Visual Basic String Shifting

[TheUnknownCoder]

Instead of StrReverse(txtString.Text) which would make 12345678 look like 87654321. I'm wanting to make it so that 12345678 would look like 78563412. Any ideas?

[Joacim Andersson]

Something like this:

Dim str As String
Dim newStr As String
Dim i As Long

str = "12345678"
For i = 1 To Len(str) Step 2
    newStr = Mid$(str, i, 2) & newStr
Next
MsgBox newStr

[LaVolpe]

Welcome to the forums.

Here is a short example. It will give you some ideas and will work only if the number of characters are even vs odd.

Private Sub Command1_Click()
    Dim sData As String, sOut As String, I As Long
    sData = "12345678"
    sOut = String$(Len(sData), vbNullChar) ' build a buffer for swapped data
    For I = 1 To Len(sData) Step 2
        Mid$(sOut, I, 2) = Mid$(sData, Len(sData) - I, 2)
    Next
    MsgBox sOut
End Sub

I see Joacim beat me to it

[Merri]

I present to you: The One Liner

MsgBox StrConv(StrReverse(StrConv("12345678", vbFromUnicode)), vbUnicode)

One Line To Rule Them All.


Disclaimer: for ANSI compatible characters only, the only perfectly safe characters are from the ASCII range, 0 - 127

[LaVolpe]

I present to you: The One Liner

MsgBox StrConv(StrReverse(StrConv("12345678", vbFromUnicode)), vbUnicode)

One Line To Rule Them All.


Disclaimer: for ANSI compatible characters only, the only perfectly safe characters are from the ASCII range, 0 - 127

Clever

[TheUnknownCoder]

Thanks guys!

[Merri]

I see no problem, the characters are the regular ones from the ASCII range: "0" = 48 ... "9" = 57, "A" = 65 ... "F" = 70 as character codes. You're not dealing with anything more magical than that. You're dealing with a String here, not with a numeric value (such as a Long).

However, if you're only dealing with hex strings and you're filling them from somewhere that are actual values (ie. you do Hex.Text = Right$("0000000" & Hex$(lngValue)) at some point or similar) it might be better to swap around the actual values.


As for why the suggestion I made works... "00F80001", as a String, takes two bytes per character. As a hex representation it looks like this (separating each hex pair with |):

30 00 30 00 | 46 00 38 00 | 30 00 30 00 | 30 00 31 00

StrConv + vbFromUnicode does this:

30 30 | 46 38 | 30 30 | 30 31

StrReverse does this:

30 31 | 30 30 | 46 38 | 30 30

StrConv + vbUnicode does this:

30 00 31 00 | 30 00 30 00 | 46 00 38 00 | 30 00 30 00

And thus we have ended up with "0100F800"

[TheUnknownCoder]

I see no problem, the characters are the regular ones from the ASCII range: "0" = 48 ... "9" = 57, "A" = 65 ... "F" = 70 as character codes. You're not dealing with anything more magical than that. You're dealing with a String here, not with a numeric value (such as a Long).

However, if you're only dealing with hex strings and you're filling them from somewhere that are actual values (ie. you do Hex.Text = Right$("0000000" & Hex$(lngValue)) at some point or similar) it might be better to swap around the actual values.


As for why the suggestion I made works... "00F80001", as a String, takes two bytes per character. As a hex representation it looks like this (separating each hex pair with |):

30 00 30 00 | 46 00 38 00 | 30 00 30 00 | 30 00 31 00

StrConv + vbFromUnicode does this:

30 30 | 46 38 | 30 30 | 30 31

StrReverse does this:

30 31 | 30 30 | 46 38 | 30 30

StrConv + vbUnicode does this:

30 00 31 00 | 30 00 30 00 | 46 00 38 00 | 30 00 30 00

And thus we have ended up with "0100F800"

Actually I found a simpler solution, by looking in my code. I beat myself up after looking at it lol. Couldn't believe I missed it.