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VB 6.0 - OpenFile Procedure
Message
De
13/11/1998 16:48:15
 
 
Information générale
Forum:
Visual Basic
Catégorie:
Autre
Divers
Thread ID:
00157504
Message ID:
00157566
Vues:
19
Joe,

I am trying to OPEN THE FILE when I click the command button or VIEWING what's in the text file. You are just storing the path of the text file.

Any other suggestions?

Thank you again

>>Can somebody tell me what's wrong with this code? I get an error message about the OPENFILE procedure (Sub or Function not defined) when I put it in the click event of a command button.
>>
>>CommonDialog1.Filter = "Text Files|*.txt"
>>CommonDialog1.ShowOpen
>>OpenFile (CommonDialog1.Filename)
>>
>>
>>Thank you in advance
>
>
>It should read "Text Files (*.txt) | *.txt"
>or for more options "Text Files (*.txt;*.text;*.bak) | *.txt;*.text;*.bak"
>
>below is an actual clip from my proj:
>'** Start OF SOURCE'
> CommonDialog1.Filter = _
> "Image Files (*.jpg;*.gif;*.jpeg;*.bmp;*.tiff;*.tif)|*.jpg;*.gif;*.*.jpeg;*.bmp;*.tiff;*.tif|" & _
> "All files (*.*)|*.*" ' *.* just in case they have some freaky format we need to allow.
> ' Specify default filter
> CommonDialog1.FilterIndex = 0
> ' Display the Open dialog box
> CommonDialog1.ShowOpen
> ' Display name of selected file
>
> txtSideBG.Text = CommonDialog1.FileName
>'** END OF SOURCE'
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