下载然后上传图像而不存储它
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【中文标题】下载然后上传图像而不存储它【英文标题】:Download then upload image without storing it 【发布时间】:2021-10-20 17:28:52 【问题描述】:我目前正在尝试在 Zapier 中构建图像上传组件,但在使其正常工作时遇到问题。我必须能够下载图像,然后将其发布到新端点而不将其存储在本地。目前,我能做的最接近的方法是将 IncomingMessage 发送到 POST,但我知道这是不对的。
有人有什么建议吗?
let FormData = require('form-data');
let http = require('https');
const makeDownloadStream = (url) =>
new Promise((resolve, reject) =>
http.request(url, resolve).on('error', reject).end();
);
const makeUploadStream = (z, bundle, options) =>
var imageRequest = options;
const promise = z.request(imageRequest);
return promise.then((response) =>
return response.data;
);
const addAttachment = async (z, bundle) =>
/*var request =
'url': bundle.inputData.attachment
;
const promiseAt = z.request(request);
return promiseAt.then((stream) => */
const form = new FormData();
var data = `"type": "records", "attributes": "form_id": $bundle.inputData.form_id`
const stream = await makeDownloadStream(bundle.inputData.attachment);
form.append(`field_$bundle.inputData.field_id`, stream);
form.append('data', data);
var request =
'url': bundle.inputData.url,
'method': 'PUT',
'headers':
'Content-Type': `multipart/form-data; boundary=$form.getBoundary()`
,
'body': form
;
const response = await makeUploadStream(z, bundle, request);
return response;
//);
【问题讨论】:
【参考方案1】:我自己想通了。对于任何需要在 Zapier 上上传图片的人,这里是:
let FormData = require('form-data');
const makeDownloadStream = (z, bundle) =>
var imageRequest =
'url': bundle.inputData.attachment,
'method': 'GET',
'raw': true
;
const promise = z.request(imageRequest);
return promise.then(async (response) =>
var buffer = await response.buffer();
return
'content-type': response.headers.get('content-type'),
'content': buffer,
'filename': response.headers.get('content-disposition').replace('attachment; filename="', '').replace('"', '')
);
const addAttachment = async (z, bundle) =>
const form = new FormData();
const content = await makeDownloadStream(z, bundle);
form.append(`field_$bundle.inputData.field_id`, Buffer.from(content.content.toString('binary'), 'binary'),
filename: content.filename
);
const request =
'url': `$bundle.inputData.url/api/records/$bundle.inputData.record_id`,
'method': 'PUT',
'headers':
'Content-Type': `multipart/form-data; boundary=$form.getBoundary()`,
'Content-Length': form.getLengthSync()
,
'body': form
;
const promise = z.request(request);
return promise.then((response) =>
return response.data;
);
【讨论】:
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