I'm trying to get the max value from a text field. All but two of the values are numbers with a single decimal. However, two of the values have something like 8.2.10. How can I pull back just the integer value? The values can go higher than 9.n, so I need to convert this field into a number so that I can get the largest value returned. So all I want to get back is the 8 from the 8.2.1.
Select cast(VERSION as int) is bombing out because of those two values with a second . in them.
You may derive by using regexp_substr with \d pattern :
with tab as
(
select regexp_substr('8.2.1', '\d', 1, 1) from dual
union all
select regexp_substr('9.0.1', '\d', 1, 1) from dual
)
select * from tab;
For Oracle you must attend the value as string for retire only the part before the dot. Ex:
SELECT NVL( SUBSTR('8.2.1',0, INSTR('8.2.1','.')-1),'8.2.1') AS SR FROM DUAL;
Check than the value is repeated 3 times in the sentence, and if the value is zero or the value didn't have decimal part then it will return the value as was set.
I had to use T-SQL rather PL/SQL, but the idea is the same:
DECLARE #s VARCHAR(10);
SELECT #s='8.2.1';
SELECT CAST(LEFT(#s, CHARINDEX('.', #s) - 1) AS INT);
returns the integer 8 - note that it won't work if there are no dots because it takes the part of the string to the left of the first dot.
If my quick look at equivalent functions was correct, then in Oracle that would end up as:
SELECT CAST(SUBSTR(VERSION, 1, INSTR(VERSION, '.') - 1) AS INT)
Related
I have a string ST0023_Lamb_Weston_2017_US in a table from particular column. While selecting the name I need to get only Lamb_Weston_2017_US. I can use
SELECT SUBSTRING('ST0023_Lamb_Weston_2017_US', 8, 20)
But there will be different names in the column. For example ,
ST0023_Lamb_Weston_2017_US
ST0053_PL_Sandbox_Dorgan_US
ST0071_EDA_Austria
ST0071_EDA_Austria
ST10338_Nestle_Soluble_Instant_Cacao_ES
So the above mentioned are the different names available. I need to remove the "ST" part and the number part till first hyphen and return name alone. Please help me with this.
Inside substring function use charindex to pick the starting position of underscore. Plus one is added with charindex to exclude the underscore position and ending position will be considered till the length of the data.
create table data
(
value varchar(100)
)
insert into data
select 'ST0023_Lamb_Weston_2017_US' union
select 'ST0053_PL_Sandbox_Dorgan_US' union
select 'ST0071_EDA_Austria' union
select 'ST0071_EDA_Austria' union
select 'ST10338_Nestle_Soluble_Instant_Cacao_ES'
go
select value, SUBSTRING(value, CHARINDEX('_',value)+1 , LEN(value)) 'Newvalue' from data
I have a query similar to this:
SELECT YEAR_CODE FROM YEAR_CODES
and it returns several records: typically 1 but sometimes 2 or 3. The returned records look like this: 2018FOO, 2019BAR
I need to get the matching previous year of the returned codes. For instance:
2018FOO becomes 2017FOO
2019BAR becomes 2018BAR
Looking for something similar to:
REGEX_REPLACE(SELECT YEAR_CODE FROM YEAR_CODES, 4th character, 4th character minus 1)
You don't need regexp_replace(), using substr() string operator with concat() function (or concatenation operators ||) is enough :
with year_codes(year_code) as
(
select '2018FOO' from dual union all
select '2019BAR' from dual
)
select concat(substr(year_code,1,4) - 1,substr(year_code,-3)) as year_code
from year_codes;
YEAR_CODE
---------
2017FOO
2018BAR
to_number() conversion is redundant, since Oracle implicitly considers a string as a number which is completely composed of digits for an arithmetic operation.
You can do use string operations:
with c as (
<your query here>
)
select
from year_code yc
where to_number(substr(yc.code, 1, 4)) = to_number(substr(c.code)) - 1 and
substr(yc.code, 5) = substr(c.code, 5)
Hi I have one doubt in sql server .
how to get first position to right side specific character position.
table : empfiles
filename:
ab_re_uk_u_20101001
ax_by_us_19991001
abc_20181002
I want output like below:
filename
ab_re_uk_u
ax_by_us
abc
I tried like below :
select SUBSTRING(filename,1,CHARINDEX('2',filename) - 1) as filename from empfiles
above query is not given expected result please tell me how to write query to achive this task in sql server .
If last position has always numeric values then you can use patindex():
select *, substring(filename, 1, patindex('%[0-9]%', filename)-2) as NewFile
from empfiles e;
If you want to get characters after than _ to right sight of string then you can use combo to reverse() and substring()
select *,
reverse(substring(reverse(filename),charindex('_', reverse(filename))+1, len(filename)))
from empfiles e;
Another way is to use reverse in combination with STUFF.
create table f(filename nvarchar(100));
insert into f values
('ab_re_uk_u_20101001')
,('ax_by_us_19991001')
,('abc_20181002');
select
filename=reverse(stuff(reverse(filename),1,charindex('_',reverse(filename)),''))
from f
Try This
CREATE TABLE #DATA([FILENAME] NVARCHAR(100));
INSERT INTO #DATA VALUES
('ab_re_uk_u_20101001')
,('ax_by_us_19991001')
,('abc_20181002');
SELECT [filename],
SUBSTRING([filename],0,PATINDEX('%[0-9]%',[filename])-1) AS ExpectedResult
FROM #Data
Result
filename ExpectedResult
--------------------------------------
ab_re_uk_u_20101001 ab_re_uk_u
ax_by_us_19991001 ax_by_us
abc_20181002 abc
Well, obviously the last position value is a date, and the format is YYYYMMDD so its 8 characters, plus, added by underscore character, so that makes its 9 character.
Assumed by the above statement applied, the following logic of the query should work
SELECT SUBSTRING(ColumnText, 1, LEN(ColumnText) - 9)
Which means, only display characters from character position 1, to character position LEN - 9, which LEN is the length of characters, and 9 is the last 9 digit of number to be removed
Try with this ..
select [filename],SUBSTRING([filename],1,PATINDEX('%_[0-9]%',[filename])-1) from empfiles
Individual Select records
SELECT SUBSTRING('ab_re_uk_u_20101001',1,PATINDEX('%_[0-9]%','ab_re_uk_u_20101001')-1)
SELECT SUBSTRING('ax_by_us_19991001',1,PATINDEX('%_[0-9]%','ax_by_us_19991001')-1)
SELECT SUBSTRING('abc_20181002',1,PATINDEX('%_[0-9]%','abc_20181002')-1)
Struggle to design a regular expression to filter field value from varchar2 to number, so that it can remove all non-digit and only left the last period in the string, so that
"about 1,000.00" return 1000.00 or 1000
"3,000,000.000" return 300000.000 or 3000000
"3.000.000.000" return return 3000000.000 or 3000000
"a^*3^%*(C4.5d*9" return 34.59
Any method just change the string into accurate convertible string that can be converted by to_number()
I use
SELECT REGEXP_REPLACE(field_value, '[^0-9\.]+', '') from dual;
but can't resolve the 3rd case....
Because the regex in oracle are somewhat limited I don't think it's possible only using regexp_replace. You could do a workaround like this:
SELECT
CASE
WHEN last_dot < 2 THEN digits_and_dots
ELSE REPLACE(SUBSTR(digits_and_dots, 1, last_dot - 1), '.') ||
SUBSTR(digits_and_dots, last_dot)
END
FROM (
SELECT
INSTR(digits_and_dots, '.', -1) last_dot,
digits_and_dots
FROM (
SELECT
REGEXP_REPLACE(field_value, '[^0-9\.]+', '') digits_and_dots
FROM DUAL
) t
) o
Here's a way to do it, assuming there is one decimal character. The value you are working with is a string so I think of the decimal that we want to keep as a separator of the string and split it into 2 parts based on that. The first part is all characters leading up to but not including the last decimal, the second part is the last decimal and all characters after it. Then apply the replace, getting rid of everything that is not a number from the first part, and everything that is not a number or a decimal from the second part, then concatenate them together. Needs more testing with varied inputs but you get the idea. All these regular expressions are kind of expensive though so I doubt this will be the fastest solution.
with tbl(str) as (
select 'about 1,000.00' from dual union
select '3,000,000.000' from dual union
select '3.000.000.000' from dual union
select 'a^*3^%*(C4.5d*9' from dual
)
select str original,
regexp_replace(regexp_substr(str, '^(.*)\.', 1, 1, NULL, 1), '[^0-9]+', '') ||
regexp_replace(regexp_substr(str, '.*(\..*)$', 1, 1, NULL, 1), '[^0-9\.]+', '') Converted
from tbl;
SQL> /
ORIGINAL CONVERTED
--------------- ---------------
3,000,000.000 3000000.000
3.000.000.000 3000000.000
a^*3^%*(C4.5d*9 34.59
about 1,000.00 1000.00
SQL>
Shortest way is as follows:
select regexp_substr('a^*3^%*(C4.5d*9s','\d+\.\d+') from dual;
or
select regexp_replace('a^*3^%*(C4.5d*9s', '[^0.0-9]', '') from dual;
I have, in my DB oracle 10g, a field that contains references.
It's stored as : name/yyyy/mm/number
The new number, is the max number found in the part mm/number.
So, for now, I have a split of my string that gives me a list of str_array like this :
str_array(name, yyyy, mm, number)
I'd like, with this, found max number, for the couple mm/number.
Is this possible to do this?
Can I have something like :
SELECT MAX(split(reference, '/').lastPartOfArray) into nb
FROM table
where lastPartOfArray-1 = sysdate.month;
Data samples :
Smith/2013/12/1
Smith/2013/11/1
Smith/2013/12/3
Jones/2013/12/6
Smith/2013/12/3
Jones/2013/11/7
Since we are in the month 12, a max on those data must give me 6 into nb.
The number part, has no limit, it can be 1000, 10000...
The part Jones/2013 doesn't really matter for the number. But I can't have the same number, for a month.
My apologies, I don't know if this is possible, so I tried to write what I want in the query.
Is this possible, or should I create more than one field in my table(name/yyyy, mm, number)?
edit : valex answer and some custom
select MAX(CAST(SUBSTR(num,INSTR(num,'/')+9,1000) as Int))
from T
where num like TO_CHAR(sysdate,'%/YYYY/MM/%')
So this, count searching first occurence.
select MAX(CAST(SUBSTR(num,INSTR(num,'/',1 ,n)+1,1000) as Int))
from T
where num like TO_CHAR(sysdate,'%/YYYY/MM/%')
This found the n occurence of the char.
This is a helpful solution in other cases.
To get a maximum you should convert this last part into INT values otherwise you can get not right results because of STRING comparing rules will be used.
As soon as /YYYY/MM/ has got a fixed length = 9 so we can find first \ position and add 9 to this position to find a last part number substring start.
Here is an example:
select MAX(CAST(SUBSTR(num,INSTR(num,'/')+9,1000) as Int))
from T
where num like TO_CHAR(sysdate,'%/YYYY/MM/%')
SQLFiddle demo
Also you can exclude wrong formatted values from this query to avoid conversion errors using the following way:
select MAX(CAST(SUBSTR(num,INSTR(num,'/')+9,1000) as Int))
from T
where num like TO_CHAR(sysdate,'%/YYYY/MM/%')
AND
LENGTH(TRIM(TRANSLATE(SUBSTR(num,INSTR(num,'/')+9,1000),
' 0123456789', ' '))) is null
SQLfiddle demo
Try this:
SELECT
MAX(SUBSTR(num, INSTR(num, '/', 1, 3) + 1))
FROM ref
WHERE
SUBSTR(num, INSTR(num, '/', 1, 2) + 1, INSTR(num, '/', 1, 3) - INSTR(num, '/', 1, 2) - 1) = TO_CHAR(sysdate, 'MM')
Sample: http://sqlfiddle.com/#!4/1b03a/1