Substring from underscore and onwards in Oracle - sql

I have a string with under score and some characters. I need to apply substring and get values to the left excluding underscore. So I applied below formula and its working correctly for those strings which have underscore (_). But for strings without (_) it is bringing NULL. Any suggestions how this can be handled in the substring itself.
Ex: ABC_BASL ---> Works correctly; ABC ---> gives null
My formula as below -
select SUBSTR('ABC_BAS',1,INSTR('ABC_BAS','_')-1) from dual;
ABC
select SUBSTR('ABC',1,INSTR('ABC','_')-1) from dual;
(NULL)

You could use a CASE expression to first check for an underscore:
WITH yourTable AS (
SELECT 'ABC_BAS' AS col FROM dual UNION ALL
SELECT 'ABC' FROM dual
)
SELECT
CASE WHEN col LIKE '%\_%' ESCAPE '\'
THEN SUBSTR(col, 1, INSTR(col, '_') - 1)
ELSE col END AS col_out
FROM yourTable;

Use regular expression matching:
SELECT REGEXP_SUBSTR('ABC_BAS', '(.*)([_]|$)?', 1, 1, NULL, 1) FROM DUAL;
returns 'ABC', and
SELECT REGEXP_SUBSTR('ABC', '(.*)([_]|$)?', 1, 1, NULL, 1) FROM DUAL;
also returns 'ABC'.
db<>fiddle here
EDIT
The above gives correct results, but I missed the easiest possible regular expression to do the job:
SELECT REGEXP_SUBSTR('ABC_BAS', '[^_]*') FROM DUAL;
returns 'ABC', as does
SELECT REGEXP_SUBSTR('ABC', '[^_]*') FROM DUAL;
db<>fiddle here

Yet another approach is to use the DECODE in the length parameter of the substr as follows:
substr(str,
1,
decode(instr(str,'_'), 0, lenght(str), instr(str,'_') - 1)
)

You seem to want everything up to the first '_'. If so, one method usesregexp_replace():
select regexp_replace(str, '(^[^_]+)_.*$', '\1')
from (select 'ABC' as str from dual union all
select 'ABC_BAS' from dual
) s
A simpler method is:
select regexp_substr(str, '^[^_]+')
from (select 'ABC' as str from dual union all
select 'ABC_BAS' from dual
) s
Here is a db<>fiddle.

I'd use
regexp_replace(text,'_.*')
or if performance was a concern,
substr(text, 1, instr(text||'_', '_') -1)
For example,
with demo(text) as
( select column_value
from table(sys.dbms_debug_vc2coll('ABC', 'ABC_DEF', 'ABC_DEF_GHI')) )
select text
, regexp_replace(text,'_.*')
, substr(text, 1, instr(text||'_', '_') -1)
from demo;
TEXT REGEXP_REPLACE(TEXT,'_.*') SUBSTR(TEXT,1,INSTR(TEXT||'_','_')-1)
------------ --------------------------- -------------------------------------
ABC ABC ABC
ABC_DEF ABC ABC
ABC_DEF_GHI ABC ABC

Ok i think i got it. Add nvl to the substring and insert the condition as below -
select nvl(substr('ABC',1,instr('F4001Z','_')-1),'ABC') from dual;

Related

Regex values after special character with empty values

I am struggle with regex to split spring into columns in Oracle database.
select (REGEXP_SUBSTR(replace('1:::9999', ' ',''), '[^: ]+', 1, 4)) from dual;
I need to obtain 4th value from that string as a column value, sometimes values at position 2,3 are empty and my query doesn't work. I am trying to figure out what regex will work
You can use
select (REGEXP_SUBSTR(replace('1:::9999', ' ',''), '([^: ]*)(:|$)', 1, 4, 'i', 1)) from dual;
Here, the ([^: ]*)(:|$) matches
([^: ]*) - Group 1: any zero or more chars other than : and space
(:|$) - Group 2, either : or end of string.
You do not need a (slower) regex for this task, use simple substr/instr functions:
with input_(val) as (
select '1:::9999' from dual
union all
select '1:2::' from dual
union all
select '1:2::3:5' from dual
)
, replaced as (
select input_.*, replace(val, ' ', '') as val_replaced
from input_
)
select
val,
substr(
val_replaced,
/*Locate the first occurrence of a colon and get a substring ...*/
instr(val_replaced, ':', 1, 3) + 1,
/*.. until the end, if the next colon is absent, or until the next colon*/
nvl(nullif(instr(val_replaced, ':', 1, 4), 0), length(val_replaced) + 1) - instr(val_replaced, ':', 1, 3) - 1
) as col
from replaced
VAL
COL
1:::9999
9999
1:2::
null
1:2::3:5
3
fiddle with performance difference.

How do I get substring after a character when the occurance of the character keeps changing

Example
123\.456.578.910.ABC
123\.456.578.910
Expected result
123\.456.578
123\.456.578
For the both the inputs I should get only the first 3
I tried the regexp and substring and instr but I’m not getting the results
We can use REGEXP_SUBSTR here with a capture group:
SELECT REGEXP_SUBSTR(col, '^(\d+(\.\d+)*)', 1, 1, NULL, 1)
FROM yourTable;
Demo
Traditional, substr + instr combination is another option:
Sample data:
SQL> with test (col) as
2 (select '123\.456.578.910.ABC' from dual union all
3 select '123\.456.578.910' from dual
4 )
Query begins here:
5 select col,
6 substr(col, 1, instr(col, '.', 1, 3) - 1) result
7 from test;
COL RESULT
-------------------- --------------------
123\.456.578.910.ABC 123\.456.578
123\.456.578.910 123\.456.578
SQL>
If you value will always have at least 3 . characters then you can use:
SELECT value,
SUBSTR(value, 1, INSTR(value, '.', 1, 3) - 1) AS expected
FROM table_name;
If it may have fewer and you want the entire string in those cases then:
SELECT value,
CASE INSTR(value, '.', 1, 3)
WHEN 0
THEN value
ELSE SUBSTR(value, 1, INSTR(value, '.', 1, 3) - 1)
END AS expected
FROM table_name;
Which, for your sample data:
CREATE TABLE table_name (value) AS
SELECT '123\.456.578.910.ABC' FROM DUAL UNION ALL
SELECT '123\.456.578.910' FROM DUAL;
Both outputs:
VALUE
EXPECTED
123.456.578.910.ABC
123.456.578
123.456.578.910
123.456.578
db<>fiddle here

SQL to find upper case words from a column

I have a description column in my table and its values are:
This is a EXAMPLE
This is a TEST
This is a VALUE
I want to display only EXAMPLE, TEST, and VALUE from the description column.
How do I achieve this?
This could be a way:
-- a test case
with test(id, str) as (
select 1, 'This is a EXAMPLE' from dual union all
select 2, 'This is a TEST' from dual union all
select 3, 'This is a VALUE' from dual union all
select 4, 'This IS aN EXAMPLE' from dual
)
-- concatenate the resulting words
select id, listagg(str, ' ') within group (order by pos)
from (
-- tokenize the strings by using the space as a word separator
SELECT id,
trim(regexp_substr(str, '[^ ]+', 1, level)) str,
level as pos
FROM test t
CONNECT BY instr(str, ' ', 1, level - 1) > 0
and prior id = id
and prior sys_guid() is not null
)
-- only get the uppercase words
where regexp_like(str, '^[A-Z]+$')
group by id
The idea is to tokenize every string, then cut off the words that are not made by upper case characters and then concatenate the remaining words.
The result:
1 EXAMPLE
2 TEST
3 VALUE
4 IS EXAMPLE
If you need to handle some other character as an upper case letter, you may edit the where condition to filter for the matching words; for example, with '_':
with test(id, str) as (
select 1, 'This is a EXAMPLE' from dual union all
select 2, 'This is a TEST' from dual union all
select 3, 'This is a VALUE' from dual union all
select 4, 'This IS aN EXAMPLE' from dual union all
select 5, 'This IS AN_EXAMPLE' from dual
)
select id, listagg(str, ' ') within group (order by pos)
from (
SELECT id,
trim(regexp_substr(str, '[^ ]+', 1, level)) str,
level as pos
FROM test t
CONNECT BY instr(str, ' ', 1, level - 1) > 0
and prior id = id
and prior sys_guid() is not null
)
where regexp_like(str, '^[A-Z_]+$')
group by id
gives:
1 EXAMPLE
2 TEST
3 VALUE
4 IS EXAMPLE
5 IS AN_EXAMPLE
Here's another solution. It was inspired by Aleksej's answer.
The idea? Get all the words. Then aggregate only fully uppercased to a list.
Sample data:
create table descriptions (ID int, Description varchar2(100));
insert into descriptions (ID, Description)
select 1 as ID, 'foo Foo FOO bar Bar BAR' as Description from dual
union all select 2, 'This is an EXAMPLE TEST Description VALUE' from dual
;
Query:
select id, Description, listagg(word, ',') within group (order by pos) as UpperCaseWords
from (
select
id, Description,
trim(regexp_substr(Description, '\w+', 1, level)) as word,
level as pos
from descriptions t
connect by regexp_instr(Description, '\s+', 1, level - 1) > 0
and prior id = id
and prior sys_guid() is not null
)
where word = upper(word)
group by id, Description
Result:
ID | DESCRIPTION | UPPERCASEWORDS
-- | ----------------------------------------- | ------------------
1 | foo Foo FOO bar Bar BAR | FOO,BAR
2 | This is an EXAMPLE TEST Description VALUE | EXAMPLE,TEST,VALUE
It is possible to achieve this thanks to the REGEXP_REPLACE function:
SELECT REGEXP_REPLACE(my_column, '(^[A-Z]| |[a-z][A-Z]*|[A-Z]*[a-z])', '') AS Result FROM my_table
It uses a regex which replaces first upper case char of the line and converts every lower case char and space with blanks.
Try this:
SELECT SUBSTR(column_name, INSTR(column_name,' ',-1) + 1)
FROM your_table;
This should do the trick:
SELECT SUBSTR(REGEXP_REPLACE(' ' || REGEXP_REPLACE(description, '(^[A-Z]|[a-z]|[A-Z][a-z]+|[,])', ''), ' +', ' '), 2, 9999) AS only_upper
FROM (
select 'Hey IF you do not know IT, This IS a test of UPPERCASE and IT, with good WILL and faith, Should BE fine to be SHOWN' description
from dual
)
I have added condition to strip commas, you can add inside that brakets other special characters to remove.
ONLY_UPPER
-----------------------------------
IF IT IS UPPERCASE IT WILL BE SHOWN
This is a function based on some of the regular expression answers.
create or replace function capwords(orig_string varchar2)
return varchar2
as
out_string varchar2(80);
begin
out_string := REGEXP_REPLACE(orig_string, '([a-z][A-Z_]*|[A-Z_]*[a-z])', '');
out_string := REGEXP_REPLACE(trim(out_string), '( *)', ' ');
return out_string;
end;
/
Removes strings of upper case letters and underscores that have lower case letters
on either end. Replaces multiple adjacent spaces with one space.
Trims extra spaces off of the ends. Assumes max size of 80 characters.
Slightly edited output:
>select id,str,capwords(str) from test;
ID STR CAPWORDS(STR)
---------- ------------------------------ ------------------
1 This is a EXAMPLE EXAMPLE
2 This is a TEST TEST
3 This is a VALUE VALUE
4 This IS aN EXAMPLE IS EXAMPLE
5 This is WITH_UNDERSCORE WITH_UNDERSCORE
6 ThiS IS aN EXAMPLE IS EXAMPLE
7 thiS IS aN EXAMPLE IS EXAMPLE
8 This IS wiTH_UNDERSCORE IS
If you only need to "display" the result without changing the values in the column then you can use CASE WHEN (in the example Description is the column name):
Select CASE WHEN Description like '%EXAMPLE%' then 'EXAMPLE' WHEN Description like '%TEST%' then 'TEST' WHEN Description like '%VALUE%' then 'VALUE' END From [yourTable]
The conditions are not case sensitive even if you write it all in uppercase.
You can add Else '<Value if all conditions are wrong>' before the END in case there are descriptions that don't contain any of the values. The example will return NULL for those cases, and writing ELSE Description will return the original value of that row.
It also works if you need to update. It is simple and practical, easy way out, haha.

Substring from string oracle

I have strings : 'A-20-1-1', 'A-10-10', 'A-10-11-1'
And result from substringing:
'A-20-1-1', 'A-10-10', 'A-10-11-1'
1 10 11
Code won't works fine:
Select Substr(string, instr(string,'-',1,2)+1, instr(string, '-',1,2)-1)
From dual;
At the beginning I find second '-', than next one if exists. If not I get string length.
create table a(b varchar2(20));
insert into a values('A-20-1-1');
insert into a values('A-10-10');
insert into a values('A-10-11-1');
Select
b,
substr(b,instr(b,'-',1,2)+1,decode(instr(b,'-',1,3),0,length(b)-instr(b,'-',1,2),instr(b,'-',1,3)-instr(b,'-',1,2)-1)) z
from a;
gives us what you need:
A-20-1-1 1
A-10-10 10
A-10-11-1 11
At first I find the second - sign position, and then get a substring of the value from this position to the rest of the value. Then I exclude the part of the string from previous step if the string has the - sign. Like this:
with t(d) as (
select 'A-20-1-1' from dual union all
select 'A-10-10-4' from dual union all
select 'A-10-11-1' from dual
)
select REPLACE(SUBSTR(d, INSTR(d, '-', 1, 2) + 1), SUBSTR(d, INSTR(d, '-', 1, 3))) from t
RES
---
1
10
11

Reg Expression in oracle?

I have a string like this '102/103/104/106'
Now if i pass 102 as input then output should be the next field that is 103. if 103 then output should be 104 and if 106 then output should be null(as for last field I don't have any further expression). I can do this using procedure by splitting the string into arrays and comparing. But can I do this through sql statement something like this
select '102/103/104/106' from dual where [expression 102 or 103].
Thanks!!
You can do it in pure SQL with something like this:
--convert your string into rows
with vals as (
select
substr('102/103/104/106',
instr('102/103/104/106', '/', 1, level)-3,
3
) col,
level lvl
from dual
connect by level <= length('102/103/104/106')-length(replace('102/103/104/106', '/'))+1
)
select *
from (
select col,
lead(col) over (order by lvl) next_val -- find the next value in the list
from vals
)
where col = :val;
Basically, convert your string into rows by parsing it. Then use the analytic lead to find the "next" value.
-- p_whole_string = '102/103/104/106'
-- p_prev = '102'
select
regexp_substr(p_whole_string, '(^|/)' || p_prev || '/([^/]+)', 1, 1, null, 2)
as next
from dual;
Added NVL to return last value if 106 is entered:
SELECT NVL(REGEXP_SUBSTR('102/103/104/106', '(^|/)' || '106' || '/([^/]+)', 1, 1, null, 2), REGEXP_SUBSTR('102/103/104/106', '[^/]+$')) as nxt
FROM dual
/
works for Oracle form 10 up.
SELECT
REGEXP_SUBSTR(
REGEXP_SUBSTR('102/103/104/106', '(^|/)102/[^/]+'), -- returns 102/103
'[^/]+',1,2) val -- takes second part
FROM DUAL;
with parameters looks like this:
-- p_string_to_search = '102/103/104/106'
-- p_string_to_match = '102'
SELECT
REGEXP_SUBSTR(
REGEXP_SUBSTR(p_string_to_search, '(^|/)' || p_string_to_match ||'/[^/]+'), -- returns 102/103
'[^/]+',1,2) val -- takes second part
FROM DUAL;