Defer formatting when only one error is returned #33
Conversation
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
|
|
mkatychev
changed the title
mkatychev
force-pushed
the
patch/single_error_format
branch
from
67058da
to
b660ff8
Compare
|
Just a note that while I would also like this behavior, it changes the output of the library and thus may break any code depending on the output to be stable. As a result, I doubt we will end up accepting it at this time unless we want to consider a major rev bump. Pinging @mitchellh for opinions. |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Add this suggestion to a batch that can be applied as a single commit.
This suggestion is invalid because no changes were made to the code.
Suggestions cannot be applied while the pull request is closed.
Suggestions cannot be applied while viewing a subset of changes.
Only one suggestion per line can be applied in a batch.
Add this suggestion to a batch that can be applied as a single commit.
Applying suggestions on deleted lines is not supported.
You must change the existing code in this line in order to create a valid suggestion.
Outdated suggestions cannot be applied.
This suggestion has been applied or marked resolved.
Suggestions cannot be applied from pending reviews.
Suggestions cannot be applied on multi-line comments.
Suggestions cannot be applied while the pull request is queued to merge.
Suggestion cannot be applied right now. Please check back later.
Testing for 1 error in unit tests and emitting only one error makes the
\n\t*prepend impractical and noisy.There should be no need to do any multiple error logic until more than one error is present.
When
multierror.Error.Error()is called withlen(multierror.Error.Errors) == 1, the formatting should be deferred (gracefully decompose) to the one and only error present:multierror.Error[0].Error()