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From: SDS2 on 9 May 2010 11:38
Hey,

I was wondering if i could create a program in Visual Studio 2008 with a
link to an executable of an application (like MS WORD) in it.

By pressing a button, the executable of the application should start.

I can't manage to find out the right code to do so.


Can anyone help me out?



Thanks in advance.

Stan

From: Family Tree Mike on 9 May 2010 11:55
On 5/9/2010 11:38 AM, SDS2 wrote:
> Hey,
>
> I was wondering if i could create a program in Visual Studio 2008 with a
> link to an executable of an application (like MS WORD) in it.
>
> By pressing a button, the executable of the application should start.
>
> I can't manage to find out the right code to do so.
>
>
> Can anyone help me out?
>
>
>
> Thanks in advance.
>
> Stan


Look at System.Diagnostics.Process.Start("some path to an exe");

--
Mike
From: Patrick A on 10 May 2010 18:24
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
From: Family Tree Mike on 10 May 2010 18:47
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
From: SDS2 on 11 May 2010 14:54
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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