSTR('0
----------------
008312100001200.
db<>fiddle demo
Cheers!!
You could try using SUBSTR() and INSTR()
select SUBSTR('121/2808B|:6081',1,INSTR('121/2808B|:6081','B', 1, 1) -1)
from DUAL
I think you forgot to mention that you wanted to use | as a field separator, but I deduced this from the expected result from the third string. As such the following should give you what you want:
WITH cteData AS (SELECT 1 AS ID, '121/2808B|:6081' AS STRING FROM DUAL UNION ALL
SELECT 2, 'OD308B|:6081_1:' FROM DUAL UNION ALL
SELECT 3, '008312100001200|:6081_1' FROM DUAL)
SELECT ID, STRING, SUBSTR(STRING, 1, CASE
WHEN INSTR(STRING, 'B') = 0 THEN INSTR(STRING, '|')-1
ELSE INSTR(STRING, 'B')-1
END) AS UP_TO_B
FROM cteData;
dbfiddle here
Assuming Bob Jarvis is correct in the assumption that "|" is also a delimiter (as seems likely) try:
-- define test data
with test as
( select '121/2808B|:6081' stg from dual union all
select 'OD308B|:6081_1:' from dual union all
select '008312100001200|:6081_1' from dual
)
-- execute extract
select regexp_substr(stg , '[^B|]+') val
from test ;
Related
I am working with table data that contains strings with decimal and back-slash like below:
info
1/2.2.2
2/1.1.1
3/1.1.11
I need to use a regular expression to replace the data like below:
info
1/2.2
2/1.1
3/1.1
Don't use a (slow) regular expression, use simple (faster) string functions instead:
SELECT info,
CASE
WHEN INSTR(info, '.', 1, 2) > 0
THEN SUBSTR(info, 1, INSTR(info, '.', 1, 2) - 1)
ELSE info
END AS part
FROM table_name;
Which, for the sample data:
CREATE TABLE table_name (info) AS
SELECT '1/2.2.2' FROM DUAL UNION ALL
SELECT '2/1.1.1' FROM DUAL UNION ALL
SELECT '3/1.1.11' FROM DUAL UNION ALL
SELECT '3/1.1' FROM DUAL;
Outputs:
INFO
PART
1/2.2.2
1/2.2
2/1.1.1
2/1.1
3/1.1.11
3/1.1
3/1.1
3/1.1
If you want to update the table then:
UPDATE table_name
SET info = SUBSTR(info, 1, INSTR(info, '.', 1, 2) - 1)
WHERE INSTR(info, '.', 1, 2) > 0
fiddle
For the sake of argument, here's a solution using REGEXP_SUBSTR(). REGEXP_SUBSTR() returns NULL if the pattern is not found. Thanks to MT0 for the CTE so I didn't have to type it up :-)
WITH table_name(ID, info) AS (
SELECT 1, '1/2.2.2' FROM DUAL UNION ALL
SELECT 2, '2/1.1.1' FROM DUAL UNION ALL
SELECT 3, '3/1.1.11' FROM DUAL UNION ALL
SELECT 4, '3/1.1' FROM DUAL UNION ALL
SELECT 5, '4/4' FROM DUAL)
SELECT ID, REGEXP_SUBSTR(info, '\d/\d\.\d') DATA
from table_name;
ID DATA
---------- --------
1 1/2.2
2 2/1.1
3 3/1.1
4 3/1.1
5
5 rows selected.
I have to test a column of a sql table for invalid values and for NULL.
Valid values are: Any number and the string 'n.v.' (with and without the dots and in every possible combination as listed in my sql command)
So far, I've tried this:
select count(*)
from table1
where column1 is null
or not REGEXP_LIKE(column1, '^[0-9,nv,Nv,nV,NV,n.v,N.v,n.V,N.V]+$');
The regular expression also matches the single character values 'n','N','v','V' (with and without a following dot). This shouldn't be the case, because I only want the exact character combinations as written in the sql command to be matched. I guess the problem has to do with using REGEXP_LIKE. Any ideas?
I guess this regexp will work:
NOT REGEXP_LIKE(column1, '^([0-9]+|n\.?v\.?)$', 'i')
Note that , is not a separator, . means any character, \. means the dot character itself and 'i' flag could be used to ignore case instead of hard coding all combinations of upper and lower case characters.
No need to use regexp (performance will increase by large data) - plain old TRANSLATE is good enough for your validation.
Note that the first translate(column1,'x0123456789','x') remove all numeric charcters from the string, so if you end with nullthe string is OK.
The second translate(lower(column1),'x.','x') removes all dots from the lowered string so you expect the result nv.
To avoid cases as n.....v.... you also limit the string length.
select
column1,
case when
translate(column1,'x0123456789','x') is null or /* numeric string */
translate(lower(column1),'x.','x') = 'nv' and length(column1) <= 4 then 'OK'
end as status
from table1
COLUMN1 STATUS
--------- ------
1010101 OK
1012828n
1012828nv
n.....v....
n.V OK
Test data
create table table1 as
select '1010101' column1 from dual union all -- OK numbers
select '1012828n' from dual union all -- invalid
select '1012828nv' from dual union all -- invalid
select 'n.....v....' from dual union all -- invalid
select 'n.V' from dual; -- OK nv
You can use:
select count(*)
from table1
WHERE TRANSLATE(column1, ' 0123456789', ' ') IS NULL
OR LOWER(column1) IN ('nv', 'n.v', 'nv.', 'n.v.');
Which, for the sample data:
CREATE TABLE table1 (column1) AS
SELECT '12345' FROM DUAL UNION ALL
SELECT 'nv' FROM DUAL UNION ALL
SELECT 'NV' FROM DUAL UNION ALL
SELECT 'nV' FROM DUAL UNION ALL
SELECT 'n.V.' FROM DUAL UNION ALL
SELECT '...................n.V.....................' FROM DUAL UNION ALL
SELECT '..nV' FROM DUAL UNION ALL
SELECT 'n..V' FROM DUAL UNION ALL
SELECT 'nV..' FROM DUAL UNION ALL
SELECT 'xyz' FROM DUAL UNION ALL
SELECT '123nv' FROM DUAL;
Outputs:
COUNT(*)
5
or, if you want any quantity of . then:
select count(*)
from table1
WHERE TRANSLATE(column1, ' 0123456789', ' ') IS NULL
OR REPLACE(LOWER(column1), '.') = 'nv';
Which outputs:
COUNT(*)
9
db<>fiddle here
REGEXP_SUBSTR(label) ,'.*_dis')
this is for sql;
my database is mysql
select REGEXP_SUBSTR(label) ,'.*_dis') as dis ,
substr(label,length(label))-1) as num
from table
table.lable col's data:
1. a_b_dis_12
2. a_dis_13
3. c_d_dis_23
3. c_dis_22
i want to get the character before '_dis' and the numeric part use regexp
1.a_b 12
2.a 13
3.c_d 23
4.c 22
thanks a lot!
You can use regexp_substr as follows:
Select regexp_substr(your_column, '^(.*)_dis_[0-9]+$',1,1,null,1) as dis,
Regexp_substr(your_column, '[0-9]+$') as num
From your table
You can use regexp_replace():
select regexp_replace(col, '^(.*)_dis.*$', '\1'),
regexp_replace(col, '^.*_dis_([0-9]+)', '\1')
from (select 'a_b_dis_12' as col from dual union all
select 'a_dis_13' as col from dual union all
select 'c_d_dis_23' as col from dual union all
select 'c_dis_22' as col from dual
) t;
Here is a db<>fiddle.
I would use regexp_replace() as follows:
select
regexp_replace(label, '_dis_.*$', '') dis,
regexp_replace(label, '^.*_dis_', '') num
from mytable
The first expression suppresses everything from '_dis_ (included) to the end of the string. The second expression removes everything from the beginning of the string until '_dis_' (included).
I need to use regexp_substr, but I can't use it properly
I have column (l.id) with numbers, for example:
1234567891123!123 EXPECTED OUTPUT: 1234567891123
123456789112!123 EXPECTED OUTPUT: 123456789112
12345678911!123 EXPECTED OUTPUT: 12345678911
1234567891123!123 EXPECTED OUTPUT: 1234567891123
I want use regexp_substr before the exclamation mark (!)
SELECT REGEXP_SUBSTR(l.id,'[%!]',1,13) from l.table
is it ok ?
You can try using INSTR() and substr()
DEMO
select substr(l.id,1,INSTR(l.id,'!', 1, 1)-1) from dual
You want to remove the exclamation mark and all following characters it seems. That is simply:
select regexp_replace(id, '!.*', '') from mytable;
Look at it like a delimited string where the bang is the delimiter and you want the first element, even if it is NULL. Make sure to test all possibilities, even the unexpected ones (ALWAYS expect the unexpected)! Here the assumption is if there is no delimiter you'll want what's there.
The regex returns the first element followed by a bang or the end of the line. Note this form of the regex handles a NULL first element.
SQL> with tbl(id, str) as (
select 1, '1234567891123!123' from dual union all
select 2, '123456789112!123' from dual union all
select 3, '12345678911!123' from dual union all
select 4, '1234567891123!123' from dual union all
select 5, '!123' from dual union all
select 6, '123!' from dual union all
select 7, '' from dual union all
select 8, '12345' from dual
)
select id, regexp_substr(str, '(.*?)(!|$)', 1, 1, NULL, 1)
from tbl
order by id;
ID REGEXP_SUBSTR(STR
---------- -----------------
1 1234567891123
2 123456789112
3 12345678911
4 1234567891123
5
6 123
7
8 12345
8 rows selected.
SQL>
If you like to use REGEXP_SUBSTR rather than regexp_replace then you can use
SELECT REGEXP_SUBSTR(l.id,'^\d+')
assuming you have only numbers before !
If I understand correctly, this is the pattern that you want:
SELECT REGEXP_SUBSTR(l.id,'^[^!]+', 1)
FROM (SELECT '1234567891123!123' as id from dual) l
Let's say i have following data:
fjflka, kdjf
ssssllkjf fkdsjl
skfjjsld, kjl
jdkfjlj, ksd
lkjlkj hjk
I want to cut out everything after ',' but in case the string doesn't contain this character, it wont do anything, if i use substr and cut everything after ',' the string which doesn't contain this character shows as null. How do i achieve this? Im using oracle 11g.
This should work. Simply use regexp_substr
with t_view as (
select 'fjflka, kdjf' as text from dual union
select 'ssssllkjf fkdsjl' from dual union
select 'skfjjsld, kjl' from dual union
select 'jdkfjlj, ksd' from dual union
select 'lkjlkj hjk' from dual
)
select text,regexp_substr(text,'[^,]+',1,1) from t_view;
Assuming your table :
SQL> desc mytable
s varchar2(100)
you may use:
select decode(instr(s,','),0,s,substr(s,1,instr(s,',')-1)) from mytable;
demo
Well the below query works as per your requirement.
with mytable as
(select 'aaasfasf wqwe' s from dual
union all
select 'aaasfasf, wqwe' s from dual)
select s,substr(s||',',1,instr(s||',',',')-1) from mytable;