Chloe 5385b5ed67
implement upload form functions
Signed-off-by: Chloe <pinkcloudvnn@gmail.com>
2024-06-25 12:17:55 +07:00

43 lines
1020 B
TypeScript

import { StagedUploadsCreatePayload } from 'lib/shopify/types';
import { createStagedUploadFiles, onUploadFile } from './actions';
export const prepareFilePayload = ({
stagedFileUpload,
file
}: {
stagedFileUpload: StagedUploadsCreatePayload;
file: File;
}) => {
const formData = new FormData();
const url = stagedFileUpload.url;
stagedFileUpload.parameters.forEach(({ name, value }) => {
formData.append(name, value);
});
formData.append('file', file);
return { url, formData };
};
export const handleUploadFile = async ({ file }: { file: File }) => {
if (!file) return;
const stagedTarget = await createStagedUploadFiles({
filename: file.name,
fileSize: String(file.size),
httpMethod: 'POST',
resource: 'FILE',
mimeType: file.type
});
if (stagedTarget) {
const data = prepareFilePayload({ file, stagedFileUpload: stagedTarget });
await onUploadFile({
...data,
fileName: file.name,
resourceUrl: stagedTarget.resourceUrl
});
}
};