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VisualizaçõesHow can I duplicate a file (completely) but change it's name?
So I need to generate the smaller previews of the image files that will be uploaded and I have to append "_preview" at the end of each file name.
Currently I'm doing this:
uploadFile.map((file) => {
if (file.type.includes('image')) {
console.log('Generating thumbnail for ' + file.name)
const fileName = file.name.split('.').slice(0, -1).join('.')
const fileExtension = file.name.split('.').pop()
const compressedFile = new File(
[file.slice(0, file.size, file.type)],
fileName + '_preview.' + fileExtension,
)
console.log('Generated file:', compressedFile)
convert({
file: compressedFile,
width: 300,
height: 300,
type: fileExtension,
})
.then((resp) => {
uploadFile.push(resp)
})
.catch((error) => {
// Error
console.error('Error compressing ', file.name, '-', error)
})
}
})
The problem is that "compressedFile" is missing some fields which were present in the original file and so the convert functoin throws the error "File type not supported". As you can see "type" and "webkitRelativePath" are not copied.
Can anybody suggest how I can retain all the information from the original file and just append _preview at the end of file name?
2 Respostas
Responde à pergunta0
I realized File API provides an option to pass "options" object as well which can specify the file type. For instance:
const file = new File(["foo"], "foo.txt", {
type: "text/plain",
});
Source: https://developer.mozilla.org/en-US/docs/Web/API/File/File
0
for copy code in js or duplicate, you can use this code
//copyfile.js
const fs = require('fs');
// destination will be created or overwritten by default.
fs.copyFile('C:\folderA\myfile.txt', 'C:\folderB\myfile.txt', (err) => {
if (err) throw err;
console.log('File was copied to destination');
});