I Used substring for this query
to_number(substr(skidno,instr(skidno,'/')+1,length(skidno)))
this i'll pass for ths query Skid no 'MMM-1718/000325 / 1' i got Invalid Number
If your need is to extract the part after the 2nd occurrence of "/"; the following query will suffice your need ---
select to_number(trim('/' FROM
substr('MMM-1718/000325 / 1',
instr('MMM-1718/000325 / 1', '/', 1, 2))))
from dual;
You may want to use regexp_substr. If you want to only get the number at the end of the value:
to_number(regexp_substr(skidno, '\d+$'))
You cannot convert it to number because you will get "/" in the result.
Result: '000325 / 1'.
If you only require only the number between '/' then try below :
to_number(substr(skidno ,instr(skidno ,'/')+1,
instr(skidno ,'/',1,2) -instr(skidno ,'/',1,1)-1 ))
Since you need part of the string that begins after second occurrence of '/' character and goes to the end of the string, the following solution is probably the easiest using substr/instr:
with t as
(
select 'MMM-1718/000325 / 1' mystring
from dual
)
select to_number(substr(mystring, instr(mystring, '/', 1, 2) + 1))
from t;
Note: in this case you just have to define starting position in substr function; length of the string is not needed.
Related
When I try to run the following function on Sybase ASE 15.7, it just spins indefinitely. Each component of the function seems to act as expected independently. This is simply a mechanism to strip all non-numeric characters from a string. Any and all thoughts appreciated.
create function dbo.mdsudf_just_digits(#str varchar(64))
returns varchar(64)
as
begin
while patindex('%[^0-9]%', #str) > 0
set #str = stuff(#str, patindex('%[^0-9]%', #str), 1, '')
return #str
end
-- A typical invocation would be like so:
select dbo.mdsudf_just_digits('hello123there456')
```
In Sybase (ASE) the empty string ('') actually translates into a single space:
select '.'+''+'.' -- period + empty string + period
go
---
. . -- we get a space between the periods
So in the current stuff(#str, ..., 1, '') you're actually replacing the first non-numeric with a single space. This 'new' space then matches the non-number test on the next pass through the loop at which point the space is replaced with ... another space. This is leading to an infinite loop of constantly replacing the first non-number character with a space.
You can get around this by using NULL as the last arg to the stuff() call, eg:
set #str = stuff(#str, patindex('%[^0-9]%', #str), 1, NULL)
I want to fetch substring from string in column between last '/' and last '.' .
Here is sample date for IMAGE_PATH column name:
sph/images/30_Fairhall_Court.jpeg
sph/images/9_Pennethorne_House.jpeg
rbkc/images/TAVISTOCK_CRESCENT.jpeg
haringey/images/399932thumb.jpg
urbanchoice/images/18190862.jpg
wandle/images/f13c10d2-2692-457d-a208-8bb9e10b27dc.png
housingmoves/images/No14_Asterid Heights_DS37620.jpg
wandle/images/f13c10d2-2692-457d-a208-8bb9e10b27dc.png
So the required output is like
30_Fairhall_Court
9_Pennethorne_House
TAVISTOCK_CRESCENT
399932thumb
18190862
f13c10d2-2692-457d-a208-8bb9e10b27dc
No14_Asterid Heights_DS37620
f13c10d2-2692-457d-a208-8bb9e10b27dc
Please suggest how to fetch. I need to update another blank column in table with this value. The table has around 10 lacks records.
One of possible solutions is to use functions substr() and instr() with negative third parameter:
select image_path,
substr(image_path,
instr(image_path, '/', -1) + 1,
instr(image_path, '.', -1)-instr(image_path, '/', -1) - 1) img
from test
SQL Fiddle
Results:
IMAGE_PATH IMG
-------------------------------------------------------- -------------------------------------
sph/images/30_Fairhall_Court.jpeg 30_Fairhall_Court
sph/images/9_Pennethorne_House.jpeg 9_Pennethorne_House
rbkc/images/TAVISTOCK_CRESCENT.jpeg TAVISTOCK_CRESCENT
haringey/images/399932thumb.jpg 399932thumb
urbanchoice/images/18190862.jpg 18190862
wandle/images/f13c10d2-2692-457d-a208-8bb9e10b27dc.png f13c10d2-2692-457d-a208-8bb9e10b27dc
housingmoves/images/No14_Asterid Heights_DS37620.jpg No14_Asterid Heights_DS37620
wandle/ima.ges/f13c10d2-2692-457d-a208-8bb9e10b27dc.png f13c10d2-2692-457d-a208-8bb9e10b27dc
This regex works with the sample data you provided:
select regexp_substr(image_path
, '(/)([a-z0-9_ \-]+)(\.)([a-z]+)$'
, 1
, 1
, 'i'
, 2)
from t23
/
We have to include all the optional parameters after pattern so we can use the subexpr parameter to select just the filename element. Find out more.
As far as the updating goes, a million row table isn't that big. Given that you have to update all the rows there's not much you can do to tune it. Just issue the UPDATE statement and let it rip.
"its not working"
Hmmm, here's a SQL Fiddle which proves it does work. You've probably introduced a typo.
"The regexp looks unnecessary complex. Why not simply"
Perhaps it is too complicated. However your simplified version doesn't produce the correct result if there's more than one dot in the IMAGE_PATH. If that's never going to happen then your solution works just fine.
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$' );
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.
I have an Oracle stored procedure, that has as one of its parameters a Social Security Number (SSN) in the hyphenated format xxx-xx-xxxx. What is the best way to remove the hyphens, so that I transform the SSN to a xxxxxxxxx format?
REPLACE('xxx-xx-xxxx', '-', '')
Or as #jitter mentions, the third argument defaults to '', so:
REPLACE('xxx-xx-xxxx', '-')
To answer your second question, if you already have the xxx-xx-xxxx version, don't overwrite it and you'll have both.
If you're expecting xxxxxxxxx and you want xxx-xx-xxxx, piece it together using:
SUBSTR('xxxxxxxxx', 0, 3) || '-' || SUBSTR('xxxxxxxxx', 3, 2) || '-' || SUBSTR('xxxxxxxxx', 5, 4)
mystring := v_mySSN;
select substr(mystring, 0, 3)||'-'||substr(mystring, 3, 2)||'-'||substr(mystring, 5, 4)
into varialble
from some_table;
This would work inside a procedure to select your value in a variable correctly.