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.
Related
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, '(^[^_]+)_.*$', '\1')
from (select 'ABC' as str from dual union all
select 'ABC_BAS' from dual
) s
A simpler method is:
select regexp_substr(str, '^[^_]+')
from (select 'ABC' as str from dual union all
select 'ABC_BAS' from dual
) s
Here is a db<>fiddle.
I'd use
regexp_replace(text,'_.*')
or if performance was a concern,
substr(text, 1, instr(text||'_', '_') -1)
For example,
with demo(text) as
( select column_value
from table(sys.dbms_debug_vc2coll('ABC', 'ABC_DEF', 'ABC_DEF_GHI')) )
select text
, regexp_replace(text,'_.*')
, substr(text, 1, instr(text||'_', '_') -1)
from demo;
TEXT REGEXP_REPLACE(TEXT,'_.*') SUBSTR(TEXT,1,INSTR(TEXT||'_','_')-1)
------------ --------------------------- -------------------------------------
ABC ABC ABC
ABC_DEF ABC ABC
ABC_DEF_GHI ABC ABC
Ok i think i got it. Add nvl to the substring and insert the condition as below -
select nvl(substr('ABC',1,instr('F4001Z','_')-1),'ABC') from dual;
I have the below structure(' ' refers to empty spaces):
name description
---------------------
a yes
b ' '
c ' '
d null
I am searching for a query that give me the rows contain empty spaces, asked for the below result .
name description
---------------------
b ' '
c ' '
this query select * from tab1 where description =' '; will give me only c, in my query I have many values have long spaces.
You can user REGEXP_LIKE:
with src as (select 'a' as name,'yes' as description from dual
union all
select 'b',' ' from dual
union all
select 'c',' ' from dual
union all
select 'd',null from dual)
select * from src where regexp_like(description,'^[ ]+$'))
Edited: added regexp_like(description,'^[ ]+$') to take into account only descriptions with spaces. If there is a description in the format ' s ', ' s' or 's ' it will not be selected.
Use TRIM function to trim the spaces.
select * from tab1 where TRIM(description) IS NULL;
I have not tested it but it should work.
with this basic query:
with sample_data(name, description) as (
select 'a', 'yes' from dual union all
select 'b', ' ' from dual union all
select 'c', ' ' from dual union all
select 'd', null from dual
)
select *
from sample_data
you can pick and choose among the following where clauses to get your desired results:
where regexp_like(description,'[ ]')); -- at least one space in the string
where regexp_like(description,'[ ]{2,')); -- two or more consecutive spaces
where regexp_like(description,'^[ ]+$')); -- just spaces of any length
where regexp_like(description,'^[ ]{2,}')); -- just paces of length 2 or more
if you want any white space character (e.g. tabs, vertical tabs, non blanking spaces, etc.) you can replace the single space character class [ ] with this [[:space:]] named character class in any of the above where clauses.
Use LIKE operator
SELECT *
FROM tab1
WHERE description LIKE ' %'
In a SELECT statement would it be possible to evaluate a Substr using CASE? Or what would be the best way to return a sub string based on a condition?
I am trying to retrieve a name from an event description column of a table. The string in the event description column is formatted either like text text (Mike Smith) text text or text text (Joe Schmit (Manager)) text text. I would like to return the name only, but having some of the names followed by (Manager) is throwing off my SELECT statement.
This is my SELECT statement:
SELECT *
FROM (
SELECT Substr(Substr(eventdes,Instr(eventdes,'(')+1),1,
Instr(eventdes,')') - Instr(eventdes,'(')-1)
FROM mytable
WHERE admintype = 'admin'
AND entrytime BETWEEN sysdate - (5/(24*60)) AND sysdate
AND eventdes LIKE '%action taken by%'
ORDER BY id DESC
)
WHERE ROWNUM <=1
This returns things like Mike Smith if there is no (Manager), but returns things like Joe Schmit (Manager if there is.
Any help would be greatly appreciated.
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE MYTABLE ( id, admintype, entrytime, eventdes ) AS
SELECT 1, 'admin', SYSDATE, 'action taken by (John Doe (Manager)) more text' FROM DUAL;
Query 1:
SELECT *
FROM ( SELECT REGEXP_SUBSTR( eventdes, '\((.*?)(\s*\(.*?\))?\)', 1, 1, 'i', 1 )
FROM mytable
WHERE admintype = 'admin'
AND entrytime BETWEEN sysdate - (5/(24*60)) AND sysdate
AND eventdes LIKE '%action taken by%'
ORDER BY id DESC
)
WHERE ROWNUM <=1
Results:
| REGEXP_SUBSTR(EVENTDES,'\((.*?)(\S*\(.*?\))?\)',1,1,'I',1) |
|------------------------------------------------------------|
| John Doe |
Edit:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE MYTABLE ( id, admintype, entrytime, eventdes ) AS
SELECT 1, 'admin', SYSDATE, 'action taken by (Doe, John (Manager)) more text' FROM DUAL;
Query 1:
SELECT SUBSTR( Name, INSTR( Name, ',' ) + 1 ) || ' ' || SUBSTR( Name, 1, INSTR( Name, ',' ) - 1 ) AS Full_Name,
REGEXP_REPLACE( Name, '^(.*?),\s*(.*)$', '\2 \1' ) AS Full_Name2
FROM ( SELECT REGEXP_SUBSTR( eventdes, '\((.*?)(\s*\(.*?\))?\)', 1, 1, 'i', 1 ) AS Name
FROM mytable
WHERE admintype = 'admin'
-- AND entrytime BETWEEN sysdate - (5/(24*60)) AND sysdate
AND eventdes LIKE '%action taken by%'
ORDER BY id DESC
)
WHERE ROWNUM <=1
Results:
| FULL_NAME | FULL_NAME2 |
|-----------|------------|
| John Doe | John Doe |
You could use INSTR to search the last ')', but I would prefer a Regular Expression.
This extracts everything between the first '(' and the last ')' and the TRIMs remove the brackets (Oracle doesn't support look around in RegEx):
RTRIM(LTRIM(REGEXP_SUBSTR(eventdes, '\(.*\)'), '('), ')')
There is probably a regex that would do this without a replace but I am not smart enough to work it out so I combined a REGEXP_SUBSTRING that replaces your multiple instr methods and a REGEXP_REPLACE to replace the ' (MANAGER)' portion is it exists.
SELECT *
FROM (
SELECT REGEXP_REPLACE(
REGEXP_SUBSTR(eventdes, '\((.*)\)', 1, 1, 'i', 1), ' \(MANAGER\)', '')
FROM mytable
WHERE admintype = 'admin'
AND entrytime BETWEEN sysdate - (5/(24*60)) AND sysdate
AND eventdes LIKE '%action taken by%'
ORDER BY id DESC
)
WHERE ROWNUM <=1
Here's a slightly different take on the problem. This regexp starts at the position of the first open paren + 1 (the first letter of the name), then returns the first group from that point in the string on that consists of 1 or more characters that is not a close paren or an open paren. The TRIM gets rid of the trailing space if there is a manager.
SQL> with tbl(id, admintype, entrytime, eventdes ) AS
2 ( SELECT 1, 'admin', SYSDATE, 'action taken by (John Doe (Manager)) more text' FROM DUAL
3 union
4 SELECT 1, 'admin', SYSDATE, 'action taken by (Jane Doe) more text' FROM DUAL
5 )
6 select trim(regexp_substr(eventdes, '([^\)\(]+)', instr(eventdes, '(')+1, 1)) name
7 from tbl;
NAME
----------------------------------------------
Jane Doe
John Doe
SQL>
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
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;