Find and replace text di vb6

Hi All,

I have some problem with vb6 code.
I have a data contain :

:25:0186350587
:25:186350587
:25:86350587
:25:6350587
:25:350587
:25:50587
:25:587
:25:87
:25:7

I want it to insert zero number after :25: , but the number after :25: is always has to be 10 digit. So it would be :

:25:0186350587
:25:0186350587
:25:0086350587
:25:0006350587
:25:0000350587
:25:0000050587
:25:0000000587
:25:0000000087
:25:0000000007

I have a code in vb6 but it only insert two digit of zero number, so it would be :

:25:000186350587 (more than 10 digit)
:25:00186350587 (more than 10 digit)
:25:0086350587 (10 digit)
:25:006350587 (less than 10 digit)
:25:00350587 (less than 10 digit)
:25:0050587 (less than 10 digit)
:25:00587 (less than 10 digit)
:25:0087 (less than 10 digit)
:25:007 (less than 10 digit)

this is my code with textbox and command :

Private Sub cmdProses_Click()
Dim txt
txt = TextBox.Text
If InStr(1, TextBox.Text, ":25:") > 0 Then
TextBox.Text = Replace(txt, ":25:", ":25:00")
End If
End Sub

Would you like to help me with the new code please

Thank you
Andi.

Oho ! try this one dear

Code:

---

Private Sub Command1_Click()
Dim txt As String

txt = TextBox.Text
  txt = (Format(Mid(txt, 5, 10), "0000000000"))    ' use format() & Mid() command
TextBox.Text = ":25:" & txt

End Sub

---

It does not work.

FYI this link is the complete of my program along with sample text that I want to edit
http://docs.google.com/leaf?id=0B8HO...ut=list&num=50

Thanks
Andi

Thread moved from the 'CodeBank VB6' forum (which is for you to post working code examples, not questions) to the 'VB6 and earlier' forum

Quote:

---

Originally Posted by andirst

It does not work.

FYI this link is the complete of my program along with sample text that I want to edit
http://docs.google.com/leaf?id=0B8HO...ut=list&num=50

Thanks
Andi

---

Andi,
Your above link is http://docs.google.com/leaf?id=0B8HO...ut=list&num=50 does not contain any sample. Please, attach your sample file here to let me know what is your motive:cool:

Somehow I can not upload the sample here, but I have another Link to my sample VB6

http://365files.com/6hd0b88aamzk/Rep...ext_4.zip.html

Thanks
Andi

Quote:

---

It does not work.

---

what does this mean?

i tried like this after copying the data (ctrl C) from the original post

vb Code:

myarr = Split(Clipboard.GetText, vbNewLine)
For i = 0 To UBound(myarr)
    myarr(i) = Left(myarr(i), 4) & Format(Mid(myarr(i), 5), "0000000000")
Next
Debug.Print Join(myarr, vbNewLine)

result

Code:

---

:25:0186350587
:25:0186350587
:25:0086350587
:25:0006350587
:25:0000350587
:25:0000050587
:25:0000000587
:25:0000000087
:25:0000000007

---

Here's another way that does not rely on the clipboard, akthough I think westconns is more elegant.

Code:

---

Dim A$(8), I As Integer, X As Integer

A$(0) = ":25:0186350587"
A$(1) = ":25:186350587"
A$(2) = ":25:86350587"
A$(3) = ":25:6350587"
A$(4) = ":25:350587"
A$(5) = ":25:50587"
A$(6) = ":25:587"
A$(7) = ":25:87"
A$(8) = ":25:7"

For I = 0 To 8
Print Left$(A$(I), 4) & String$(14 - Len(A$(I)), "0") & Mid$(A$(I), 5)
Next I

---

i assumed the data comes from some file, so substituting an array from file input is quite similar to the result from clipboard

and i was lazy to type in 8 lines of data

Dear All,

Finally I know why my code and your code can never be executed, the user does not ever mention that in the search string that exists between two strings. I will be working on this code to start all over again. Thanks for all your help, I really appreciate it. I'll be back again if need your help. Once again thank you all.

Warm regards
Andi

Hi again All,

This is what I've figure out :

If between ":25:" and ":28:" there is 10 character e.g 0123456789 then do nothing
If between ":25:" and ":28:" there is 9 character e.g 12345678 then replace ":25:" with ":25:0"
If between ":25:" and ":28:" there is 8 character e.g 12345678 then replace ":25:" with ":25:00"
If between ":25:" and ":28:" there is 7 character e.g 1234567 then replace ":25:" with ":25:000"
and so on.

is it possible to do it wit InStr?

Code:

---

Dim A$(8), I As Integer

A$(0) = ":25:0186350587"
A$(1) = ":28:186350587"
A$(2) = ":29:86350587"
A$(3) = ":30:6350587"
A$(4) = ":25:350587"
A$(5) = ":25:50587"
A$(6) = ":25:587"
A$(7) = ":25:87"
A$(8) = ":25:7"

For I = 0 To 8
    If val(Mid$(A$(I),2))>24 and val(Mid$(A$(I),2))<29 then
        Debug.Print Left$(A$(I), 4) & String$(14 - Len(A$(I)), "0") & Mid$(A$(I), 5)
    end if
Next I

---

I have tested the code above and I hope I've understood your requirements.:)

Quote:

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the user does not ever mention that in the search string that exists between two strings.

---

i do not understand this at all

and still I bet OP has not tried this code...

Code:

---

Dim My1stArray() As String, My2ndArray() As String, S As String
Dim LB As Integer, UB As Integer, LoopCnt As Integer
 
S = ":25:0186350587,:25:186350587,:25:86350587,:25:6350587,:25:350587,:25:50587,:25:587,:25:87,:25:7"
 
My1stArray = Split(S, ",")
 
LB = LBound(My1stArray)
UB = UBound(My1stArray)
 
For LoopCnt = LB To UB
 
  My2ndArray = Split(My1stArray(LoopCnt), ":25:")
  Debug.Print ":25:" & Format(My2ndArray(1), "0000000000")
 
Next LoopCnt

---

Quote:

---

the user does not ever mention that in the search string that exists between two strings.

---

I mean They don't even mention that I have to find the length of the text between two string.

For example I have a data contain :

:25:0123456789:28:abcdef:25:123456789:28:defgh:25:12345678:28:ijklm:25:1234567:28:

if between ":25:" and ":28:" there is 10 char e.g 0123456789 then do nothing or
if between ":25:" and ":28:" there is 9 char e.g 123456789 then insert 0 after ":25:" or
if between ":25:" and ":28:" there is 8 char e.g 12345678 then insert 00 after ":25:" or
if between ":25:" and ":28:" there is 7 char e.g 1234567 then insert 000 after ":25:"

so the result will be

:25:0123456789:28:abcdef:25:0123456789:28:defgh:25:0012345678:28:ijklm:25:0001234567:28:

Your code works fine but how about my new problem?

Thanks for your kindness
Andi