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vbscript - instrRev reduce a string down

does anyone know how to reduce a string down with the instrRev or instr..dynamically...

Logical5=6015\:USD\:5\:6050@5@6017@6025@5@0@0\:0\:0\:0@false@-1_1@44@0@0@-1_1;5;x;0\:0\:0\:0;0;0

var1 = @5@6017@6025@5@0@0\:0\:0\:0@false@-1_1@44@0@0@-1_1;5;
var2 = ;0\:0\:0\:0;0;0

so that I can now modify the x value in that row...

newRowVar = "Logical5=6015\:USD\:5\:6050" & var1 & "900" & var2

thanks!!

Guy Hengel [angelIII / a3]

is "x" the real value in there?
if yes, you might consider doing a
newRowVar = REPLACE(Logical5, ";x;", "900")

apart from that, it's far from obvious otherwise how to "locate" the x...
you might want to clarify the "rule(s)" ?

GlobaLevel

ASKER

hi angelIII...sorry x is not a real value I only put that in there so you can see where the value is located...its a number that change ...just like any other value in this row...

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Guy Hengel [angelIII / a3]

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GlobaLevel

ASKER

I know ...thats what im asking.. ;)

var1 = Mid(origVar, InStrRev(origVar, "=") +1)
msgbox(var1)
var2 = right(var1, InStrRev(var1, "\") +1)
msgbox(var2)
var3 = right(var2, InStrRev(var2, "\") +1)
msgbox(var3)
var4 = right(var3, InStrRev(var3, "\") +1)
msgbox(var4)
var5 = (left(right(var4, InStrRev(var4, "@") +1)-1))
msgbox(var5)

Guy Hengel [angelIII / a3]

you might want to check split() function, but still not sure how you locate this exactly?

RobSampson

Something like this might work. It's reasonably long winded, as opposed to using Split, but does the job. It also works on the assumption that there is exactly the same amount of semi-colons in each line.

Regards,

Rob.

strLine = "Logical5=6015\:USD\:5\:6050@5@6017@6025@5@0@0\:0\:0\:0@false@-1_1@44@0@0@-1_1;5;x;0\:0\:0\:0;0;0"
var1 = ""
var2 = ""
strValue = ""
' Find the first ;
intPos = InStr(strLine, ";")
If intPos > 0 And intPos < Len(strLine) Then
	' Find the second ;
	intPos = InStr(intPos + 1, strLine, ";")
	var1 = Left(strLine, intPos)
	If intPos > 0 And intPos < Len(strLine) Then
		' Now find the next ; and get the value of x in between
		intPos2 = InStr(intPos + 1, strLine, ";")
		If intPos2 > 0 Then
			strValue = Mid(strLine, intPos + 1, (intPos2 - 1) - intPos)
			var2 = Mid(strLine, intPos2)
		End If
	End If
End If
MsgBox "var1: " & var1 & VbCrLf & "var2: " & var2 & VbCrLf & "strValue: " & strValue

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RobSampson

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GlobaLevel

ASKER

turns out the '6050' is an unknown value and needs to stay dynamic...
was this:
newRowVar = "Logical5=6015\:USD\:5\:6050" & var1 & "900" & var2
needs to be this:
newRowVar = "Logical5=6015\:USD\:5\:" & var1 & "900" & var2

RobSampson

OK, what about this?

Regards,

Rob.

strLine = "Logical5=6015\:USD\:5\:6050@5@6017@6025@5@0@0\:0\:0\:0@false@-1_1@44@0@0@-1_1;5;x;0\:0\:0\:0;0;0"
var1 = ""
var2 = ""
var3 = ""
strValue2 = ""
arrBits = Split(strLine, ";")
For intBit = 0 To 1
	var1 = var1 & arrBits(intBit) & ";"
Next
strValue2 = arrBits(2)
For intBit = 3 To 5
	var3 = var3 & ";" & arrBits(intBit)
Next
' Now find the first @ in var1
intEnd = InStr(var1, "@")
intStart = InStrRev(var1, ":", intEnd)
strValue1 = Mid(var1, intStart + 1, intEnd - 1 - intStart)
var2 = Mid(var1, intEnd)
var1 = Left(var1, intStart)

strNewRow = var1 & strValue1 & var2 & strValue2 & var3

MsgBox "var1: " & var1 & VbCrLf & "strValue1: " & strValue1 & VbCrLf & "var2: " & var2 & VbCrLf & "strValue2: " & strValue2 & VbCrLf & "var3: " & var3 & VbCrLf & VbCrLf & "strNewRow: " & VbCrLf & strNewRow

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tv0085

Maybe it's better to do this with regular expressions.
You can use http://www.txt2re.com to create the regex to select x and the replace method to replace x with the new value. Select VBScript at the bottom to get the code generated.
replace method: http://msdn.microsoft.com/en-us/library/k9z80300(VS.85).aspx

GlobaLevel

ASKER

tv0085...can you give feedback as to what this regular expression generator actually does..thanks...

SOLUTION

tv0085

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