2
I would like to take the last paragraph 1 from this sequence:
0495747500000049908275289000100040000000041
How do I do in sql server?
2
I would like to take the last paragraph 1 from this sequence:
0495747500000049908275289000100040000000041
How do I do in sql server?
7
Can combine the function left()
to pick up all characters starting left until the limit is set by len()
(which returns the string size) -1
. Something like:
select left('0495747500000049908275289000100040000000041',
len('0495747500000049908275289000100040000000041') - 1)
Or
select left(campo, len(campo) - 1)
what if it’s a variable? type @test =
It worked. Thank you
@Felipemichaeldafonseca if any answer solved, mark as chosen by clicking on the green V next to the points
3
You can use both LEFT and SUBSTRING.
DECLARE @numero VARCHAR(MAX)
SET @numero = '0495747500000049908275289000100040000000041'
SELECT LEFT(@numero, LEN(@numero) - 1) AS 'LEFT', SUBSTRING(@numero, 0, LEN(@numero)) AS 'SUBSTRING';
Returns the left part of a string with the number of specified characters.
Returns part of a character, binary, text or image in SQL Server.
2
You can use the function RIGHT
, would look like this:
SELECT RIGHT(CAMPO, 1)
If your field is a number you need to convert to varchar:
SELECT RIGHT(CAST(CAMPO AS VARCHAR(50)))
Don’t forget to change the size of the Varchar
for your need.
I hope I’ve helped.
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Will always remove the last character or will remove the last one only be for
1
?– rray
not any number
– Felipe Michael da Fonseca
Is the size fixed? 43 characters?
– rray
To take, to extract, to obtain, I think would change the perception of the question presented somewhat. But, for what you want, the LEFT, RIGHT and SUBSTRING functions work very well
– Alberto Cláudio Mandlate