I am trying to get a query where I get all characters from a string before the 'n'the occurence of a character.
Say I could have the following strings:
'123456,123456,123456'
'123456'
'123456,123456,123456,123456,123456,123456'
'123456,123456,123456,123456,123456,123456,123456'
Now I want my query to always return everything before the 5th occurence of the comma,
Result:
'123456,123456,123456'
'123456'
'123456,123456,123456,123456,123456'
'123456,123456,123456,123456,123456'
I've been trying with some substr or regexes, but I can't get my head around this.
INSTR function has exactly what you need to find the position of n-th substring - see the occurrence parameter.
To get the part of a string till this location use SUBSTRING.
To avoid the case when there is no Nth symbol, use NVL (or COALESCE).
For example (replace 5 with N and insert your columns):
SELECT NVL(
SUBSTR(YOUR_COLUMN, 1,
INSTR(YOUR_COLUMN,',',1,5) -1),
YOUR_COLUMN)
FROM YOUR_TABLE;
You can do that:
define string_value='123456,123456';
select CASE
WHEN (length('&string_value') - length(replace('&string_value',',',null))) >=5
THEN SUBSTR('&string_value',0,INSTR('&string_value',',',1,5)-1)
ELSE '&string_value'
END as output
from dual;
output:
123456,123456
define string_value='123456,123456,123456,123456,123456,123456';
select CASE
WHEN (length('&string_value') - length(replace('&string_value',',',null))) >=5
THEN SUBSTR('&string_value',0,INSTR('&string_value',',',1,5)-1)
ELSE '&string_value'
END as output
from dual;
output:
123456,123456,123456,123456,123456
This will work event if the number of character between the commas is not always the same.
Related
I am trying to do substring in Oracle using last index of a character '_'.
Not able to get it rite. Please see the query which does a substring of second occurrence of "-". But it does not work if the string only has one occurrence of "-".
SELECT NVL(SUBSTR('TEMP_ABC', 0, INSTR('TEMP_ABC', '_',1,2)-1), 'TEMP_ABC')
FROM DUAL
Result - TEMP_ABC
Expected Result - TEMP
SELECT NVL(SUBSTR('TEMP_ABC_XYZ', 0, INSTR('TEMP_ABC_XYZ', '_',1,2)-1), 'TEMP_ABC_XYZ')
FROM DUAL
Result - TEMP_ABC
Expected Result - TEMP_ABC
Any clue on what I am doing wrong here?
In instr function, if you use -1 at the last parameter, it means last occurrence of the char string.
instr(string, '_', -1) = last occurrence of _
Thus:
select substr('TEMP_ABC',1,instr('TEMP_ABC','_',-1)-1)
from dual;
Result: TEMP
select substr('TEMP_ABC_XYZ',1,instr('TEMP_ABC_XYZ','_',-1)-1)
from dual;
Result: TEMP_ABC
Can anyone help me, I have a problem regarding on how can I get the below result of data. refer to below sample data. So the logic for this is first I want delete the letters before the number and if i get that same thing goes on , I will delete the numbers before the letter so I can get my desired result.
Table:
SALV3000640PIX32BLU
SALV3334470A9CARBONGRY
TP3000620PIXL128BLK
Desired Output:
PIX32BLU
A9CARBONGRY
PIXL128BLK
You need to use a combination of the SUBSTRING and PATINDEX Functions
SELECT
SUBSTRING(SUBSTRING(fielda,PATINDEX('%[^a-z]%',fielda),99),PATINDEX('%[^0-9]%',SUBSTRING(fielda,PATINDEX('%[^a-z]%',fielda),99)),99) AS youroutput
FROM yourtable
Input
yourtable
fielda
SALV3000640PIX32BLU
SALV3334470A9CARBONGRY
TP3000620PIXL128BLK
Output
youroutput
PIX32BLU
A9CARBONGRY
PIXL128BLK
SQL Fiddle:http://sqlfiddle.com/#!6/5722b6/29/0
To do this you can use
PATINDEX('%[0-9]%',FieldName)
which will give you the position of the first number, then trim off any letters before this using SUBSTRING or other string functions. (You need to trim away the first letters before continuing with the next step because unlike CHARINDEX there is no starting point parameter in the PATINDEX function).
Then on the remaining string use
PATINDEX('%[a-z]%',FieldName)
to find the position of the first letter in the remaining string. Now trim off the numbers in front using SUBSTRING etc.
You may find this other solution helpful
SQL to find first non-numeric character in a string
Try this it may helps you
;With cte (Data)
AS
(
SELECT 'SALV3000640PIX32BLU' UNION ALL
SELECT 'SALV3334470A9CARBONGRY' UNION ALL
SELECT 'SALV3334470A9CARBONGRY' UNION ALL
SELECT 'SALV3334470B9CARBONGRY' UNION ALL
SELECT 'SALV3334470D9CARBONGRY' UNION ALL
SELECT 'TP3000620PIXL128BLK'
)
SELECT * , CASE WHEN CHARINDEX('PIX',Data)>0 THEN SUBSTRING(Data,CHARINDEX('PIX',Data),LEN(Data))
WHEN CHARINDEX('A9C',Data)>0 THEN SUBSTRING(Data,CHARINDEX('A9C',Data),LEN(Data))
ELSE NULL END AS DesiredResult FROM cte
Result
Data DesiredResult
-------------------------------------
SALV3000640PIX32BLU PIX32BLU
SALV3334470A9CARBONGRY A9CARBONGRY
SALV3334470A9CARBONGRY A9CARBONGRY
SALV3334470B9CARBONGRY NULL
SALV3334470D9CARBONGRY NULL
TP3000620PIXL128BLK PIXL128BLK
I have a sql that returns the example string below:
Input PKIND:BCMOX:10048301-
output BCMOX:10048301
I need to write code the first substring the string on - then split it on : and return the 2 & 3 item (BCMOX:10048301)
If the string format is consistent and you want to extract everything after the first : until the first occurrence of -, use a combination of substr and instr.
select substr(col, instr(col,':')+1, instr(col,'-')-instr(col,':')-1)
from yourtable
where instr(col,':') > 0 and instr(col,'-') > 0 --to get the rows which have these 2 characters
The REGEXP_SUBSTR version. Return everything between the first colon and the first hyphen.
select regexp_substr('PKIND:BCMOX:10048301-', ':(.*)-', 1, 1, NULL, 1) from dual;
I have a string something like this 'SERO02~~~NA_#ERO5'. I need to sub string it using delimiter ~~~. So can get SERO02 and NA_#ERO5 as result.
I create an regex experession like this:
select regexp_substr('SERO02~~~NA_#ERO5' ,'[^~~~]+',1,2) from dual;
It worked fine and returns : NA_#ERO5
But if I change the string to ERO02~NA_#ERO5 the result is still same.
But I expect the expression to return nothing since delimiter ~~~ is not found in that string. Can someone help me out to create correct expression?
[^~~~] matches a single character that is not one of the characters following the caret in the square brackets. Since all those characters are identical then [^~~~] is the same as [^~].
You can match it using:
SELECT REGEXP_SUBSTR(
'SERO02~~~NA_#ERO5',
'~~~(.*?)(~~~|$)',
1,
1,
NULL,
1
)
FROM DUAL;
Which will match ~~~ then store zero-or-more characters in a capture group (the round brackets () indicates a capture group) until it finds either ~~~ or the end-of-string. It will then return the first capture group.
You can do it without regular expressions, with a bit of logics:
with test(text) as ( select 'SERO02~~~NA_#ERO5' from dual)
select case
when instr(text, '~~~') != 0 then
substr(text, instr(text, '~~~') + 3)
else
null
end
from test
This will give the part of the string after '~~~', if it exists, null otherwise.
You can edit the ELSE part to get what you need when the input string does not contain '~~~'.
Even using regexp,to match the string '~~~', you need to write it exactly, without []; the [] is used to list a set of characters, so [aaaaa] is exactly the same than [a],while [abc] means 'a' OR 'b' OR 'c'.
With regexp, even if not necessary, one way could be the following:
substr(regexp_substr(text, '~~~.*'), 4)
In case you want all elements. Handles NULL elements too:
SQL> with tbl(str) as (
select 'SERO02~~~NA_#ERO5' from dual
)
select regexp_substr(str, '(.*?)(~~~|$)', 1, level, null, 1) element
from tbl
connect by level <= regexp_count(str, '~~~') + 1;
ELEMENT
-----------------
SERO02
NA_#ERO5
SQL>
I have the following problem.
There is a String:
There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235
I need to show only just the last date from this string: 2015.06.07.
I tried with regexp_substr with insrt but it doesn't work.
So this is just test, and if I can solve this after it with this solution I should use it for a CLOB query where there are multiple date, and I need only the last one. I know there is regexp_count, and it is help to solve this, but the database what I use is Oracle 10g so it wont work.
Can somebody help me?
The key to find the solution of this problem is the idea of reversing the words in the string presented in this answer.
Here is the possible solution:
WITH words AS
(
SELECT regexp_substr(str, '[^[:space:]]+', 1, LEVEL) word,
rownum rn
FROM (SELECT 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 2015.06.08 2015.06.17. 2015.07.01. 12345678999 12125235' str
FROM dual) tab
CONNECT BY LEVEL <= LENGTH(str) - LENGTH(REPLACE(str, ' ')) + 1
)
, words_reversed AS
(
SELECT *
FROM words
ORDER BY rn DESC
)
SELECT regexp_substr(word, '\d{4}\.\d{2}\.\d{2}', 1, 1)
FROM words_reversed
WHERE regexp_like(word, '\d{4}\.\d{2}\.\d{2}')
AND rownum = 1;
From the documentation on regexp_substr, I see one problem immediately:
The . (period) matches any character. You need to escape those with a backslash: \. in order to match only a period character.
For reference, I am linking this post which appears to be the approach you are taking with substr and instr.
Relevant documentation from Oracle:
INSTR(string , substring [, position [, occurrence]])
When position is negative, then INSTR counts and searches backward from the end of string. The default value of position is 1, which means that the function begins searching at the beginning of string.
The problem here is that your regular expression only returns a single value, as explained here, so you will be giving the instr function the appropriate match in the case of multiple dates.
Now, because of this limitation, I recommend using the approach that was proposed in this question, namely reverse the entire string (and your regular expression, i.e. \d{2}\.\d{2}\.\d{4}) and then the first match will be the 'last match'. Then, perform another string reversal to get the original date format.
Maybe this isn't the best solution, but it should work.
There are three different PL/SQL functions that will get you there.
The INSTR function will identify where the first "period" in the date string appears.
SUBSTR applied to the entire string using the value from (1) as the start point
TO_DATE for a specific date mask: YYYY.MM.DD will convert the result from (2) into a Oracle date time type.
To make this work in procedural code, the standard blocks apply:
DECLARE
v_position pls_integer;
... other variables
BEGIN
sql code and function calls;
END
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE finddate
(column1 varchar2(11), column2 varchar2(39))
;
INSERT ALL
INTO finddate (column1, column2)
VALUES ('row1', '1234567 242424 2015.06.07. 12125235')
INTO finddate (column1, column2)
VALUES ('string2', '1234567 242424 2015.06.07. 12125235')
SELECT * FROM dual
;
Query 1:
select instr(column2,'.',1) from finddate
where column1 = 'string2'
select substr(column2,(20-4),10) from finddate
select to_date('2015.06.07','YYYY.MM.DD') from finddate
Results:
| TO_DATE('2015.06.07','YYYY.MM.DD') |
|------------------------------------|
| June, 07 2015 00:00:00 |
| June, 07 2015 00:00:00 |
Here's a way using regexp_replace() that should work with 10g, assuming the format of the lines will be the same:
with tbl(col_string) as
(
select 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235'
from dual
)
select regexp_replace(col_string, '^.*(\d{4}\.\d{2}\.\d{2})\. \d*$', '\1')
from tbl;
The regex can be read as:
^ - Match the start of the line
. - followed by any character
* - followed by 0 or more of the previous character (which is any character)
( - Start a remembered group
\d{4}\.\d{2}\.\d{2} - 4 digits followed by a literal period followed by 2 digits, etc
) - End the first remembered group
\. - followed by a literal period
- followed by a space
\d* - followed by any number of digits
$ - followed by the end of the line
regexp_replace then replaces all that with the first remembered group (\1).
Basically describe the whole line as a regular expression, group around what you want to return. You will most likely need to tweak the regex for the end of the line if it could be other characters than digits but this should give you an idea.
For the sake of argument this works too ONLY IF there are 2 occurrences of the date pattern:
with tbl(col_string) as
(
select 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235' from dual
)
select regexp_substr(col_string, '\d{4}\.\d{2}\.\d{2}', 1, 2)
from tbl;
returns the second occurrence of the pattern. I expect the above regexp_replace more accurately describes the solution.