Determining the required column width of a ListView control

[Geert-Pieter Hof]

Hello,

I use a ListView control (report view) to show the user a summary of error
messages. The width of the column that holds the error messages is such that
no horizontal scrollbar is needed. However, in some cases, the length of the
error message text is wider than the width of the column. In this case, the
text is hyphened with three dots.

I want to check the width of the error message text and if necessary,
increase the width of the ListView column. By trial and error, I know that
the amount of text that fits in a column of width 5620 twips is exactly
"XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX ".

So, I created the following piece of code (the string Message holds a
certain error message):

With Printer

If .TextWidth(Message) >
..TextWidth("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXX") Then

ListView1.ColumnHeaders.Item(3).Width = .TextWidth(Message) /
..TextWidth("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXX") * 5620

End If

End With

Although this code functions properly, I don't think it's an outstanding
solution. Does somebody knows a better piece of code?

Thanks in advance,

Geert-Pieter

[Asperamanca]

"Geert-Pieter Hof" <g.*****@wbmt.tudelft.nl.blablabla> wrote in message news:<bg**********@news.tudelft.nl>...

Hello,

I use a ListView control (report view) to show the user a summary of error
messages. The width of the column that holds the error messages is such that
no horizontal scrollbar is needed. However, in some cases, the length of the
error message text is wider than the width of the column. In this case, the
text is hyphened with three dots.

I want to check the width of the error message text and if necessary,
increase the width of the ListView column. By trial and error, I know that
the amount of text that fits in a column of width 5620 twips is exactly
"XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX ".

So, I created the following piece of code (the string Message holds a
certain error message):

With Printer

If .TextWidth(Message) >
.TextWidth("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXX") Then

ListView1.ColumnHeaders.Item(3).Width = .TextWidth(Message) /
.TextWidth("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXX") * 5620

End If

End With

Although this code functions properly, I don't think it's an outstanding
solution. Does somebody knows a better piece of code?

Thanks in advance,

Geert-Pieter

You can use the TextWidth function of some controls to "test" the true
width of a certain text. Don't forget to assign the same font that you
currently use in the listview.
I use this in a loop over all my texts of a column, then select the
widest text, add some pixels for good measure, and resize the column
to that.

Robert

[kharma]

Private Declare Function SendMessage Lib "user32" Alias _
"SendMessageA" (ByVal hWnd As Long, ByVal wMsg As Long, _
ByVal wParam As Long, lParam As Any) As Long

Private Const LVM_FIRST = &H1000
Public Sub LV_AutoSizeColumn(LV As ListView, Optional Column _
As ColumnHeader = Nothing)

Dim C As ColumnHeader
If Column Is Nothing Then
For Each C In LV.ColumnHeaders
SendMessage LV.hWnd, LVM_FIRST + 30, C.Index - 1, -1
Next
Else
SendMessage LV.hWnd, LVM_FIRST + 30, Column.Index - 1, -1
End If
LV.Refresh
End Sub