请问如何判断字符串中全部是英文字母?

[wangyangcheng]

什么办法效率最高？
谢谢

[wangyangcheng]

I used ISNUMERIC() function.

[ynmc]

Learning...
顺顶

[leon_kin]

declare @word nchar(1)
set @word='中'
if unicode(@word) between 19968 and 19968+20901
print '汉字'
else
print '不是汉字'

[MorningTea]

SQL中有没有办法区分数字和中文字符
URL:http://community.csdn.net/Expert/topic/4218/4218786.xml?temp=.9425012

[天地客人]

帮UP

[wangyangcheng]

又如何判断全是阿拉伯数字呢？

[lisiyong]

select * from tablename where Columnname1 like '[a-z]' or Columnname1 like '[A-Z]'

[wgsasd311]

--假设此表主键为id
select a.* from tbname a where not exists(
select * from tbname where id=a.id and 字段 not like '%[a-z,A-Z]%')

[lisiyong]

判断全是阿拉伯数字:

select * from tablename where Columnname1 not like '%[^0-9]%'

[wgsasd311]

又如何判断全是阿拉伯数字呢？=====>
--假设此表主键为id
select a.* from tbname a where not exists(
select * from tbname where id=a.id and 字段 not like '%[0-9]%')

[vivianfdlpw]

/*
判断指定字符串中是否全是英文字母(可区分全角字符)
　　是，返回１；反之０

　　半角英文字母Unicode范围65-90,97-122
全角英文字母Unicode范围65313-65338,65345-65370
*/
create function isletter(@str varchar(8000))
returns int
as
begin
while len(@str)>0
begin
if (unicode(left(@str,1)) not between 65 and 90)
and
(unicode(left(@str,1)) not between 97 and 122)
and
(unicode(left(@str,1)) not between 65313 and 65338)
and
(unicode(left(@str,1)) not between 65345 and 65370)
return 0

set @str=stuff(@str,1,1,'')
end

return 1
end
go

--测试
select dbo.isletter('a1')
select dbo.isletter('1')
select dbo.isletter('ａ')
select dbo.isletter('a')

[iwl]

select * from tablename where Lower(Columnname1) like '[a-z]'

或者

用vivianfdlpw() 的方法