I'm trying to find special string that has a format like ##-%, start from two character and '-'
select *
from temp
where name like '##-%'
Give me some advice, or example how I can get this?
And how can I replace this format on empty string ""
If would you find the special character which is start from two character and '-'
select * from temp where name like '[a-z][a-z]-%'
if would you want to find the special string have format like '##-%
select * from temp where name like '##-[%]%'
If would you want to find the third letter is '-'
select * from temp where name like '__-%'
Add underscores for how many characters you need at start.In your case its two __
select * from temp where name like '__-%'
It's quite simple. Just use [ ] around '%' to escape it.
Select * From temp
Where name like '##-[%]%'
This will return any record where the 3rd character in the 'name' column is '-'
SELECT *
FROM temp
WHERE SUBSTRING(name, 3, 1) = '-'
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
select translate('flipflop','fl','h') from dual;
select replace('flipflop','fl','h') from dual;
These two queries give the same output,then why do we have different commands for carrying out the same operation?
Translate translate the string byte by byte
The a is replaced with an e, the h has no complement and is dropped.
SELECT TRANSLATE('So What', 'ah', 'e')
FROM dual;
Replace only occurance of word with specified word if match found like o with ay
SELECT REPLACE('So What', 'o', 'ay')
FROM dual;
UG
see https://community.oracle.com/thread/676170
REPLACE TRANSLATE
Replaces entire string at a time Replaces character one-to-one basis
Returns string if no match found Returns null if no match found
I have a simple table like the following
id, target
-----------
1, test_1
2, test_2
3, test_3
4, testable
I have a simple query like so:
select * from my_table where target like 'test_%'
What I'm expecting are the first 3 records but I'm getting all 4 records
See SQLFiddle example here
Underscore is a pattern matching character. Try this:
select * from my_table where target like 'test[_]%'
_ is also a wildcard. You can escape it like:
... like 'test\_%' escape '\'
The underscore character _ as you've used it is a wildcard for a single character, hence it returns 4 rows. Try using [_] instead of _.
To illustrate..
CREATE TABLE #tmp (val varchar(10))
INSERT INTO #tmp (val)
VALUES ('test_1'), ('test_2'), ('test_3'), ('testing')
-- This returns all four
SELECT * FROM #tmp WHERE val LIKE 'test_%'
-- This returns the three test_ rows
SELECT * FROM #tmp WHERE val LIKE 'test[_]%'
The underscore is a wildcard character that says "match any character single character", just like the % is a wildcard that says "match any 0 or more characters". If you're familiar with Regular Expressions, the underscore character is equivalent to the dot there. You'll need to properly escape the underscore to match that character literally.