When I call cdo some-dir echo --version I would expect to see --version as the output but instead, I see cdo's version output. This is true for all the following:
$ cdo some-dir echo --version
cdo 0.1.0
$ cdo some-dir echo -V
cdo 0.1.0
$ cdo some-dir echo --help
Runs a command in the given directory
Usage: cdo <DIRECTORY> <COMMAND> [ARGS]...
Arguments:
<DIRECTORY> The directory to run the command in
<COMMAND> The command to run
[ARGS]... The arguments to pass to the command
Options:
-h, --help Print help
-V, --version Print version
$ cdo some-dir echo -h
Runs a command in the given directory
Usage: cdo <DIRECTORY> <COMMAND> [ARGS]...
Arguments:
<DIRECTORY> The directory to run the command in
<COMMAND> The command to run
[ARGS]... The arguments to pass to the command
Options:
-h, --help Print help
-V, --version Print version
What's interesting is that when I pass a flag before one of those, it behaves as expected. So this works fine:
$ cdo src echo --foo --version
--foo --version
When I call
cdo some-dir echo --versionI would expect to see--versionas the output but instead, I seecdo's version output. This is true for all the following:What's interesting is that when I pass a flag before one of those, it behaves as expected. So this works fine:
$ cdo src echo --foo --version --foo --version