This example illustrates the implementation of File API with GraphQL Server pattern. The best example for GraphQL Server basic usage can be found here: https://github.com/graphcool/graphql-server-example .
graphcool deploy # choose local cluster
# copy API endpoint into the `PRISMA_ENPOINT` env var in .envTo get GRAPHCOOL_SECRET visit http://jwtbuilder.jamiekurtz.com and scroll to the bottom where you can hash your secret from graphcool.yml and get the hashed output. (sssh is used in the example.)
- Head over to the AWS console and navigate to the
S3section. - Click create bucket and follow the instructions on screen.
- Once you have created a
bucket, add bucket name that you've picked to .envS3_BUCKETproperty. - Head back to the AWS Console and open
Identity and Access Management (IAM)console. Navigate toUsersand clickAdd user. - Under Access type check Programmatic access and press
Next. From options, select Attach existing policies directly and a table below will open. Search for AmazonS3FullAccess and check it. PressNextto review everything and submit by pressingCreate user. - Once done, copy the Access key ID to .env
S3_KEYproperty and Secret access key to .envS3_SECRETproperty. - You are all set to start the server!
yarn install
yarn start
# Open http://localhost:5000/You can upload files to a project by doing a multipart/form-data HTTP request to the File API http://localhost:5000/upload.
It's important to use the form parameter data as seen in the example below.
Everytime you upload a file to Prisma, a new File node is created that contains information about that file.
id: the familiar system fieldsecret: a unique, unguessable secret that allows access to the filename: the file namesize: the file sizeurl: the url of the file where it can be accessed. The url contains of the project id and the filesecret, so is unguessable as well.contentType: the contentType of the file. It is determined based on the file name (extension in the name is required!).
If you want to connect the File node to another node in a relation, you can use the id in the response.
With curl you could execute:
curl -X POST 'http://localhost:5000/upload' -F "data=@example.png; filename=coolimage.png"
This uploads the local file example.png under coolimage.png name. The response could look something like this:
[{
"id": "cjbqvp4ii00390181b1q0dq6h",
"name": "coolimage.png",
"secret": "43de4b08-78b2-4b5c-a5b7-05ee350ee09a",
"contentType": "image/png",
"size": 36625,
"url": "https://__S3_BUCKET__.s3-eu-west-1.amazonaws.com/43de4b08-78b2-4b5c-a5b7-05ee350ee09a"
}]If there's no filename provided, the original name of the file is used instead.
MIT