Quick String Parsing of a Comma Delimited String
This example will show you how parse a list a variables quickly with minimal coding.
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Codice sorgente
Sub TestMe()
'Run this sub to demonstate how the parsing functions work.
'This is an example to show begining vb programmers how to parse strings
'quickly with out having to use mid/instr/right/left every time then need
'the next item in a string of items.
'Declare our working strings
Dim myString As String
Dim curShiz As String, curVar As String
'Declare our name containers
Dim Name1 As String
Dim Name2 As String, Name3 As String, Name4 As String
myString = "Tom,Debbie,Mark,Joanie"
'Get tom
curVar = GetFirstVar(myString)
myString = GetRestOfVars(myString)
Name1 = curVar
'Get Debbie
curVar = GetFirstVar(myString)
myString = GetRestOfVars(myString)
Name2 = curVar
'Get MArk
curVar = GetFirstVar(myString)
myString = GetRestOfVars(myString)
Name3 = curVar
'Get joanie
curVar = GetFirstVar(myString)
myString = GetRestOfVars(myString)
Name4 = curVar
MsgBox "Name1 = " & Name1
MsgBox "Name2 = " & Name2
MsgBox "Name3 = " & Name3
MsgBox "Name4 = " & Name4
End Sub
Public Function GetFirstVar(sStr As String)
'Given a string like: "choad,flap,blah"
'this returns "choad"
f = InStr(1, sStr, ",")
If f = 0 Then
GetFirstVar = sStr
Else
GetFirstVar = Left(sStr, f - 1)
End If
End Function
Public Function GetRestOfVars(sStr As String)
'Given a string like: "choad,flap,blah"
'this returns "flap,blah"
f = InStr(1, sStr, ",")
GetRestOfVars = Right(sStr, Len(sStr) - f)
End Function
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