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)
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 column with the given values
MRN
1946
456
27
557
The column values length is fixed.
If at all any value is less than 6characters,then it should concate 0's to the left and make it 6characters length.
The desired output is
MRN
001946
000456
000027
000557
This is called left paddings. In SQL Server, this is typically done with more basic string operations:
select right(replicate('0', 6) + mrn, 6)
If mrn is a number, then use the concat() function:
select right(concat(replicate('0', 6), mrn), 6)
You can also use the FORMAT function for this. (Demo)
SELECT FORMAT(MRN ,'D6')
FROM YourTable
Change the number 6 to whatever your total length needs to be:
SELECT REPLICATE('0',6-LEN(EmployeeId)) + EmployeeId
If the column is an INT, you can use RTRIM to implicitly convert it to a VARCHAR
SELECT REPLICATE('0',6-LEN(RTRIM(EmployeeId))) + RTRIM(EmployeeId)
And the code to remove these 0s and get back the 'real' number:
SELECT RIGHT(EmployeeId,(LEN(EmployeeId) - PATINDEX('%[^0]%',EmployeeId)) + 1)
We can achieve this by adding leading zero's
select RIGHT('0000'+CAST(MRN AS VARCHAR(10)),6)
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)
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
Using T-SQL, how would I go about getting the last 3 characters of a varchar column?
So the column text is IDS_ENUM_Change_262147_190 and I need 190
SELECT RIGHT(column, 3)
That's all you need.
You can also do LEFT() in the same way.
Bear in mind if you are using this in a WHERE clause that the RIGHT() can't use any indexes.
You can use either way:
SELECT RIGHT(RTRIM(columnName), 3)
OR
SELECT SUBSTRING(columnName, LEN(columnName)-2, 3)
Because more ways to think about it are always good:
select reverse(substring(reverse(columnName), 1, 3))
declare #newdata varchar(30)
set #newdata='IDS_ENUM_Change_262147_190'
select REVERSE(substring(reverse(#newdata),0,charindex('_',reverse(#newdata))))
=== Explanation ===
I found it easier to read written like this:
SELECT
REVERSE( --4.
SUBSTRING( -- 3.
REVERSE(<field_name>),
0,
CHARINDEX( -- 2.
'<your char of choice>',
REVERSE(<field_name>) -- 1.
)
)
)
FROM
<table_name>
Reverse the text
Look for the first occurrence of a specif char (i.e. first occurrence FROM END of text). Gets the index of this char
Looks at the reversed text again. searches from index 0 to index of your char. This gives the string you are looking for, but in reverse
Reversed the reversed string to give you your desired substring
if you want to specifically find strings which ends with desired characters then this would help you...
select * from tablename where col_name like '%190'