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From: SDS2 on 11 May 2010 15:00
The System.Diagnostics.Process.Start
works perfectly!

Thanks a lot!


"SDS2" <nomailtoSDS(a)collegewaregemdeletethis.com> schreef in bericht
news:eavtktT8KHA.1316(a)TK2MSFTNGP02.phx.gbl...
> Isn't there a method that is able to open any path? (an application?)
>
>
> "Family Tree Mike" <FamilyTreeMike(a)ThisOldHouse.com> schreef in bericht
> news:uBgZ3KJ8KHA.3880(a)TK2MSFTNGP04.phx.gbl...
>> On 5/10/2010 6:24 PM, Patrick A wrote:
>>> Stan,
>>>
>>> Do you want to open a file while you're at it?
>>>
>>> Here's what I'm doing...not sure it's the best method.
>>>
>>> Imports Microsoft.Office.Interop
>>>
>>> Private Sub ReportsToolStripMenuItem_Click(ByVal sender As
>>> System.Object, ByVal e As System.EventArgs) Handles
>>> ReportsToolStripMenuItem.Click
>>> Dim MyAccess As Access.Application
>>> MyAccess = GetObject("P:\Production Resources\Legal55\Reports
>>> \Timekeeper Reports.mdb", "Access.Application")
>>> MyAccess.Visible = True
>>> End Sub
>>>
>>> Patrick
>>
>> If you just want to launch access (or word) with a given file, then you
>> should use:
>>
>> dim p as new Process()
>> p.StartInfo.FileName = "somefilename.mdb" ' or "somefilename.doc"
>> p.StartInfo.Verb = "Open"
>> p.Start()
>>
>> --
>> Mike
>

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