如何通过修剪最后 8 个字符来返回字符串中的前几个字符？（SQL Server 2008）
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9891912/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to return first few characters in a string by trimming last 8? (SQL Server 2008)
提问by eek
I have a bunch of strings like this:
我有一堆这样的字符串：
'ABCD99991000'
'XYZ79991000'
'E2493991039'
What I really care about is everything before the last 8 characters. The problem is that the characters I care about at the start of the string are of variable length. In the above examples, what I'd want to return is:
我真正关心的是最后8个字符之前的所有内容。问题是我在字符串开头关心的字符是可变长度的。在上面的例子中，我想要返回的是：
'ABCD'
'XYZ'
'E24'
The RIGHT() function would be perfect if I could get it to return everything BEFORE the results. For example, RIGHT(E2499991039,8) returns the numbers I don't want! How can I return the variable length strength before the last 8 characters in my query?
如果我可以让它在结果之前返回所有内容，则 RIGHT() 函数将是完美的。例如，RIGHT(E2499991039,8) 返回我不想要的数字！如何在查询中的最后 8 个字符之前返回可变长度强度？
SQL Server 2008
SQL Server 2008
回答by Irfy
回答by Mitch Wheat
declare @str varchar(50)
set @str = 'ABCD99991000'
select LEFT(@str, len(@str) - 8)
(No error checking for Len(@str) < 8)
（不对 Len(@str) < 8 进行错误检查）
so,
所以，
select LEFT(colname, LEN(colname) - 8)
from table_name
回答by em3ricasforsale
Left(column, length(column - 8))
回答by Louis Ricci
BEGIN
declare @s varchar(16)
select @s = 'abcdefghij'
--
SELECT LEFT(@s, LEN(@s) - 8)
END
回答by Adam
LEFT(E2499991039,LEN(E2499991039)-8)
