Address problem

I need to split up the address data.
so
12345  mystreet  
or
123 mystreet
 
they are in one field in the database  and not broken up i need to grab the number  so I'm thinking if there is some way to go to the first space of 123 mystreet. not sure how to code this to pull just the street number.
Thanks for any help
cdb424ttmAsked:
Who is Participating?
 
bkthompson2112Commented:
Hi cdb424ttm,

Try this:

Dim AddrParts() as string
AddrParts = Split("12345 mystreet", " ")

AddrParts(0) will contain "12345"
AddrParts(1) will contain "mystreet"

bkt
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JR2003Commented:
Val("12345  mystreet ")
will give the number 12345

JR
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Mike TomlinsonMiddle School Assistant TeacherCommented:
Split may not be such a good idea since if you have "12345 E Highland Drive" you will end up with "E", "Highland" and "Drive" all in seperate elements of the array.

Private Sub Command1_Click()
    Dim address As String
    Dim firstSpace As Integer
    Dim number As String
    Dim street As String
   
    address = "12345  mystreet"
    firstSpace = InStr(address, " ")
    number = Trim(Left(address, firstSpace))
    street = Trim(Mid(address, firstSpace))
End Sub
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bkthompson2112Commented:
>Split may not be such a good idea

That's true.  It will depend on the data and how it's used.

Another example:

"12345 1/2 E Highland Drive"
number = 123456
street = 1/2 E Highland Drive

The point is, neither solution is perfect.  It will depend on the data format and how the program intends to use it.
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GrahamSkanRetiredCommented:

Perhaps you should look for a comma first, I used to work at "Computer House, Boston Manor Road", If missing, go for the first space.

v = Split(Address, ",")
If UBound(v) = 0 Then
    v = Split(Address, " ")
End If
building = v(0)
street = v(1)
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Mike TomlinsonMiddle School Assistant TeacherCommented:
Good point bkthompson2112.

Idle_Mind
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ShauliCommented:
cdb424ttm ,

I did not participate in this paq, however, I think you have selected the wrong answer.
Lets see:
JR's solution will fail if the address is "12345 72nd Street" because it will result in 1234572
bkthompson2112's (accepted answer) will fail because the result would be AddrParts(0) will contain "12345" AND AddrParts(0) will contain "72nd", so what about the rest of the address?

Idle_Mind's solution is the right solution, even if in rare cases of "12345 1/2 E Highland Drive" it does not give a solution.

You realize that the solution you have chosen will result, in this case, like this:
AddrParts(0) will contain "12345"
AddrParts(1) will contain "1/2"

Just to let you know :)

And my appology for interupting,

S
0
 
bkthompson2112Commented:
Shauli,  you're exactly right.  As I pointed out, none of the proposed solutions is perfect.

I've always found address parsing to be problematic.
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