I am trying to use LRTIM to obtain part of a string.
This is the code that I am using.
SELECT
C.MANUFACTURER,
C.MODEL_GROUP,
LTRIM(C.VARIANT, C.MANUFACTURER || ' ' || C.MODEL_GROUP), "VAR DESC"
C.VARIANT
FROM STD_BI.RL2_CONTRACTS_VW C
This is what I have field wise
VARIANT
AUDI A3 DIESEL SPORTBACK 2.0 TDI SE Technik 5dr Manual 150
MANUFACTURER
AUDI
MODEL_GROUP
A3
This is the result that I am getting
MANUFACTURER MODEL_GROUP VAR DESC
AUDI A3 ESEL SPORTBACK 2.0 TDI SE Technik 5dr Manual 150
It is cutting off the first 2 characters of DIESEL, the problem is that this is not happening in all cases. Sometimes it is cutting off 1 character and sometimes none!
I have considered other methods of achieving this, i.e. looking for the spaces, but have other manufacturers and model groups that have multiple spaces so was trying to subtract the MANUFACTURER and MODEL_ROUP from the VARIANT field (string)
Any ideas as to why this might be happening or other suggestions for achieving this would be greatly appreciated.
Many thanks in advance,
Keith
Apologies seems like the editor is reformating the results that I am getting
Oracle query result
You can use REGEXP_REPLACE with '^' - Indicating start of string.
SELECT
C.MANUFACTURER,
C.MODEL_GROUP,
REGEXP_REPLACE(C.VARIANT, '^'||C.MANUFACTURER || ' +' || C.MODEL_GROUP) "VAR DESC",
C.VARIANT
FROM STD_BI.RL2_CONTRACTS_VW C
The second parameter to LTRIM is a set of single characters, not a sequence of characters. In your example the database removes characters from the left until it encounters the first character not in ('A','U','D','I','A','3'), which is the E of DIESEL.
In order to remove a prefix p from a string s use something like
case when SUBSTR(s,1,LENGTH(p))=p then SUBSTR(s,LENGTH(p)+1) else s end
which in your case expands to
case when SUBSTR(C.VARIANT, 1, LENGTH(C.MANUFACTURER || ' ' || C.MODEL_GROUP))
= C.MANUFACTURER || ' ' || C.MODEL_GROUP
then SUBSTR(C.VARIANT, 1+ LENGTH(C.MANUFACTURER || ' ' || C.MODEL_GROUP))
else s
end
Related
I am working on a project where I need to check if the employee enter *done* in a text field, though employee enters '* done *' or '*done *' or '* done*' in similar fashion. As you see they are putting trailing and leading blank or both at a time.I have to check the column for all three/four possible entry in like statement, I tried trim,rtrim nothing seems like working.
case when
col like ('*done*')
or col like ('* done*')
or col like ('*done *')
or col like ('* done *')
end as work_status
doesn't seems a smart way to do it. What is the best way to to check this. Any help will be appreciated. Thank you.
Remove spaces:
case when replace(col, ' ') = '*done*' then 'done'
else 'not'
end as work_status
You can look for the done substring with anything preceding and following using the LIKE operator and % wildcards:
CASE WHEN col LIKE '%done%' THEN 'done' END AS work_status
Or you can trim the leading and trailing space characters:
CASE WHEN TRIM(col) = 'done' THEN 'done' END AS work_status
Or you can replace all the leading/trailing white spaces (in case the users have entered new lines, tabs, etc. rather than space characters) using a regular expression:
CASE
WHEN REGEXP_REPLACE(col, '^[[:space:]]+|[[:space:]]+$') = 'done'
THEN 'done'
END AS work_status
fiddle
I have a pl sql string as follows :
String := 'ctx_ddl.add_stopword(''"SHARK_IDX19_SPL"'',''can'');
create index "SCOTT"."SHARK_IDX2"
on "SCOTT"."SHARK2"
("DOC")
indextype is ctxsys.context
parameters(''
datastore "SHARK_IDX2_DST"
filter "SHARK_IDX2_FIL"
section group "SHARK_IDX2_SGP"
lexer "SHARK_IDX2_LEX"
wordlist "SHARK_IDX2_WDL"
stoplist "SHARK_IDX2_SPL"
storage "SHARK_IDX2_STO"
sync (every "SYSDATE+(1/1)" memory 67108864)
'')
/
';
I have to get search the final occurrence of '/' and add ';' to it. Also I need to escape the quotes preset in parameters ('') to have extra quotes. I need output like
String := 'ctx_ddl.add_stopword(''"SHARK_IDX19_SPL"'',''can'');
create index "SCOTT"."SHARK_IDX2"
on "SCOTT"."SHARK2"
("DOC")
indextype is ctxsys.context
parameters(''''
datastore "SHARK_IDX2_DST"
filter "SHARK_IDX2_FIL"
section group "SHARK_IDX2_SGP"
lexer "SHARK_IDX2_LEX"
wordlist "SHARK_IDX2_WDL"
stoplist "SHARK_IDX2_SPL"
storage "SHARK_IDX2_STO"
sync (every "SYSDATE+(1/1)" memory 67108864)
'''')
/;
';
Any help.
There's an age-old saying: "Some people, when confronted with a problem, think "I know, I'll use regular expressions." Now they have two problems"
Unless you are confronted by a problem that truly requires regular expressions, I'd recommend working with basic string manipulation functions.
Semi-colon:
Use INSTR to find last occurence of '/', call this P1.
Result = Substr from position 1 through P1||';'||substr from P1+1 through to end-of-string
Parameters substitution:
Use INSTR to find where parameter list starts (i.e. find "parameters(" in your string) and ends (presumably the last closing parenthesis ")" in your string). Call these P2 and P3.
Result = substr from 1 through P2 || REPLACE(substr from P2+1 through P3-1,'''','''''''') || substr from P3 to end-of-string
In Oracle I want to check whether the string has "=' sign at the end. could you please let me know how to check it. If it has '=' sign at the end of string, I need to trailing that '=' sign.
for eg,
varStr VARCHAR2(20);
varStr = 'abcdef='; --needs to trailing '=' sign
I don't think you need "pattern matching" here. Just check if the last character is the =
where substr(varstr, -1, 1) = '='
substr when called with a negative position will work from the end of the string, so substr(varstr,-1,1) extracts the last character of the given string.
Use the REGEX_EXP function. I'm putting a sql command since you didn't specify on your question.:
select *
from someTable
where regexp_like( someField, '=$' );
The pattern $ means that the precedent character should be at the end of the string.
see it here on sql fiddle: http://sqlfiddle.com/#!4/d8afd/3
It seems that substr is the way to go, at lease with my sample data of about 400K address lines this returns 1043 entries that end in 'r' in an average of 0.2 seconds.
select count(*) from addrline where substr(text, -1, 1) = 'r';
On the other hand, the following returns the same results but takes 1.1 seconds.
select count(*) from addrline where regexp_like(text, 'r$' );
I am trying to remove certain characters from a VARCHAR2 using translate. Characters 160 (some kind of space) and 243 (paragraph control character?), however, appear to be "phantom" characters that are undetectable by both INSTR and TRANSLATE. LENGTH works, but only if it's the only character in a string. LENGTH(CHR(160)) returns 1, but LENGTH(CHR(160) || CHR(110)) also returns 1 when you'd think it would return 2. I've found that REPLACE works in stripping these phantom characters from a string, but I like translate better because it's easier to read and maintain whereas a long nesting or REPLACE functions is just cumbersome.
Is there some other way to strip these characters from a VARCHAR2 without using replace?
EDIT: It appears that character 243 elsewhere registers as ≤. However, Oracle has no problem displaying this character when I selected it explicitly. When I select CHR(243), it just displays the block replacement character. Plus, this source points 243 to the paragraph character which makes more sense since that's a control code.
What about using a regular expression? This removes your trouble characters by replacing char 160 or 243 with nothing:
SQL> select regexp_replace('abc' || chr(160) || chr(243) || 'def', '(' || chr(160) || '|' || chr(243) || ')', '') from dual;
REGEXP
------
abcdef
SQL>
hello i have i want to do something like this.
i have 4 rows with unique id 1,2,3,4 all four rows contains some string like
option1,option2,option3,option4
now i want to add "a ) " to the option1, "b ) " to the option2 and so on so is there a way i can do this with a query.currently i am adding these to a lots of rows manually
It's not clear exactly by what logic you want to select the letter to prepend to field somestring, but if for example it's a "Caesar's cypher" (1 gives 'a', 2 gives 'b' etc) based on the id field, as your question suggests, then this should work:
UPDATE sometable
SET somestring = (
substr('abcdefghijklmnopqrstuvwxyz', id, 1) ||
' ) ' || somestring)
WHERE id <= 26;
...for no more than 26 rows of course, since beyond that the logic must change and obviously we can't guess just how you want to extend it (use id modulo 26 + 1, use more characters than just lowercase letters, or ...?) since you give no clue on why you want to do this.