Is there a way to search a list for an item, then remove it?

[Justin M]

Hi there everyone. I am working on a math game where students can make their own custom game patterns. I am working on a way for students to be able to delete custom patterns down the road.

So far I have this..

Dim iczg As Integer
   Dim srch_stringzg As String
   srch_stringzg = LoadGameNameTXT.text    'or whatever you want to search for
   For iczg = 0 To CustomGameList.ListCount - 1
      If CustomGameList.List(iczg) = srch_stringzg Then
         CustomGameList.RemoveItem srch_stringzg
 
      End If
   Next iczg

Basically the name of the pattern is in the LoadGameNameTXT.text. I am trying to get the code to search for it in the CustomGameList listbox, and if found remove it from the listbox, but I am getting a type mismatch error on this line..

CustomGameList.RemoveItem srch_stringzg

Can anyone tell me what I did wrong?

Thanks!!

[Rattled_Cage]

remove the item by index and not the string it looks like

change yours to

CustomGameList.RemoveItem iczg

[Arnoutdv]

Also process the List backwards, otherwise the indexing goes wrong!

[Honduras 2811]

Also process the List backwards, otherwise the indexing goes wrong!

Honest question. No harm intended.

This is another area I don't have much experience in.

Does the conflict arise because a member of the list has been deleted, but the list hasn't been re-indexed and the program thinks that member is still there?

[Arnoutdv]

No problem Honduras!

The problem arises when you have 2 items in a row which need to be removed.
If you need to remove item 5 + 6, then after you remove item 5, item 6 will become the new item 5.
But the For Next loop sets your Counter to 6, so you skip the new item 5

Private Sub Command2_Click()
  Dim cCol As Collection
  Dim i As Long
  
  Set cCol = New Collection
  
  cCol.Add "AAA"
  cCol.Add "BBB"
  cCol.Add "BBB"
  cCol.Add "CCC"

  Debug.Print "---"
  Debug.Print "Forwards"
  Debug.Print "---"

  ' We want to remove all items with "BBB"
  For i = 1 To cCol.Count - 1
    If cCol(i) = "BBB" Then cCol.Remove i
  Next i
  
  For i = 1 To cCol.Count
    Debug.Print cCol(i)
  Next i
  
  Debug.Print "---"
  Debug.Print "Backwards"
  Debug.Print "---"
  
  Set cCol = New Collection
  
  cCol.Add "AAA"
  cCol.Add "BBB"
  cCol.Add "BBB"
  cCol.Add "CCC"
    
  ' We want to remove all items with "BBB"
  For i = cCol.Count - 1 To 1 Step -1
    If cCol(i) = "BBB" Then cCol.Remove i
  Next i
  
  For i = 1 To cCol.Count
    Debug.Print cCol(i)
  Next i
  
End Sub

[Honduras 2811]

So you have to take an action after ever deletion. Re-index or re-create the array? That's good to know and may explain some problems I have had in the past.

It's good to keep in mind that a program does things, and usually well. But there are some it doesn't do. Knowing the difference is what makes us all millionaires.

[Arnoutdv]

The action is, loop backwards.
Just test my example, it does both and shows the results.

[DataMiser]

Yep backwards is the ticket, this prevents the unchecked entries from changing their index numbers

So if you loop forwards and have 10 items in the list when you remove item 2 then item 3 becomes item 2 and you move on to item 3 which used to be item 4 so 3 is skipped over and not checked and then of course you will get an index out of bounds when you reach item 10 as it no longer exists.

Looping backwards the indexes change only on those items that have already been tested so you do not run into these issues

btw wasn't this same question asked by the Op and answered just a few days ago?

[Justin M]

Thank you guys, I just have a small follow up question.

Is there a way to search for blank list items? Like nothing is there, and if there is a blank item, remove it?



I tried using the above code, but it gives me a type mismatch error when trying to find "nothing" lol.

[jmsrickland]

If List1.List(n) = "" Then List1.RemoveItem n

or if you are really checking for blank spaces (not null as above)

Dim Lgth As Integer

Lgth = Len(List1(n))

If List1.List(n) = Space(Lgth) Then.......

[Justin M]

Awesome thanks! Would n be dimmed as a string in this case?

[jmsrickland]

No, n is an Integer which I used as the index I think you are using iczg

[4x2y]

One go

If Trim$(List1.List(n)) = "" Then List1.RemoveItem n

[jmsrickland]

Good show, 4x2y

[Justin M]

Thank you for the help.

Just as a follow up question. Lets say I have a textbox instead of listbox

Would

If Trim$(List1.List(n)) = "" Then List1.RemoveItem n

become

Dim
If Trim$(Text1.text(n)) = "" Then
Text1.text = Text1.text - n
end if

???

[Arnoutdv]

The Texbox.Text holds a single string. Not a string array.

[Justin M]

Ok, to make things more complicated lets say the stuff I needed to remove was always in brackets (), and they are always infront/on the left side of the text I want to keep, is there a way to get rid of the text in the () including the brackets from the textbox?

[4x2y]

s = "(123)45678"
s = Mid$(s, InStr(1, s, ")") + 1)
MsgBox s

[Justin M]

Thanks, would "s" in this case be a string or the textbox with the original text?

[4x2y]

Thanks, would "s" in this case be a string or the textbox with the original text?

Yes of course

[Justin M]

Oh this is awesome!

So now my code looks like this.

Dim s As String
s = LoadGameNameTXT.text
s = Mid$(s, InStr(1, s, ")") + 1)
LoadGameNameTXT.text = s

Would there be a way for it to check for cases where there could be two sets of brackets? like ()()

[Justin M]

For example, if I run the code twice would it get rid of the first set of brackets (), then if it runs again would it get rid the second set?

[4x2y]

If you are sure the part of the string you want to keep doesn't contain ")" then you can use InStrRev to get rid of all brackets in one step.

    LoadGameNameTXT.Text = Mid$(LoadGameNameTXT.Text, InStrRev(LoadGameNameTXT.Text, ")") + 1)