Excel VBA十六进制到ascii错误 [英] Excel VBA hex to ascii error
问题描述
朋友
我发现以下功能将十六进制转换为文本。它适用于大部分十六进制文字转换,但会产生异常结果。 p>
例如:Hex值为:
050003d40201414c4552542d42656c6f772038353435206966204e462054726164657320666f722031352d3230206d696e75746573202c77617463682050414e49432050414e4943207570746f20353439332d2d383437360d0a0d0a53656c6c204549434845522061742032343931302e2e2e73746f706c6f7373732032353
结果,我获得使用hex2text功能= <强> | 强>
正确的结果应该是这样的:
ALERT-低于8545如果NF交易15-20分钟,观看PANIC PANIC至5493--8476 ........
(注意:我使用hex2ascii转换网站 http:// www。 rapidtables.com/c转换/数字/十六进制到ascii.htm 以获得正确的结果)
我的Hex2text函数是:
公共函数HexToText(文本作为范围)As String
Dim i As Integer
Dim DummyStr As String
对于i = 1到Len(文本)步骤2
DummyStr = DummyStr& Chr(Val(& H&(Mid(text,i,2))))
DoEvents
Next i
HexToText = DummyStr
结束功能
请帮助/提示
正如Alex所说,前6个字节正在引起问题。
您可以从第13个字符开始,通过更改ForNext循环:
For i = 13 To Len(text)步骤2
pre>
或者您可以过滤ASCII码低于32的任何字符。
code>如果Val(& H&(Mid(Text,i,2)))> 31然后DummyStr = DummyStr& Chr(Val(& H&(Mid(Text,i,2))))
或者替换(例如)一个空格。
如果Val(& H& (Text,i,2)))> 31然后
DummyStr = DummyStr& Chr(Val(& H&(Mid(Text,i,2))))
Else
DummyStr = DummyStr&
结束如果
Friends,
I have found below stated function for conversion of hex to text.It works fin for most of hex to text conversions but give abnormal result.
For Example: Hex value is:
050003d40201414c4552542d42656c6f772038353435206966204e462054726164657320666f722031352d3230206d696e75746573202c77617463682050414e49432050414e4943207570746f20353439332d2d383437360d0a0d0a53656c6c204549434845522061742032343931302e2e2e73746f706c6f7373732032353
Result i get on using hex2text Function = |
Correct Result should have been something like:
ALERT-Below 8545 if NF Trades for 15-20 minutes ,watch PANIC PANIC upto 5493--8476........
(Note: I used hex2ascii conversion website http://www.rapidtables.com/convert/number/hex-to-ascii.htm to obtain correct results)
My Hex2text Function is:
Public Function HexToText(text As Range) As String Dim i As Integer Dim DummyStr As String For i = 1 To Len(text) Step 2 DummyStr = DummyStr & Chr(Val("&H" & (Mid(text, i, 2)))) DoEvents Next i HexToText = DummyStr End Function
PLEASE HELP/SUGGEST
解决方案As Alex has stated, the first 6 bytes are causing the issue.
You could either start at the 13th character by changing the ForNext loop:
For i = 13 To Len(text) Step 2
Or you could filter any characters with ASCII code lower than 32 out..
If Val("&H" & (Mid(Text, i, 2))) > 31 Then DummyStr = DummyStr & Chr(Val("&H" & (Mid(Text, i, 2))))
Or replace them with (for instance) a space..
If Val("&H" & (Mid(Text, i, 2))) > 31 Then DummyStr = DummyStr & Chr(Val("&H" & (Mid(Text, i, 2)))) Else DummyStr = DummyStr & " " End If
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