• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 142
  • Last Modified:

Display correct string from Type...End Type

I'm using Type...End Type with a ton of strings and would like to be able to call the right one using a changing variable.  Like below:

Type ThisIsATest
  Temp as String
End Type

Dim Test as ThisIsATest

'This will give you an idea of how I want to use it

Temp2 = "Temp" 'temp2 will always be different

'temp will of course have a value from somewhere else in the 'program

text1.Text = test.temp2
--------------------------------------------

I've tried it as many different ways that I could think of and none of them worked.  I'm trying to display the text that is in a string when a user selects a specific one.
0
dokken
Asked:
dokken
  • 4
  • 3
1 Solution
 
clifABBCommented:
This will not work:
  text1.Text = test.temp2
Because temp2 isn't described in then ThisIsATest structure.

From the what I gather in your question, what you want is an array.
0
 
dokkenAuthor Commented:
Right I knew Temp2 wouldn't work, it was just given to make things easier to understand.  How would I use an Array?  I don't have a lot of experience working with them.
0
 
clifABBCommented:
You would dimension an array as such:

Dim MyArray(10) As String

Then you would assign and read values just like any other variable, except you must include the index:

MyArray(1) = "Value 1"
MyArray(2) = "Value 2"

Text1.Text = MyArray(2)
0
Free Tool: ZipGrep

ZipGrep is a utility that can list and search zip (.war, .ear, .jar, etc) archives for text patterns, without the need to extract the archive's contents.

One of a set of tools we're offering as a way to say thank you for being a part of the community.

 
dokkenAuthor Commented:
Not quite what I want to do.  I'm reading in over 300 strings and using the treeview component.  I want to be able to use the treeview click event to display the right string which is named the same as the treeview's key.  That's why I'm using the Type...End Type.  I was just doing:
If tree.Nodes("string1").Selected Then txtInfo.Text = temp.string1

I was doing it over and over again changing the key for each one, so I was ending up with a ton of "if" lines and a lot of typing.  So I was hoping their would be a quicker way of doing it.
0
 
clifABBCommented:
Not necessarily quicker, but less coding:

Using the type and array:
Type MyType
  Key As String
  Value As String
End Type
Dim MyArray(10) As MyType
MyArray(1).Key = "String 1"
MyArray(1).Value = "Text 1"
MyArray(2).Key = "String 2"
MyArray(2).Value = "Text 2"
'etc

Then to select the correct array element:
For nCnt1 = 0 To tree.Nodes.Count - 1
  If tree.Nodes(nCnt1).Selected Then
    For nCnt2 = LBound(MyArray) to UBound(MyArray)
      If MyArray(nCnt2).Key = tree.Nodes(nCnt1).Key Then
        txtInfo.Text = MyArray(nCnt2).Value
        Exit For
      End If
    Next nCnt2
  End If
Next nCnt1
0
 
dokkenAuthor Commented:
Cool, that worked... a lot less typing too :)  post something as an answer so I can give you credit.
0
 
clifABBCommented:
Thanks.  :)


Using the type and array:
Type MyType
  Key As String
  Value As String
End Type
Dim MyArray(10) As MyType
MyArray(1).Key = "String 1"
MyArray(1).Value = "Text 1"
MyArray(2).Key = "String 2"
MyArray(2).Value = "Text 2"
'etc

Then to select the correct array element:
For nCnt1 = 0 To tree.Nodes.Count - 1
  If tree.Nodes(nCnt1).Selected Then
    For nCnt2 = LBound(MyArray) to UBound(MyArray)
      If MyArray(nCnt2).Key = tree.Nodes(nCnt1).Key Then
        txtInfo.Text = MyArray(nCnt2).Value
        Exit For
      End If
    Next nCnt2
  End If
Next nCnt1
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

Join & Write a Comment

Featured Post

Get your problem seen by more experts

Be seen. Boost your question’s priority for more expert views and faster solutions

  • 4
  • 3
Tackle projects and never again get stuck behind a technical roadblock.
Join Now