I have following issue when I used the following command on visual studio cmd window. also I have visual studio 2015 installed on windows 7
asked 01 Dec ‘15, 04:58 Tony_2013 edited 01 Dec ‘15, 05:19 grahamb ♦ |
One Answer:
You have specified Visual Studio 2013 in the CMake generator parameter `-G Visual Studio 12 Win64" and CMake couldn't find it. For VS2015, which isn't currently supported by the Wireshark build, you would need to pass the parameter You can see all the CMake generators supported by passing a plain answered 01 Dec '15, 05:09 grahamb ♦ |
Thanks for coming back
Now I have following issue after I went through as you have suggested.
Please kindly advise me.
It appears that you didn’t follow the Developers Guide correctly. Section 2.3.2 asks you to:
From the error you show, it looks like the env vars aren’t set, and from the path in the prompt it looks like you haven’t created a build directory. Instead you’re attempting to run from the VC directory.
What have you set for the required env vars?
I did set following vars as it shown below.
Now i have changed the directory before I execute generate cmd. But I still get the following error.
The PowerShell script win-setup.ps1 did not like the Destination argument of “C:\Development”.
This argument is generated by CMake from the WIRESHARK_BASE_DIR and WIRESHARK_LIB_DIR env vars. If you have defined WIRESHARK_LIB_DIR that value takes priority and defines the absolute path to the location for the 3rd party libs. If you haven’t set WIRESHARK_LIB_DIR, then you must define WIRESHARK_BASE_DIR and CMake will add on the appropriate lib path, in your case
wireshark-win64-libs
. The win-setup.ps1 script checks the argument to ensure it has the form\wireshark--libs
, and as can be seen from your output, it doesn’t.I suspect you have defined the env var WIRESHARK_LIB_DIR and set it to “C:\Development”. If this is the case, then undefine it.
Arguably win-setup.ps1 is being a bit pernickety in “requiring” the lib path to be of the form"
\wireshark--libs
, a patch to relax that would be considered.