I am trying to get last numeric part in the given string.
For Example, below are the given strings and the result should be last numeric part only
SB124197 --> 124197
287276ACBX92 --> 92
R009321743-16 --> 16
How to achieve this functionality. Please help.
Try this:
select right(#str, patindex('%[^0-9]%',reverse(#str)) - 1)
Explanation:
Using PATINDEX with '%[^0-9]%' as a search pattern you get the starting position of the first occurrence of a character that is not a number.
Using REVERSE you get the position of the first non numeric character starting from the back of the string.
Edit:
To handle the case of strings not containing non numeric characters you can use:
select case
when patindex(#str, '%[^0-9]%') = 0 then #str
else right(#str, patindex('%[^0-9]%',reverse(#str)) - 1)
end
If your data always contains at least one non-numeric character then you can use the first query, otherwise use the second one.
Actual query:
So, if your table is something like this:
mycol
--------------
SB124197
287276ACBX92
R009321743-16
123456
then you can use the following query (works in SQL Server 2012+):
select iif(x.i = 0, mycol, right(mycol, x.i - 1))
from mytable
cross apply (select patindex('%[^0-9]%', reverse(mycol) )) as x(i)
Output:
mynum
------
124197
92
16
123456
Demo here
Here is one way using Patindex
SELECT RIGHT(strg, COALESCE(NULLIF(Patindex('%[^0-9]%', Reverse(strg)), 0) - 1, Len(strg)))
FROM (VALUES ('SB124197'),
('287276ACBX92'),
('R009321743-16')) tc (strg)
After reversing the string, we are finding the position of first non numeric character and extracting the data from that position till the end..
Result :
-----
124197
92
16
Related
In my local table, I am try to check if an Oracle Number column called JOBNUMBER has a value that exists in a string parameter. Technically I am passing in the string as a stored procedure nvarchar2 parameter, but for simplicity, I hardcoded the string in my Query below:
SELECT FIRST_NAME, JOB_NUMBER
FROM JOBTABLE
WHERE TO_CHAR(JOB_NUMBER) IN ('00052, 00048');
When Oracle runs the query above, it returns no values even though 00052 is a number value in the table column for JOB_NUMBER. I'm thinking that it checks for the whole string ('00052, 00048') in JOB_NUMBER and can't find it, so it returns no values. The string will contain different values each time, and there will several numbers (of type string) in that string.
Does anyone know how to do this?
The trick is to keep the leading zeroes of the number when comparing to the string, then looping through the string to compare. Here a CTE is used is to simulate creating a numeric job number and a string to search. The TO_CHAR function makes sure to preserve the leading zeroes and the FM format removes the leading space that TO_CHAR leaves for the sign. CONNECT BY loops through the elements for the count of the delimiter + 1 times, keeping the count in the value in 'LEVEL'. This value is used in REGEXP_SUBSTR to iterate through the elements to compare the converted numeric value to each element to see if a match is found. Note this regular expression allows for NULL elements should you need to know which item in the list is your match.
SQL> with tbl(job_nbr_in, job_str_in) as (
select 00052, '00052, 00048' from dual
)
select --level element_nbr,
to_char(job_nbr_in, 'FM00000') search_for, job_str_in in_string,
regexp_substr(job_str_in, '(.*?)(, |$)', 1, level, NULL, 1) found
from tbl
where to_char(job_nbr_in, 'FM00000') = regexp_substr(job_str_in, '(.*?)(, |$)', 1, level, NULL, 1)
connect by level <= regexp_count(job_str_in, ',')+1;
SEARCH_FOR IN_STRING FOUND
---------- ------------ ------------
00052 00052, 00048 00052
If you are not sure if you will always have a space after the comma, remove spaces with REPLACE and adjust the delimiter in REGEXP_SUBSTR:
with tbl(job_nbr_in, job_str_in) as (
select 00052, '00052, 00048' from dual
)
select to_char(job_nbr_in, 'FM00000') search_for, job_str_in in_string,
regexp_substr(replace(job_str_in, ' '), '(.*?)(,|$)', 1, level, NULL, 1) found
from tbl
where to_char(job_nbr_in, 'FM00000') = regexp_substr(replace(job_str_in, ' '), '(.*?)(,|$)', 1, level, NULL, 1)
connect by level <= regexp_count(job_str_in, ',')+1;
I want to make a substring of a column value using a specific delimiter.I tried SUBSTRING_INDEX,but it doesn't work for SQL.Is there any way to achieve this??
Column values are:
ARTCSOFT-1111
ARTCSOFT-1112
ARTCSOFT-1113
and I want to achieve the same thing in SQL:
SUBSTRING_INDEX(Code,'SOFT-',1))
i.e I want the number after SOFT- in my substring.I can't use only - because before SOFT- there is chance that - may occur(rare case,but I don't want to take a chance)
Try using just SUBSTRING . For example
SELECT
SUBSTRING(code, CHARINDEX('SOFT-', code) + 5, LEN(code)) AS [name] from dbo.yourtable
hope this helps.
Tested Result:
SELECT RIGHT(Code , CHARINDEX ('-' ,REVERSE(Code))-1)
Read this as: Get the rightmost string after the first '-' in a reversed string - which is the same as the string after the last '-' character.
Try This Query:
select substring(col,charindex('-',col)+1,len(col)-charindex('-',col)) from #Your_table
Explanation of Query:
Here Charindex find the '-' delimeter [length] IN Given String and now that Result[length+1] is our starting point and ending length is [len(col)-starting length] gives ending point and then use substring Function to split a string according to our requirement.
Result of Query:
Required_col
1111
1112
1113
hie am trying to select the integer value before the char C in my SQL database table which contains the information below.
240mm2 X 15C WIRING CABLE
150mm2 X 3C flex
10mm2 x 4C swa
so far i have used the query
select left ('C',CHARINDEX ('C',product_name)) from product
and i get 'C' on my results which is correct. Now am stuck does anyone know how i can modify the above select query to get a result which only lists the integers for eg
15
3
4
Two observations: the integer before "C" has a space before it and there is no space between the integer and "C".
If these are generally true, then you can do what you want using substring_index():
select substring_index(substring_index(product_name, 'C', 1), ' ', -1) + 0 as thenumber
The + 0 simply converts the value to a number.
If you're doing this in SQL Server you could try the following:
Select Substring(product_name,
PATINDEX('% [0-9]%',product_name) + 1,
PATINDEX('%[0-9]C%',product_name) - PATINDEX('% [0-9]%',product_name)
) as num
from Product
This assumes that there is a space before the number and always a C after the number with no space.
It works out the starting point and then the length based on the start and end and performs a substring with the results.
You could use a combination of instring and substring.
First get the position of the C
Then substring till C
It goes like this:
SELECT INSTR('foobarbar', 'bar');
= 4
And then you select substring from 1 to 4.
Someone asked here how to get only values which are a number :
So , if the table is :
DECLARE #Table TABLE(
Col nVARCHAR(50)
)
INSERT INTO #Table SELECT 'ABC'
INSERT INTO #Table SELECT '234.62'
INSERT INTO #Table SELECT '10:10:10:10'
INSERT INTO #Table SELECT 'France'
INSERT INTO #Table SELECT '2'
then - the desired results are :
234.62
2
But when I tested this query :
SELECT * FROM #Table WHERE Col LIKE '%[0-9.]%' --expected to see only 234.62
it showed :
234.62
10:10:10:10
2
Question #1
How come 10:10:10:10 , 2 satisfies the condition ?
Question #2
I saw this answer here which does work
SELECT * FROM #Table WHERE Col NOT LIKE '%[^0-9.]%'
But I don't understand why this works. AFAIU - it selects all values which are not like (not(has number) and not( has dot)) which is ===>(de morgan)===> not like ( has number or has dot)
Can someone please shed light ?
nb I already know that isnumeric can be used also , but it's unsafe (+). also valid wildcards are %,_,[],[^]
Any particular use of [set] within a LIKE expression is a check against one character in the target string.
So, LIKE '%[0-9.]%' says - % - match 0-to-many arbitrary characters, then [0-9.] match one character in the set 0-9., and then % match 0-to-many arbitrary characters. Paraphrased, it says "match any string that contains at least one character in the set 0-9.". So, 10:10:10:10 can be matched as 0 arbitrary characters, then 1 matches [0-9.], and then 0:10:10:10 matches the final %.
LIKE '%[^0-9.]%' says - % - match 0-to-many arbitrary characters, then [^0-9.] match one character not in the set 0-9., and then % match 0-to-many arbitrary characters. Paraphrased, it says "match any string that contains at least one character outside of the set 0-9.. So when we apply the NOT to the front of that, we are saying "match any string that doesn't contain at least one character outside of the set 0-9." or "match strings that only contain characters in the set 0-9..
Essentially, the double-negative is a way to make an assertion about all characters in the string.
I have a column in a SQL Server table that has the following rows:
MyColumn : C1_xxx1,C2_xxx1,C3_xxx1,C1_xxx2,C1_xxx3,C3_xxx2 etc
It is a text column that contains strings that have the following format: CY_mystring where Y is a number from 1 to 5, followed by the '_' character then mystring that can have any value.
Is there a way to make a select return this column ordered as following:
C1_xxx1
C1_xxx2
C1_xxx3
......
C1_xxxn
C2_xxx1
......
C2_xxxn
C3_xxx1
.......
C3_xxxn
etc
Ordered by the CY_ substring.
thank you
This should do it .. (order first by the first two chars, and then by the last char (assuming that the final n is always one digit long))
SELECT
Column1
FROM
TABLENAME
ORDER BY
LEFT(Column1,2) ASC,
RIGHT(Column1,1) ASC
You say that Y is a number from 1 to 5 it's always one character long. Assuming the format is xY_xxxZ, you can order on Y then Z like:
order by
substring(MyColumn,2,1) -- Second digit
, right(MyColumn,1) -- Last digit
If Z can be longer than one character (i.e. 10 or higher) you can use pathindex to determine the number of digits at the end:
order by
substring(MyColumn,2,1) -- Second digit
, right(MyColumn, patindex('%[^0-9]%', reverse(MyColumn))-1) -- Digits at end