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return uploaded files information in s3 example

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Rob Ede
2022-08-01 03:19:39 +01:00
parent 4840cfdb68
commit c8a4544ea1
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31
forms/multipart-s3/README.md
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@ -1,7 +1,6 @@
# Multipart + AWS S3
Upload a file in multipart form to AWS S3 using [AWS S3 SDK](https://crates.io/crates/aws-sdk-s3).
Receive multiple data in multipart form in JSON format and receive it as a struct.
Upload a file in multipart form to AWS S3 using [AWS S3 SDK](https://crates.io/crates/aws-sdk-s3).
# Usage
@ -9,13 +8,31 @@ Receive multiple data in multipart form in JSON format and receive it as a struc
cd forms/multipart-s3
```
1. copy .env.example .env
1. edit .env AWS_ACCESS_KEY_ID=your_key
1. edit .env AWS_SECRET_ACCESS_KEY=your_key
1. edit .env AWS_S3_BUCKET_NAME=your_chosen_region
1. copy `.env.example` to `.env`
1. edit `.env` key `AWS_REGION` = your_bucket_region
1. edit `.env` key `AWS_ACCESS_KEY_ID` = your_key_id
1. edit `.env` key `AWS_SECRET_ACCESS_KEY` = your_key_secret
1. edit `.env` key `AWS_S3_BUCKET_NAME` = your_bucket_name
```sh
cargo run
```
<http://localhost:8080>
Go to <http://localhost:8080> in you browser.
Or, start the upload using [HTTPie]:
```sh
http --form POST :8080/ file@Cargo.toml
http --form POST :8080/ file@Cargo.toml file@README.md meta='{"namespace":"foo"}'
```
Or, using cURL:
```sh
curl -X POST http://localhost:8080/ -F 'file=@Cargo.toml'
curl -X POST http://localhost:8080/ -F 'file=@Cargo.toml' -F 'file=@README.md' -F 'meta={"namespace":"foo"}'
```
[httpie]: https://httpie.org
[curl]: https://curl.haxx.se