Oracle - Update string to replace only the last character - replace

I have the following string in an Oracle 9i database:
A,B,C,
I need to replace all instances of ',' when it is the last item in the string. I have come up with the following statement but it deletes everything in the field not just the comma. Any suggestions?
UPDATE table SET column = REPLACE(SUBSTR(column, -1, 1), ',', '');

rtrim(column, ',') is both efficient and much shorter

You forgot to add the condition: WHERE SUBSTR(column, -1, 1) = ','
Quassnoi caught another issue - REPLACE returns null - you can't use it inside the "set"
Full sql:
UPDATE table SET column = SUBSTR(column, 0, length(column)-1)
WHERE SUBSTR(column, -1, 1) = ',';
This will make sure you're doing the substitute only in rows that has values that ends with ","

UPDATE mytable
SET column = SUBSTR(column, 1, LENGTH(column) - 1)
WHERE SUBSTR(column, -1, 1) = ','

If you want to refer 'column' only 1 time in your query, just do as following:
UPDATE table SET column = REVERSE(SUBSTR(REVERSE(column), 2));

Related

Oracle PLSQL regexp_substr separate comma separated string with double quotes

I've seen examples of how to separate comma-separated strings into rows like this:
select distinct id, trim(regexp_substr(value,'[^,]+', 1, level) ) value, level
from tbl1
connect by regexp_substr(value, '[^,]+', 1, level) is not null
order by id, level;
but, my question is, how do I do this on double quote and comma delimited strings?
Ex: the above works for strings like "1,2,3,4,5,6,7", but what about "1","2","3","4,5","6,7,8","9" so that the rows end up like:
1
2
3
4,5
6,7,8
9
edit: I'm on Oracle 11.2.0.4, 11gR2.
There is a hack. Replace the pattern "," with # and use it in regular expression.It works like a charm.
Input String : "1","2","3","4,5","6,7,8","9","Ant,B","Gradle","E,F","G"(Can be number/Character doesn't matter)
with temp as (
select replace(replace('"1","2","3","4,5","6,7,8","9","Ant,B","Gradle","E,F","G"','","','#'),'"') Colmn from dual
)
SELECT trim(regexp_substr(str, '[^#]+', 1, level)) str
FROM (SELECT Colmn str FROM temp) t
CONNECT BY instr(str, '#', 1, level - 1) > 0
Output :
STR
1
2
3
4,5
6,7,8
9
Ant,B
Gradle
E,F
G
10 rows
Refer DBFiddle link for demo.
https://dbfiddle.uk/?rdbms=oracle_18&fiddle=d09c326f614d10f5d3c407fdfd3a44c5
here is the another solution.it must be working all string with have number.
i used number values index as base.
with temp as (
select '"1","2","3",x"4,5","6,73,8","9"' Colmn from dual
)
SELECT regexp_substr(Colmn, '\d{1}', REGEXP_INSTR(Colmn, '\d{1}', REGEXP_INSTR(Colmn, '\d{1}') ,level),1 ) from temp
CONNECT BY REGEXP_COUNT (Colmn,'\d{1}')+1> level

Eliminate first and second number after a symbol in stata

I have a variable cn that is a string variable that includes court case names.
For example, 1:2013-cv-10153 and 0:1979-cv-06704.
I would like to remove the first two numbers after the : so that it appears as:
1:13-cv-10153 and 0:79-cv-06704
I've tried substr, but I keep getting an error.
Could someone please advise the best coding strategy in Stata?
This works for me:
display substr("1:2013-cv-10153", 1, 2) + substr("1:2013-cv-10153", 5, .)
Similarly:
. generate string = "1:2013-cv-10153"
. display string
1:2013-cv-10153
. generate new_string = substr(string, 1, 2) + substr(string, 5, .)
. display new_string
1:13-cv-10153
You could also focus on the colon without worrying about its exact character position in the string.
You can do this using the strpos() function in addition to substr():
generate string = "1:2013-cv-10153"
generate new_string = substr(string, 1, strpos(string, ":")) + ///
substr(string, strpos(string, ":") + 3, .)

PL SQL regular expression substring

I have a long string.
message := 'I loooove my pet animal';
This string in 23 chars long. If message is greater that 15 chars, I need to find the length of message where I can break the string into 2 strings. For example, in this case,
message1 := 'I loove my'
message2 := 'pet animal'
Essentially it should find the position of a whole word at the previous to 15 chars and the break the original string into 2 at that point.
Please give me ideas how I can do this.
Thank you.
Here is a general solution - with possibly more than one input string, and with inputs of any length. The only assumption is that no single word may be more than 15 characters, and that everything between two spaces is considered a word. If a "word" can be more than 15 characters, the solution can be adapted, but the requirement itself would need to state what the desired result is in such a case.
I make up two input strings in a CTE (at the top) - that is not part of the solution, it is just for testing and illustration. I also wrote this in plain SQL - there is no need for PL/SQL code for this type of problem. Set processing (instead of one row at a time) should result in much better execution.
The approach is to identify the location of all spaces (I append and prepend a space to each string, too, so I won't have to deal with exceptions for the first and last substring); then I decide, in a recursive subquery, where each "maximal" substring should begin and where it should end; and then outputting the substrings is trivial. I used a recursive query, that should work in Oracle 11.1 (or 11.2 with the syntax I used, with column names in CTE declarations - it can be changed easily to work in 11.1). In Oracle 12, it would be easier to rewrite the same idea using MATCH_RECOGINZE.
with
inputs ( id, str ) as (
select 101, 'I loooove my pet animal' from dual union all
select 102, '1992 was a great year for - make something up here as needed' from dual
),
positions ( id, pos ) as (
select id, instr(' ' || str || ' ', ' ', 1, level)
from inputs
connect by level <= length(str) - length(replace(str, ' ')) + 2
and prior id = id
and prior sys_guid() is not null
),
r ( id, str, line_number, pos_from, pos_to ) as (
select id, ' ' || str || ' ', 0, null, 1
from inputs
union all
select r.id, r.str, r.line_number + 1, r.pos_to,
( select max(pos)
from positions
where id = r.id and pos - r.pos_to between 1 and 16
)
from r
where pos_to is not null
)
select id, line_number, substr(str, pos_from + 1, pos_to - pos_from - 1) as line_text
from r
where line_number > 0 and pos_to is not null
order by id, line_number
;
Output:
ID LINE_NUMBER LINE_TEXT
---- ----------- ---------------
101 1 I loooove my
101 2 pet animal
102 1 1992 was a
102 2 great year for
102 3 - make
102 4 something up
102 5 here as needed
7 rows selected.
First you reverse string.
SELECT REVERSE(strField) FROM DUAL;
Then you calculate length i = length(strField).
Then find the first space after the middle
j := INSTR( REVERSE(strField), ' ', i / 2, i)`
Finally split by i - j (maybe +/- 1 need to test it)
DEMO
WITH parameter (id, strField) as (
select 101, 'I loooove my pet animal' from dual union all
select 102, '1992 was a great year for - make something up here as needed' from dual union all
select 103, 'You are Supercalifragilisticexpialidocious' from dual
), prepare (id, rev, len, middle) as (
SELECT id, reverse(strField), length(strField), length(strField) / 2
FROM parameter
)
SELECT p.*, l.*,
SUBSTR(strField, 1, len - INSTR(rev, ' ', middle)) as first,
SUBSTR(strField, len - INSTR(rev, ' ', middle) + 2, len) as second
FROM parameter p
JOIN prepare l
ON p.id = l.id
OUTPUT

Regex: How to remove English words from sentences using Regex?

I've number of rows in SQLite, each row has one column that contains data like this:
prosperکامیاب شدن ، موفق شدن ، رونق یافتن
As you can see, the sentence starts with English words, Now I want to remove English words at first of each sentence. Is there any way to do that via T-SQL query(using Regex)?
you may try this :) I have made it as a function to call upon
create function dbo.RemoveEngChars (#Unicode_string nvarchar(max))
returns nvarchar(max) as
begin
declare #i int = 1; -- must start from 1, as SubString is 1-based
declare #OriginalString nvarchar(100) = #Unicode_string collate SQL_Latin1_General_Cp1256_CS_AS
declare #ModifiedString nvarchar(100) = N'';
while #i <= Len(#OriginalString)
begin
if SubString(#OriginalString, #i, 1) not like '[a-Z]'
begin
set #ModifiedString = #ModifiedString + SubString(#OriginalString, #i, 1);
end
set #i = #i + 1;
end
return #ModifiedString
end
--To call the function , you can run the following script and pass the Unicode in N' prefix
select dbo.RemoveEngChars(N'prosperکامیاب شدن ، موفق شدن ، رونق یافتن')

Query to replace a string with matching pattern

I have a string like below.
'comp' as "COMPUTER",'ms' as "MOUSE" ,'keybr' as "KEYBOARD",'MONT' as "MONITOR",
Is it possible to write a query so that i will get the result as
'comp' ,'ms' ,'keybr' ,'MONT' ,
I can replace the string "as" with empty string using REPLACE query.
But how do I remove the string inside double quote?
Can anyone help me doing this?
Thanks in advance.
select replace(
regexp_replace('''comp'' as "COMPUTER"'
, '(".*")'
,null)
,' as '
,null)
from dual
The regex '(".*")' selects text whatever in double quotes.
EDIT:
the regex replaced the entire length of the matching pattern. So, we might need to tokenise the string first using comma as delimiter and apply the regex. Later join it.(LISTAGG)
WITH str_tab(str1, rn) AS
(SELECT regexp_substr(str, '[^,]+', 1, LEVEL), -- delimts
LEVEL
FROM (SELECT '''comp'' as "computer",''comp'' as "computer"' str
FROM dual) tab
CONNECT BY LEVEL <= LENGTH(str) - LENGTH(REPLACE(str, ',')) + 1)
SELECT listagg(replace(
regexp_replace(str1
, '(".*")'
,null)
,' as '
,null), ',') WITHIN GROUP (ORDER BY rn) AS new_text
FROM str_tab;
EDIT2:
A cleaner approach from #EatAPeach
with x(y) as (
select q'<'comp' as "COMPUTER",'ms' as "MOUSE" ,'keybr' as "KEYBOARD",'MONT' as "MONITOR",>'
from dual
)
select y,
regexp_replace(y, 'as ".*?"' ,null)
from x;