Hi I have a string like this
ABCDEFGH I want the output to be ABCDEF.GH
If it's a number like 1234567 then i want the output to be 12345.67
Basically i want the delimeter (.) before last 2 characters.
You can use regular expressions for this:
with v_data(val) as (
select '123456' from dual union all
select 'abcdef' from dual union all
select '678' from dual
)
select
val,
regexp_replace(val, '(\d+)(\d{2})', '\1.\2')
from v_data
This matches
one or more digits (\d+) (capturing them in group #1)
followed by exactly two digits (\d{2}) (capturing them in group #2)
and replaces this with the contents of group #1 followed by a . followed by the contents of group #2: \1.\2
Related
Table name: TEST
Column name: ID [VARCHAR(200)]
The format of ID is ‘XXXXXXXX-X’, where ‘X’ is a number from 0 to 9.
Additional operations in case above format is not satisfied:
if the ID consists of 9 digits and there is a double dash between eighth and ninth digit , the extra dash is removed (e.g. 08452142--6 -> 08452142-6)
if the ID consists of 9 digits and there is/are space(s) between eighth and ninth digit and/or non-digits and/or non-letter symbol(s) then replace them to dash (e.g. 08452142 - . 3 -> 08452142-3)
if the ID consists 9 digits and starts/ends with non-digits and/or non-letter symbol(s) then delete that symbol(s) up to digit (e.g. 08452142-2.. -> 08452142-2)
if the ID contains only 9 digits then put a dash before the last digit (e.g. 123456789 -> 12345678-9)
I have achieved the necessary format by using the below snippet.
UPDATE TEST
SET ID = (SELECT REGEXP_REPLACE(ID,'^\d{8}-\d{1}$','') AS "ID"
from TEST
WHERE PK = 11;
)
What are the possible ways to add transformations as mentioned in points[1-4] above in a single query?
Using REGEXP_REPLACE, I can achieve ID in above format. But in case format is incorrect, and ID needs to be transformed[like removing extra dash, or adding dash in case 9 digits are received] to achieve satisfactory format, how can that be achieved in a single UPDATE query?
In any case, you need to extract 9 digits from your string in the first step. And then
add a hyphen before the last character. For both steps use regexp_replace() function
with test(id) as
(
select '08452142--6' from dual union all
select '08452142 - . 3' from dual union all
select '08452142-2..' from dual union all
select '123456789' from dual union all
select '1234567890' from dual
)
select case when length(regexp_replace(id,'(\D)'))=9 then
regexp_replace(regexp_replace(id,'(\D)'),
'(^[[:digit:]]{8})(.*)([[:digit:]]{1}$)','\1-\3')
end as id
from test;
ID
----------
08452142-6
08452142-3
08452142-2
12345678-9
<null>
Demo
You can use the following I think:
UPDATE TEST
SET ID = REGEXP_REPLACE(ID,'^\D*(\d{8})\D*(\d)\D*$','\1-\2')
WHERE REGEXP_LIKE(ID,'^\D*(\d{8})\D*(\d)\D*$')
This way you ignore all non-digit charcters and search for a 8-digit number and then an 1-digit number. Take these 2 numbers and put a single '-' in between.
This is a little more generous as you might need but should work with all your provided examples.
I think you want the first 8 digits, then a hyphen, then the 9th digit:
select ( substr(regexp_replace(id, '[^0-9]', ''), 1, 8) ||
'-' ||
substr(regexp_replace(id, '[^0-9]', ''), 9, 1)
)
I tried an approach based on the suggestion by #BarbarosÖzhan:
with source as (
select '02426467--6' id from dual union all
select '02426467-6' id from dual union all
select '02597718 -- . 3' id from dual union all
select '02597718 --dF5 . 3' id from dual union all
select '00120792-2..' id from dual union all
select '..00120792-2..' id from dual union all
select '123456789' id from dual union all
select '1234567890' id from dual
)
select
case
when regexp_like(id, '\d{8}-\d{1}')
then id
else
case
when regexp_like(id, '\d{8}-\d{1}')
then id
else
case
when regexp_count(id, '\d') = 9
then
case
when
regexp_like(
regexp_replace(
regexp_replace(
id, '(\d{8}-)(-)(\d{1})', '\1\3'
), '(\d{8})([^A-Za-z1-9])(\d{1})', '\1-\3'
)
, '\d{8}-\d{1}')
then
regexp_replace(
regexp_replace(
id, '(\d{8}-)(-)(\d{1})', '\1\3'
), '(\d{8})([^A-Za-z1-9])(\d{1})', '\1-\3'
)
else id
end
else id
end
end id_tr
from source
However, in cases 3 and 4, I cannot get rid of the space, dot and alphabets. I think something wrong with the logic in case length is more than 9. I end with "id" as it is so the result is the same without any modifications.
Any suggestions to impprove this?
I need get a substring from the below example
luvi.luci#gma
and i want to return luci. So basically i need to remove all the information before '.' and after '#'
more examples:
pd.prd#gded
You can do this with regexp_substr(). Here is an example:
select translate(regexp_substr(email, '[.].*#', 1, 1), 'x.#', 'x')
from (select 'luvi.luci#gma' as email from dual) x
with data (val) as
(
select null from dual union all
select 'luvi.luci' from dual union all
select 'luvi.luci#gma' from dual union all
select 'pd.prd#gded' from dual
)
-- step:1
-- find the second group (\2) within the match
-- ie. (any word/sequence of characters (\w+) flanked by a dot and a #)
-- step:2
-- |. OR any other character not matched in step:1 - will be ignored
-- step:3
-- \2 for each match found while parsing, for the entire match,
-- replace it with the second group - so the dot and the # are dropped from the match
select val, regexp_replace (val, '(\.(\w+)#)|.', '\2') ss from data;
I have a table in oracle in which we have one column having data as B12345, means first alphabet always B and followed by numeric. I want to replace all such instances with BH that will become BH12345
So if already there is a value called BH45678 in that column don't update.
Only where find B followed by numeric need updates.
Get the rows which have B followed by digits using regexp_like. Then use replace to replace B with BH for those rows.
select replace(col,'B','BH')
from tablename
where regexp_like(col,'^B\d+$')
with
inputs( str ) as (
select 'B123' from dual union all
select 'BONE' from dual union all
select 'BH55' from dual union all
select 'Z123' from dual union all
select 'B13H' from dual
)
select str, regexp_replace(str, '^B(\d)', 'BH\1') as new_str
from inputs
;
STR NEW_STR
---- -------
B123 BH123
BONE BONE
BH55 BH55
Z123 Z123
B13H BH13H
5 rows selected.
I have a sample source string like below, which was in pipe delimited format in that the value obr can be at anywhere. I need to get the second value of the pipe from the first occurrence of obr. So for the below source strings the expected would be,
Source string:
select 'asd|dfg|obr|1|value1|end' text from dual
union all
select 'a|brx|123|obr|2|value2|end' from dual
union all
select 'hfv|obr|3|value3|345|pre|end' from dual
Expected output:
value1
value2
value3
I have tried the below regexp in oracle sql, but it is not working fine properly.
with t as (
select 'asd|dfg|obr|1|value1|end' text from dual
union all
select 'a|brx|123|obr|2|value2|end' from dual
union all
select 'hfv|obr|3|value3|345|pre|end' from dual
)
select text,to_char(regexp_replace(text,'*obr\|([^|]*\|)([^|]*).*$', '\2')) output from t;
It is working fine when the string starts with OBR, but when OBR is in the middle like the above samples it is not working fine.
Any help would be appreciated.
Not sure of how Oracle handles regular expressions, but starting with an asterisk usually implies that you're looking for zero or more null characters.
Have you tried '^.*obr\|([^|]*\|)([^|]*).*$' ?
This handles null elements and is wrapped in a NVL() call which supplies a value if 'obr' is not found or occurs too far toward the end of a record so a value 2 away is not possible:
SQL> with t(id, text) as (
select 1, 'asd|dfg|obr|1|value1|end' from dual
union
select 2, 'a|brx|123|obr|2|value2|end' from dual
union
select 3, 'hfv|obr|3|value3|345|pre|end' from dual
union
select 4, 'hfv|obr||value4|345|pre|end' from dual
union
select 5, 'a|brx|123|obriem|2|value5|end' from dual
union
select 6, 'a|brx|123|obriem|2|value6|obr' from dual
)
select
id,
nvl(regexp_substr(text, '\|obr\|[^|]*\|([^|]*)(\||$)', 1, 1, null, 1), 'value not found') value
from t;
ID VALUE
---------- -----------------------------
1 value1
2 value2
3 value3
4 value4
5 value not found
6 value not found
6 rows selected.
SQL>
The regex basically can be read as "look for a pattern of a pipe, followed by 'obr', followed by a pipe, followed by zero or more characters that are not a pipe, followed by a pipe, followed by zero or more characters that are not a pipe (remembered in a captured group), followed by a pipe or the end of the line". The regexp_substr() call then returns the 1st captured group which is the set of characters between the pipes 2 fields from the 'obr'.
I need help with this problem:
I have a column named phone_number and I wanted to query this column to get the the string right of the last occurrence of '.' for all kinds of numbers in one single sql query.
example #:
515.123.1277
011.44.1345.629268
I need to get 1277 and 629268 respectively.
I have this so far:
select phone_number,
case when length(phone_number) <= 12
then
substr(phone_number,-4)
else
substr (phone_number, -6) end
from employees;
This works for this example, but I want it for all kinds of # formats.
Would be great to get some input.
Thanks
It should be as easy as this regex:
SELECT phone_number, REGEXP_SUBSTR(phone_number, '[^.]*$')
FROM employees;
With the end anchor $ it should get everything that is not a . character after the final .. If the last character is . then it will return NULL.
Search for a pattern including the period, [.] with digits, \d, followed by the end of the string, $.
Associate the digits with a character group by placing the pattern, \d, in parenthesis (see below). This is referenced with the subexpr parameter, 1 (last parameter).
Here is the solution:
SCOTT#dev> list
1 WITH t AS
2 ( SELECT '414.352.3100' p_number FROM dual
3 UNION ALL
4 SELECT '515.123.1277' FROM dual
5 UNION ALL
6 SELECT '011.44.1345.629268' FROM dual
7 )
8* SELECT regexp_substr(t.p_number, '[.](\d+)$', 1, 1, NULL, 1) end_num FROM t
SCOTT#dev> /
END_NUM
========================================================================
3100
1277
629268
You can do something like this in oracle:
select regexp_substr(num,'[^\.]+',1,regexp_count(num,'\.')+1) last_number from
(select '515.123.1277' num from dual union all
select '011.44.1345.629268' from dual );
Previous to 11gR2 you can use regexp_replace instead regexp_count:
select regexp_substr(num,'[^\.]+',1,length(regexp_replace (num , '[^\.]+'))+1) last_number from
(select '515.123.1277' num from dual union all
select '011.44.1345.629268' from dual );