CODEWITHSUNDEEP
excelvbaif-statementvb-like-operator
obizues
obizues

Reputation: 1473

VBA Excel Not Like String If Statement

I'm trying to write an "If" statement that checks to see if a the Key that corresponds to a Player object in a (Scripting.)Dictionary called Players has a String value that contains the letter "P."

I must be missing some VBA syntactical sugar, but I can't seem to find the answer to what I'm doing wrong anywhere.

My statement right now is:

For Each Key In Players.Keys()
    If Not (Players(Key).Position Like "*P*") Then 'something'
    End If
Next Key

However, it selects the first dictionary entry it loops through even though the Position property has a P in it.

In this case Player(Key).Position = "RP," which I would like to then skip over the "Then" statement. It currently does not work.

Any help would be appreciated.

Edit

The Player class is: 

'Class Module: Player
Public Name As String
Public Position As String
Public WAR As Double

Upvotes: 1

Views: 40344

Answers (1)

JNevill
JNevill

Reputation: 50034

This doesn't solve your problem, but... it may help. I was not able to reproduce this behavior:

Player Class Module:

'Class Module: Player
Public Name As String
Public Position As String
Public WAR As Double

Subroutine:

Sub test()
    Dim players As Scripting.Dictionary
    Set players = New Scripting.Dictionary

    Dim pers As Player

    Set pers = New Player
    pers.Position = "RP"
    players.Add "1", pers

    Set pers = New Player
    pers.Position = "What"
    players.Add "2", pers

    Set pers = Nothing

    For Each pkey In players.Keys
        If Not (players(pkey).Position Like "*P*") Then
            Debug.Print players(pkey).Position, "Not a P Player"
        Else
            Debug.Print players(pkey).Position, "Its a P Player"
        End If
    Next

End Sub

Results in Immediate Pane:

RP            Its a P Player
What          Not a P Player

Like I said in the comments, I don't know why this isn't working for you, but hopefully seeing this in simplified code may point out some problem with your class implementation, or your dictionary iterations, or your like condition... or something that isn't obvious in the bit of code you have shared.

Upvotes: 2

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