I have a string like
"Pros: Cuts WellCons: Not Durable"
I want to get this sub string
Pros: Cuts Well
A sub string before Cons.
You can use PATINDEX to get the index of Cons. Then use LEFT to extract the required substring:
SELECT LEFT(#mystring, PATINDEX('%Cons%', #mystring) - 1)
If you're always going to use the word "Cons" as the delimiter, you can use this:
declare #string as varchar(100)
set #string = 'Pros: Cuts WellCons: Not Durable'
select substring(#string,0, patindex('%Cons%', #string))
Pat Index finds the first occurrence of the pattern "Cons" in your string.
Then you just use the substring function using that as the length of the substring.
Related
I have a simple string; for example,'01023201580001'.
I would like to replace the last two characters of this string; '01', with '00'.
I could extract the last two characters from this string as RIGHT(columname,2) and then use
REPLACE([columname], RIGHT([columname], 2), '00') as newColumnString
But in the result, it replaces the first two characters as well?
Expected result: 01023201580000
Result I get: 00023201580000
What am I doing wrong?
The second argument to the replace() function defines a pattern to match. The function will look for all instances of that pattern in the target string (first argument) and replace them with the replacement text (third argument).
If you know you only need to change the last two characters, you can take the value excluding those characters and then append the characters you want:
select left(columname, len(columname) - 2) + '00';
If you are doing this for an entire column and some of the rows might not end with '01', you can filter those out:
update MyTable
set columname = left(columname, len(columname) - 2) + '00'
where columname like '%01';
You could also use stuff() in a similar way.
In SQL server, you can use substring like so:
DECLARE #s NVARCHAR(20) = N'01023201580001';
DECLARE #ReplaceWith NVARCHAR(20) = N'00';
SELECT SUBSTRING(#s, 0, LEN(#s) - 1) + #ReplaceWith;
Output: 01023201580000
I have a VARCHAR value that looks like this:
5.95 $ Additional fees
How can I remove everything left from character '$' (including that character) ? So that I get the following result:
Additional fees
The '$' is always present.
STUFF and CHARINDEX would be the simpliest way, in my opinion:
SELECT STUFF(YourColumn,1, CHARINDEX('$',YourColumn),'')
FROM (VALUES('5.95 $ Additional fees'))V(YourColumn);
Note that as $ has a whitespace afterwards, the value returned will have a leading whitespace (' Additional fees'). You could use TRIM (or LTRIM and RTRIM on older versions of SQL Server) to remove this, if it isn't wanted.
I haven't assumed that the portion string to be replaced is CHARINDEX('$',YourColumn)+1, as we have one sample. As far as we know, you could also have values such as '10.99$Base Cost'. If the +1 was used, it would return 'ase Cost' for such a value.
Hello do it like below syntax
declare #temp nvarchar(max)='5.95 $ Additional fees'
select SUBSTRING(#temp,charindex('$',#temp)+1,len(#temp)-1)
You can use SUBSTRING get the particular string and CHARINDEX function to get index of special character, in your case $.
DECLARE #Var VARCHAR(100)
SET #Var = '5.95 $ Additional fees'
SELECT SUBSTRING(#Var, CHARINDEX('$', #Var) + 1, LEN(#Var) - LEN(LEFT(#Var, CHARINDEX('$', #Var))))
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
I need to validate in my query if the value of a string (the first part) is equal to a definited value, for instance:
String
----------
F11-EDEDED
F1-SAFSDA
F455-ADADD
F11-ASDA-FAFA
And validate when the string is F11, i was searching something like split on vba, but i can't find it.
Im working with :
Case when ("splitted string") =F11 then X)
Use a Left() and Charindex() to grab the beginning of your strings.
Declare #str varchar(100)='F11-ASDA-FAFA'
Select #str,Case When left(#str,charindex('-',#str)-1)='F11' Then 1 Else 0 End
I have some sql function that returns character varying type. The output is something like this:
'TTFFFFNN'. I need to get this characters by index. How to convert character varying to array?
Use string_to_array() with NULL as delimiter (pg 9.1+):
SELECT string_to_array('TTFFFFNN'::text, NULL) AS arr;
Per documentation:
In string_to_array, if the delimiter parameter is NULL, each character
in the input string will become a separate element in the resulting array.
In older versions (pg 9.0-), the call with NULL returned NULL. (Fiddle.)
To get the 2nd position (example):
SELECT (string_to_array('TTFFFFNN'::text, NULL))[2] AS item2;
Alternatives
For single characters I would use substring() directly, like #a_horse commented:
SELECT substring('TTFFFFNN'::text, 2, 1) AS item2;
SQL Fiddle showing both.
For strings with actual delimiters, I suggest split_part():
Split comma separated column data into additional columns
Only use regexp_split_to_array() if you must. Regular expression processing is more expensive.
In addition to the solutions outlined by Erwin and a horse, you can use regexp_split_to_array() with an empty regexp instead:
select regexp_split_to_array('TTFFFFNN'::text, '');
With an index, that becomes:
select (regexp_split_to_array('TTFFFFNN'::text, ''))[2];