How to download an image from Python 3 / Selenium if the URL starts with "blob:"?

Question

When using web.whatsapp.de, it can be seen that the link to the resulting image may look like this:

blob:https://web.whatsapp.com/3565e574-b363-4aca-85cd-2d84aa715c39

If the link is copied to the address window, it will open the image, however, if the "blob" is ignored, it will simply open a new window on the whatsapp website.

I am trying to download the image displayed on this link.

But using common methods, such as using a request, or urllib.request, or even BeautifulSoup, will always work at one point: "blob" at the beginning of the URL will throw an error.

These answers Downloading a file from a Blob URL using Python will result in an Error error

URLError: <urlopen error unknown url type: blob>

or error

InvalidSchema: No connection adapters were found for 'blob:https://web.whatsapp.com/f50eac63-6a7f-48a4-a2b8-8558a9ffe015'

(using BeatufilSoup)

Using your own approach, for example:

import requests

url = 'https://web.whatsapp.com/f50eac63-6a7f-48a4-a2b8-8558a9ffe015'
fileName = 'test.png'
req = requests.get(url)
file = open(fileName, 'wb')
for chunk in req.iter_content(100000):
    file.write(chunk)
file.close()

, BeautifulSoup.

Chrome, Selenium Python, , .

Answer 1

- , .

chrome://blob-internals/

script. , , script /, blob:

def get_file_content_chrome(driver, uri):
  result = driver.execute_async_script("""
    var uri = arguments[0];
    var callback = arguments[1];
    var toBase64 = function(buffer){for(var r,n=new Uint8Array(buffer),t=n.length,a=new Uint8Array(4*Math.ceil(t/3)),i=new Uint8Array(64),o=0,c=0;64>c;++c)i[c]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/".charCodeAt(c);for(c=0;t-t%3>c;c+=3,o+=4)r=n[c]<<16|n[c+1]<<8|n[c+2],a[o]=i[r>>18],a[o+1]=i[r>>12&63],a[o+2]=i[r>>6&63],a[o+3]=i[63&r];return t%3===1?(r=n[t-1],a[o]=i[r>>2],a[o+1]=i[r<<4&63],a[o+2]=61,a[o+3]=61):t%3===2&&(r=(n[t-2]<<8)+n[t-1],a[o]=i[r>>10],a[o+1]=i[r>>4&63],a[o+2]=i[r<<2&63],a[o+3]=61),new TextDecoder("ascii").decode(a)};
    var xhr = new XMLHttpRequest();
    xhr.responseType = 'arraybuffer';
    xhr.onload = function(){ callback(toBase64(xhr.response)) };
    xhr.onerror = function(){ callback(xhr.status) };
    xhr.open('GET', uri);
    xhr.send();
    """, uri)
  if type(result) == int :
    raise Exception("Request failed with status %s" % result)
  return base64.b64decode(result)

bytes = get_file_content_chrome(driver, "blob:https://developer.mozilla.org/7f9557f4-d8c8-4353-9752-5a49e85058f5")

Answer 2

Blobs - , URI. psuedo-URL-, , . <img> , blob src.

MDN, :

Blob - FileReader. Blob .

var reader = new FileReader();
reader.addEventListener("loadend", function() {
   // reader.result contains the contents of blob as a typed array
});
reader.readAsArrayBuffer(blob);

Answer 3

, node , . .

var script = function (blobUrl) {
    console.log(arguments);
    var uri = arguments[0];
    var callback = arguments[arguments.length - 1];
    var toBase64 = function(buffer) {
        for(var r,n=new Uint8Array(buffer),t=n.length,a=new Uint8Array(4*Math.ceil(t/3)),i=new Uint8Array(64),o=0,c=0;64>c;++c)
            i[c]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/".charCodeAt(c);for(c=0;t-t%3>c;c+=3,o+=4)r=n[c]<<16|n[c+1]<<8|n[c+2],a[o]=i[r>>18],a[o+1]=i[r>>12&63],a[o+2]=i[r>>6&63],a[o+3]=i[63&r];return t%3===1?(r=n[t-1],a[o]=i[r>>2],a[o+1]=i[r<<4&63],a[o+2]=61,a[o+3]=61):t%3===2&&(r=(n[t-2]<<8)+n[t-1],a[o]=i[r>>10],a[o+1]=i[r>>4&63],a[o+2]=i[r<<2&63],a[o+3]=61),new TextDecoder("ascii").decode(a)
    };
    var xhr = new XMLHttpRequest();
    xhr.responseType = 'arraybuffer';
    xhr.onload = function(){ callback(toBase64(xhr.response)) };
    xhr.onerror = function(){ callback(xhr.status) };
    xhr.open('GET', uri);
    xhr.send();
}
driver.executeAsyncScript(script, imgEleSrc).then((result) => {
    console.log(result);
})

. . https://medium.com/@anoop.goudar/how-to-get-data-from-blob-url-to-node-js-server-using-selenium-88b1ad57e36d