Conda channels

Conda-build supports standard conda channel behavior.

Identical channel and package name problem

If the channel and package name are identical, it's possible to encounter a build problem if the short channel name is used.

Let's say your Anaconda.org username or an organization name is example. And suppose you created a package example, whose files' layout is similar to:

setup.py
conda/meta.yaml
example/

If your build depends on some other packages inside your channel, you will need to add -c example, however, the following code:

conda-build ./conda/ -c example

will fail with the following error message (the path will be different):

requests.exceptions.HTTPError: 404 Client Error: None for url:
file:///path/to/your/local/example/noarch/repodata.json
[...]
The remote server could not find the noarch directory for the requested channel with
url: file:///path/to/your/local/example/noarch/repodata.json
[...]
As of conda 4.3, a valid channel must contain a `noarch/repodata.json` and
associated `noarch/repodata.json.bz2` file, even if `noarch/repodata.json`
is empty. please request that the channel administrator create
`noarch/repodata.json` and associated `noarch/repodata.json.bz2` files.

This happens because conda-build will consider the directory ./example/ in your project as a channel. This is by design due to conda's CI servers, where the build path can be long, complicated, and not predictable prior to build.

There are several ways to resolve this issue.

1. Use the URL of the desired channel:

   conda-build ./conda/ -c https://conda.anaconda.org/example/
   

2. Run the build from inside the conda recipe directory:

   cd conda
   conda-build . -c example
   

3. Use the label specification workaround:

conda-build ./conda/ -c example/label/main

which technically is the same as -c example, since main is the default label, but now it won't mistakenly find a channel example/label/main on the local filesystem.