3 ms·

Yeah, the WebCrypto API has a built-in function for this. I hope the author fixes the code, because this is otherwise a dangerous tutorial. window.crypto.getRa

Yeah, the WebCrypto API has a built-in function for this. I hope the author fixes the code, because this is otherwise a dangerous tutorial. window.crypto.getRandomValues(new Uint8Array(12)); If the author followed the MDN guides for it, they would have noticed that the example code also uses a random array. https://developer.mozilla.org/en-US/docs/Web/API/SubtleCrypto/encrypt#AES-GCM_2 https://developer.mozilla.org/en-US/docs/Web/API/SubtleCrypt...

dchest 7y ago
No, it doesn't need to be random if the key is used only once. PS The author didn't just copy something from a tutorial without knowing what he was doing, he actually asked for advice https://github.com/excalidraw/excalidraw/issues/610 https://github.com/excalidraw/excalidraw/issues/610
- prophesi 7y ago
  Hm, I stand corrected! Though I have two thoughts on that: 1. It wouldn't add very much complexity to implement random IV's. You can send it with the ciphertext, or in the generated URL. In my own project, I use btoa(JSON.stringify({ key: generatedKey, iv: randomIv })) and reverse the operation when the URL is accessed: const fragmentData = window.location.hash.substr(1); const keyParameters = JSON.parse(atob(fragmentData)); const key = HexMix.hexToUint8(keyParameters.key); const iv = HexMix.hexToUint8(keyParameters.iv); Edit: I'm also converting the key/iv to hexadecimal. Don't remember why, but that's not necessary lol. Edit #2: Ah-ha, and now I've learned something else from the Excalidraw team; I can skip my hex/uint & base64 conversions if I export the key as jwk. 2. There are a _lot_ of tutorials out there that incorrectly reuse the key with a fixed IV. I was more concerned about this being yet another tutorial that someone blindly follows and ends up reusing the key regardless.

3 ms·

Yeah, the WebCrypto API has a built-in function for this. I hope the author fixes the code, because this is otherwise a dangerous tutorial. window.crypto.getRa

by prophesi 7y ago

dchest 7y ago
No, it doesn't need to be random if the key is used only once. PS The author didn't just copy something from a tutorial without knowing what he was doing, he actually asked for advice https://github.com/excalidraw/excalidraw/issues/610 https://github.com/excalidraw/excalidraw/issues/610
- prophesi 7y ago
  Hm, I stand corrected! Though I have two thoughts on that: 1. It wouldn't add very much complexity to implement random IV's. You can send it with the ciphertext, or in the generated URL. In my own project, I use btoa(JSON.stringify({ key: generatedKey, iv: randomIv })) and reverse the operation when the URL is accessed: const fragmentData = window.location.hash.substr(1); const keyParameters = JSON.parse(atob(fragmentData)); const key = HexMix.hexToUint8(keyParameters.key); const iv = HexMix.hexToUint8(keyParameters.iv); Edit: I'm also converting the key/iv to hexadecimal. Don't remember why, but that's not necessary lol. Edit #2: Ah-ha, and now I've learned something else from the Excalidraw team; I can skip my hex/uint & base64 conversions if I export the key as jwk. 2. There are a _lot_ of tutorials out there that incorrectly reuse the key with a fixed IV. I was more concerned about this being yet another tutorial that someone blindly follows and ends up reusing the key regardless.