2017-08-17 59 views
2

这里是我的十六进制输入如何十六进制转换为字符串在SQL Server

0x3c0x3c0x5bIMG0x5d0x5bSIZE0x5dHALF0x5b0x2fSIZE0x5d0x5bID0x5d540x5b0x2fID0x5d0x5b0x2fIMG0x5d0x3e0x3e 

预期成果是:

<<[IMG][SIZE]HALF[/SIZE][ID]54[/ID][/IMG]>> 
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https://dba.stackexchange.com/questions/132996/convert-hexadecimal-to-varchar –

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SELECT CONVERT(VARCHAR(MAX),0x3c0x3c0x5bIMG0x5d0x5bSIZE0x5dHALF0x5b0x2fSIZE0x5d0x5bID0x5d540x5b0x2fID0x5d0x5b0x2fIMG0x5d0x3e0x3e); –

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此链接doenst帮助我。需要得到确切的结果<< [IMG] [SIZE] HALF [/ SIZE] [ID] 54 [/ ID] [/ IMG] >>这个结果。 –

回答

1

如果它是只有一小集,总是需要更换一个变量,那么你也可以替换他们这样的ASCII码:

declare @string varchar(max) = '0x3c0x3c0x5bIMG0x5d0x5bSIZE0x5dHALF0x5b0x2fSIZE0x5d0x5bID0x5d540x5b0x2fID0x5d0x5b0x2fIMG0x5d0x3e0x3e'; 

select @string = replace(@string,hex,chr) 
from (values 
('0x3c','<'), 
('0x3e','>'), 
('0x5b','['), 
('0x5d',']'), 
('0x2f','/') 
) hexes(hex,chr); 

select @string as string; 

返回:

string 
------ 
<<[IMG][SIZE]HALF[/SIZE][ID]54[/ID][/IMG]>> 

如果有更多的字符,或者硬编码被压低了?
然后循环的替代也将得到这一结果:

declare @string varchar(max) = '0x3c0x3c0x5bIMG0x5d0x5bSIZE0x5dHALF0x5b0x2fSIZE0x5d0x5bID0x5d540x5b0x2fID0x5d0x5b0x2fIMG0x5d0x3e0x3e'; 

declare @loopcount int = 0; 
declare @hex char(4); 
while (patindex('%0x[0-9][a-f0-9]%',@string)>0 
     and @loopcount < 128) -- just safety measure to avoid infinit loop 
begin 
    set @hex = substring(@string,patindex('%0x[0-9][a-f0-9]%',@string),4); 
    set @string = replace(@string, @hex, convert(char(1),convert(binary(2), @hex, 1))); 
    set @loopcount = @loopcount + 1; 
end; 

select @string as string; 

如果你想在一个UDF包,那么你甚至可以在查询中使用它。

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这是我正在寻找的,真的很感谢你的帮助! –

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@NileshBankar谢谢。如果有更多的代码需要替换,我已经包含了非硬编码版本。 – LukStorms

2

你的字符串是混合十六进制和字符的数据,所以你需要对它进行解析用代码。一个棘手的部分是将0xCC子字符串转换为它表示的字符。首先假设它是二进制的,然后转换为char。使用递归所有0xCC子迭代

declare @imp nvarchar(max) = '0x3c0x3c0x5bIMG0x5d0x5bSIZE0x5dHALF0x5b0x2fSIZE0x5d0x5bID0x5d540x5b0x2fID0x5d0x5b0x2fIMG0x5d0x3e0x3e'; 

with cte as (
select replace(col, val, cast(convert(binary(2), val, 1) as char(1))) as col 
from (
    -- sample table 
    select @imp as col 
    ) tbl 
cross apply (select patindex('%0x__%',tbl.col) pos) p 
cross apply (select substring(col,pos,4) val) v 
union all 
select replace(col, val, cast(convert(binary(2), val, 1) as char(1))) as col 
from cte 
cross apply (select patindex('%0x__%',col) pos) p 
cross apply (select substring(col,pos,4) val) v 
where pos > 0 
) 
select * 
from cte 
where patindex('%0x__%',col) = 0; 

返回

col 
<<[IMG][SIZE]HALF[/SIZE][ID]54[/ID][/IMG]>> 
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感谢您的帮助! –

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