ZIP file not work

[sal21]

Please check my shell string:

dim file as tring, retval as avriant
FILE="tipo.xls"
retval = Shell("C:\Program Files\Winzip\WINZIP32.EXE -min -a " & "'C:\TEST\" & Replace(FILE, ".xls", "") & ".zip C:\TEST\'" & FILE)

[HansV]

Try changing

dim file as tring, retval as avriant

to

Dim file As String, retval As Variant

[sal21]

Try changing

dim file as tring, retval as avriant

to

Dim file As String, retval As Variant

ok changed...
but the erro is on the shell command i think is a quote and single quote and double quote not correct....

[HansV]

Try

Code: Select all

    retval = Shell("C:\Program Files\Winzip\WINZIP32.EXE -min -a ""C:\TEST\" & _
        Replace(FILE, ".xls", ".zip") & """ ""C:\TEST\" & FILE & """")

[sal21]

Try

Code: Select all

    retval = Shell("C:\Program Files\Winzip\WINZIP32.EXE -min -a ""C:\TEST\" & _
        Replace(FILE, ".xls", ".zip") & """ ""C:\TEST\" & FILE & """")

now on C:\... wotk perfect.

But i need to use a server dir!
And in a server dir are a blank spce in the dir:

similar:
\\server\dir 021 14\14 23\...

exists a method to eleiminate this error!!!!!

in debug print the shell command get only the first part of server string:
\\server\dir\

[HansV]

The following doesn't work?

Code: Select all

    retval = Shell("C:\Program Files\Winzip\WINZIP32.EXE -min -a ""\\server\dir 021 14\14 23\" & _
        Replace(FILE, ".xls", ".zip") & """ ""\\server\dir 021 14\14 23\" & FILE & """")

[sal21]

The following doesn't work?

Code: Select all

    retval = Shell("C:\Program Files\Winzip\WINZIP32.EXE -min -a ""\\server\dir 021 14\14 23\" & _
        Replace(FILE, ".xls", ".zip") & """ ""\\server\dir 021 14\14 23\" & FILE & """")

ops!
The last dir in server path is a var!!!

Similar:

"\\server\dir 021 14\14 23\" & vardir

How to modify the your last string?

[HansV]

You should know how to do that by now, Sal.