Open novalbayusetiawan opened 1 month ago
Hi @novalbayusetiawan You will only need to pass the part after the _capacitor_file_
to the Filesystem.readFile method.
The videoUrl you receive is constructed such that you can directly pass it to a html video element.
If you are looking to access the file path on the native side, all you need to do is split the videoUrl by _capacitor_file_
and access the 2nd item in the array.
Describe the bug I Cannot get recorded video file with Capacitor plugin Filesystem?
To Reproduce Steps to reproduce the behavior:
capacitor://localhost/_capacitor_file_/private/var/mobile/Containers/Data/Application/D143EA6E-D71D-47D1-9897-4211FC42B076/tmp/9F33AED5-9F9F-437A-AAD0-8DCAF54409CC.mp4
let result = await Filesystem.readFile({ path: videoUrl.value!, })
[error] - {"errorMessage":"The file “9F33AED5-9F9F-437A-AAD0-8DCAF54409CC.mp4” couldn’t be opened."}
Expected behavior can read the recorded video file
Smartphone :