MS Access VBA get result of WinSCP transfer
1
I have code that uploads a folder to an FTP server using WinSCP.
Dim strSFTPDir As String
strSFTPDir = "C:Program Files (x86)WinSCP"
Dim UploadsDir As String
UploadsDir = DLookup("PathToPics", "Admin") & "Uploads*"
Call Shell(strSFTPDir & "winscp.com /command ""open ftp://***/"" ""put -filemask=|*.ldb """"" & UploadsDir & """"" "" ""exit"" ", vbNormalFocus)
However I need to let Access know whether it's succeeded or failed.
How can I do that?
vba ms-access ftp access-vba winscp
edited Nov 28 '18 at 10:37
Martin Prikryl
90.4k22179384
asked Nov 28 '18 at 10:27
Tristan KingTristan King
346
add a comment |
1
I have code that uploads a folder to an FTP server using WinSCP.
Dim strSFTPDir As String
strSFTPDir = "C:Program Files (x86)WinSCP"
Dim UploadsDir As String
UploadsDir = DLookup("PathToPics", "Admin") & "Uploads*"
Call Shell(strSFTPDir & "winscp.com /command ""open ftp://***/"" ""put -filemask=|*.ldb """"" & UploadsDir & """"" "" ""exit"" ", vbNormalFocus)
However I need to let Access know whether it's succeeded or failed.
How can I do that?
vba ms-access ftp access-vba winscp
edited Nov 28 '18 at 10:37
Martin Prikryl
90.4k22179384
asked Nov 28 '18 at 10:27
Tristan KingTristan King
346
add a comment |
1
1
1
1
I have code that uploads a folder to an FTP server using WinSCP.
Dim strSFTPDir As String
strSFTPDir = "C:Program Files (x86)WinSCP"
Dim UploadsDir As String
UploadsDir = DLookup("PathToPics", "Admin") & "Uploads*"
Call Shell(strSFTPDir & "winscp.com /command ""open ftp://***/"" ""put -filemask=|*.ldb """"" & UploadsDir & """"" "" ""exit"" ", vbNormalFocus)
However I need to let Access know whether it's succeeded or failed.
How can I do that?
vba ms-access ftp access-vba winscp
edited Nov 28 '18 at 10:37
Martin Prikryl
90.4k22179384
asked Nov 28 '18 at 10:27
Tristan KingTristan King
346
I have code that uploads a folder to an FTP server using WinSCP.
Dim strSFTPDir As String
strSFTPDir = "C:Program Files (x86)WinSCP"
Dim UploadsDir As String
UploadsDir = DLookup("PathToPics", "Admin") & "Uploads*"
Call Shell(strSFTPDir & "winscp.com /command ""open ftp://***/"" ""put -filemask=|*.ldb """"" & UploadsDir & """"" "" ""exit"" ", vbNormalFocus)
However I need to let Access know whether it's succeeded or failed.
How can I do that?
vba ms-access ftp access-vba winscp
vba ms-access ftp access-vba winscp
edited Nov 28 '18 at 10:37
Martin Prikryl
90.4k22179384
asked Nov 28 '18 at 10:27
Tristan KingTristan King
346
edited Nov 28 '18 at 10:37
Martin Prikryl
90.4k22179384
asked Nov 28 '18 at 10:27
Tristan KingTristan King
346
edited Nov 28 '18 at 10:37
Martin Prikryl
90.4k22179384
edited Nov 28 '18 at 10:37
Martin Prikryl
90.4k22179384
edited Nov 28 '18 at 10:37
Martin Prikryl
90.4k22179384
90.4k22179384
asked Nov 28 '18 at 10:27
Tristan KingTristan King
346
asked Nov 28 '18 at 10:27
Tristan KingTristan King
346
asked Nov 28 '18 at 10:27
Tristan KingTristan King
346
346
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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WinSCP indicates a result with its exit code.
See WinSCP FAQ How do I know that script completed successfully?
So run WinSCP and make sure it returns 0 (success).
See VBA Shell and Wait with Exit Code
answered Nov 28 '18 at 10:37
Martin PrikrylMartin Prikryl
90.4k22179384
Thanks for the answer, but the code in that link is long and confusing and I'm not sure what parts exactly I need? Does my original Call Shell() code need to be changed entirely?
– Tristan King
Nov 28 '18 at 13:33
You need to replace yourShellcall with what theRunApplicationfunction does in the linked code. There's no easier way in VBA. - If you do not explain us, what you find "confusing" about the code, we cannot help you on that.
– Martin Prikryl
Nov 28 '18 at 13:52
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
WinSCP indicates a result with its exit code.
See WinSCP FAQ How do I know that script completed successfully?
So run WinSCP and make sure it returns 0 (success).
See VBA Shell and Wait with Exit Code
answered Nov 28 '18 at 10:37
Martin PrikrylMartin Prikryl
90.4k22179384
Thanks for the answer, but the code in that link is long and confusing and I'm not sure what parts exactly I need? Does my original Call Shell() code need to be changed entirely?
– Tristan King
Nov 28 '18 at 13:33
You need to replace yourShellcall with what theRunApplicationfunction does in the linked code. There's no easier way in VBA. - If you do not explain us, what you find "confusing" about the code, we cannot help you on that.
– Martin Prikryl
Nov 28 '18 at 13:52
add a comment |
2
WinSCP indicates a result with its exit code.
See WinSCP FAQ How do I know that script completed successfully?
So run WinSCP and make sure it returns 0 (success).
See VBA Shell and Wait with Exit Code
answered Nov 28 '18 at 10:37
Martin PrikrylMartin Prikryl
90.4k22179384
Thanks for the answer, but the code in that link is long and confusing and I'm not sure what parts exactly I need? Does my original Call Shell() code need to be changed entirely?
– Tristan King
Nov 28 '18 at 13:33
You need to replace yourShellcall with what theRunApplicationfunction does in the linked code. There's no easier way in VBA. - If you do not explain us, what you find "confusing" about the code, we cannot help you on that.
– Martin Prikryl
Nov 28 '18 at 13:52
add a comment |
2
2
2
WinSCP indicates a result with its exit code.
See WinSCP FAQ How do I know that script completed successfully?
So run WinSCP and make sure it returns 0 (success).
See VBA Shell and Wait with Exit Code
answered Nov 28 '18 at 10:37
Martin PrikrylMartin Prikryl
90.4k22179384
WinSCP indicates a result with its exit code.
See WinSCP FAQ How do I know that script completed successfully?
So run WinSCP and make sure it returns 0 (success).
See VBA Shell and Wait with Exit Code
answered Nov 28 '18 at 10:37
Martin PrikrylMartin Prikryl
90.4k22179384
edited Nov 28 '18 at 11:04
answered Nov 28 '18 at 10:37
Martin PrikrylMartin Prikryl
90.4k22179384
answered Nov 28 '18 at 10:37
Martin PrikrylMartin Prikryl
90.4k22179384
answered Nov 28 '18 at 10:37
Martin PrikrylMartin Prikryl
90.4k22179384
90.4k22179384
Thanks for the answer, but the code in that link is long and confusing and I'm not sure what parts exactly I need? Does my original Call Shell() code need to be changed entirely?
– Tristan King
Nov 28 '18 at 13:33
You need to replace yourShellcall with what theRunApplicationfunction does in the linked code. There's no easier way in VBA. - If you do not explain us, what you find "confusing" about the code, we cannot help you on that.
– Martin Prikryl
Nov 28 '18 at 13:52
add a comment |
Thanks for the answer, but the code in that link is long and confusing and I'm not sure what parts exactly I need? Does my original Call Shell() code need to be changed entirely?
– Tristan King
Nov 28 '18 at 13:33
You need to replace yourShellcall with what theRunApplicationfunction does in the linked code. There's no easier way in VBA. - If you do not explain us, what you find "confusing" about the code, we cannot help you on that.
– Martin Prikryl
Nov 28 '18 at 13:52
Thanks for the answer, but the code in that link is long and confusing and I'm not sure what parts exactly I need? Does my original Call Shell() code need to be changed entirely?
– Tristan King
Nov 28 '18 at 13:33
Thanks for the answer, but the code in that link is long and confusing and I'm not sure what parts exactly I need? Does my original Call Shell() code need to be changed entirely?
– Tristan King
Nov 28 '18 at 13:33
You need to replace your
Shell call with what the RunApplication function does in the linked code. There's no easier way in VBA. - If you do not explain us, what you find "confusing" about the code, we cannot help you on that.– Martin Prikryl
Nov 28 '18 at 13:52
You need to replace your
Shell call with what the RunApplication function does in the linked code. There's no easier way in VBA. - If you do not explain us, what you find "confusing" about the code, we cannot help you on that.– Martin Prikryl
Nov 28 '18 at 13:52
add a comment |
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