Get the image as file in angular

You can use FormData to attach the read file and send to the API.

 onSelectFile(event) {
    if (event.target.files && event.target.files[0]) {

      this.uploadToServer(event.target.files[0]);
      ... rest of the code
  }

  uploadToServer(file) {
    let formData: FormData = new FormData();
    formData.append('fileName', file);
    // call your api service to send it to server, send formData
  }

EDIT:

Try this out if you have no option to touch onSelectFile() or trigger a different function when you upload the file.

_url = ''
set url(val) {
  this._url = val;
  if (val) {
    this.dataURLtoFile(val);
  }
}

get url() {
    return this._url;
}
uploadedImage: File ;

dataURLtoFile(dataurl) {
    const arr = dataurl.split(',');
    const mime = arr[0].match(/:(.*?);/)[1];
    const imageExtension = mime.split('/')[1];
    const bstr = atob(arr[1]);
    let n = bstr.length;
    const u8arr = new Uint8Array(n);
    while (n--) {
        u8arr[n] = bstr.charCodeAt(n);
    }
    this.uploadedImage = new File([u8arr], `uploaded.${imageExtension}`);
  }

On your API call, maybe when you click on a button,

uploadPersonaImage(e) {
  // this.apiService.someMethod(this.uploadedImage);
}

If you want to trigger the API call just when you upload the image, add the code of dataURLtoFile() to uploadPersonaImage() and call uploadPersonaImage() from url setter

Clarification

Do you understand what does event.target.src mean (considering e as event)?

1. Here event means the click/change event you triggered when you clicked onto upload photo.

2. event.target means the DOM element on which the event took place.

3. event.target.src will give you the src attribute value of the DOM element on which you triggered the change event.

Now, you say won't it work? No, it won't because the element which you clicked is an HTMLInputElement but the src resides under the image in under the label tag. And how are you intending to call uploadPersonaImage()? what calls your method? You haven't answered that even after asking so many times.

In my last edit, I have added code under the setter of the url which will convert the dataUrlFile to an actual File, It completely depends on your server how you want to store the file. As a file or as a dataUrl? If you want to send it as a file then follow the conversions I added in the answer if you want to save as dataUrl then directly save the content of this.url on your API call.