Execute Command for 7-zip command line application
Posted: Thu Apr 30, 2020 11:13 am
I am trying Execute Command for 7-zip command line application.
1. So I write a batch file with direct file names as below first.
cd "C:\01-Job Folder\7za920"
7za a -t7z test-a.7z readme.txt
It can work when I run in Windows 10, the "readme.txt" file can be zip with filename "test-a.7z"
2. Then I re-write the batch file for Execute Command use.
set arg1=&1
set arg2=&2
cd "C:\01-Job Folder\7za920"
7za a -t7z %arg2% %arg1%
(In the 7z command, the 7-zip file name comes first, then the source file name after. So I guess it is "%arg2" first, then "%arg1")
3. In the Execute Command Setting,
Execute Mode: Concurrent
Command or Path: (batch file location)
Argument": "%2" "%1"
Output: File at Path
3. When I run the flow, the incoming file has been put into Problem jobs. So may I ask how to correct this?
Many Thanks in advance.
1. So I write a batch file with direct file names as below first.
cd "C:\01-Job Folder\7za920"
7za a -t7z test-a.7z readme.txt
It can work when I run in Windows 10, the "readme.txt" file can be zip with filename "test-a.7z"
2. Then I re-write the batch file for Execute Command use.
set arg1=&1
set arg2=&2
cd "C:\01-Job Folder\7za920"
7za a -t7z %arg2% %arg1%
(In the 7z command, the 7-zip file name comes first, then the source file name after. So I guess it is "%arg2" first, then "%arg1")
3. In the Execute Command Setting,
Execute Mode: Concurrent
Command or Path: (batch file location)
Argument": "%2" "%1"
Output: File at Path
3. When I run the flow, the incoming file has been put into Problem jobs. So may I ask how to correct this?
Many Thanks in advance.