Thanks buddy for the response.
But I have tried all the possible combinations with w and y variables.
var w = vars.filename;
execCommand("zip '" + w + "'.zip '" + "/" + y + "'");
execCommand("zip '" + w + "'.zip '" + "/" + w + "'");
execCommand("zip '" + y + "'.zip '" + "/" + w + "'");
execCommand("zip '" + y + "'.zip '" + "/" + y + "'");
In the above command "y" represent the the filename only)
We checked the folder where we are trying to place the zip file but its not there.
var w = vars.filename; //file path + file name: /usr/local/neolane/outgoing/TEST_ABCI_00000100_20171025193028.txt
var x = w.split("/");
var y = x[5]; //fileName: TEST_ABCI_00000100_20171025193028.txt
So it would be helpful if we can have some another set of code for zipping the files or updates in this command.Amit Kumar Adhiyan, Could you guys please have a look over my issue and help me out.