图像 url 到 file() 对象使用 js

Question

For a registration module in my vue app I let users upload images in a form. I upload these images to my storage and save the download url of this photo in the registration. When editing a registration I need to get the photo's out of the storage which is simple since I got the url. But I need it to be a file() object. I have found ways to turn it into a blob but it needs to be a file. How can I do this?

Answer 1

It can be done by requesting a blob and generating a File object. It is necessary to specify the MIME type of the blob.

const urlToObject= async()=> {
  const response = await fetch(image);
  // here image is url/location of image
  const blob = await response.blob();
  const file = new File([blob], 'image.jpg', {type: blob.type});
  console.log(file);
}

Answer 2

Since you are letting the user upload a file, you already have the file as a File object.

But if you wanna convert it to a blob for making some edits and convert it back to a File object, you can use the File() constructor for converting a blob to a File.

const file = new File([blob], "imagename.png");

Also, notice that the File() constructor takes an array of blobs as argument and not a single blob.

Answer 3

The ES6 way, with Promise :

const blobUrlToFile = (blobUrl:string): Promise<File> => new Promise((resolve) => {
    fetch(blobUrl).then((res) => {
      res.blob().then((blob) => {
        // please change the file.extension with something more meaningful
        // or create a utility function to parse from URL
        const file = new File([blob], 'file.extension', {type: blob.type})
        resolve(file)
      })
    })
  })