fix: csharp get file name from content disposition#17183
Merged
wing328 merged 2 commits intoOpenAPITools:masterfrom Nov 27, 2023
Merged
fix: csharp get file name from content disposition#17183wing328 merged 2 commits intoOpenAPITools:masterfrom
wing328 merged 2 commits intoOpenAPITools:masterfrom
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Contributor
Author
|
@wing328 I appreciate the extra details you added. |
Member
|
@fghpdf thanks for the PR. Have you tested the fix locally to confirm it works for your use cases? |
Contributor
Author
Sure, I tried this case locally: [Fact]
public void UpdatePetTest()
{
FileParameter file = instance.DownloadFile(Guid.NewGuid());
_output.WriteLine(file.Name);
Assert.Equal("example.txt", file.Name);
}And this PR can solve this issue. |
Member
|
ok. let's give it a try |
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Summary
Download a file can't get the file name from the header, content-disposition.
The generator will generate the
FileParameterclass, and the code doesn't handle this type.And also, according to RFC 2616, the HTTP headers are case insensitive.
OpenAPI Example
OpenAPI CLI Command
Expect
Get file name:
example.txtActual
Get file name: 'no_name_provided'

Server
If you want to have a quick test, you can use this Golang code:
PR checklist
Commit all changed files.
This is important, as CI jobs will verify all generator outputs of your HEAD commit as it would merge with master.
These must match the expectations made by your contribution.
You may regenerate an individual generator by passing the relevant config(s) as an argument to the script, for example
./bin/generate-samples.sh bin/configs/java*.IMPORTANT: Do NOT purge/delete any folders/files (e.g. tests) when regenerating the samples as manually written tests may be removed.
master(upcoming 7.1.0 minor release - breaking changes with fallbacks),8.0.x(breaking changes without fallbacks)