whisperfile server: convert files without ffmpeg #568
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jart
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Sep 28, 2024
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Thanks for doing this. Several people have requested this feature. read_wav() isn't ready to be used in this manner, but I can make it ready after merging this. I anticipate that it won't work if concurrent requests are sent to the server until I get rid of the global variable in common.cpp. This will obviously be fixed before the next release. So please don't publish any whisperfiles until we've had a chance to fix that.
jart
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We now have a new function slurp_audio_file() which replaces read_wav(). This function has simpler code, and allows us to avoid a temporary file. See #568
jart
reviewed
Sep 28, 2024
| { | ||
| fprintf(stderr, "error: failed to read WAV file\n"); | ||
| const std::string error_resp = "{\"error\":\"failed to read WAV file\"}"; | ||
| res.set_content(error_resp, "application/json"); | ||
| return; | ||
| } | ||
| } | ||
| // remove temp file | ||
| std::remove(temp_filename.c_str()); |
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The function you want here is unlink().
jart
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Sep 28, 2024
We now have a new function slurp_audio_file() which replaces read_wav(). This function has simpler code, and allows us to avoid a temporary file. See #568
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This PR allows the
whisperfileserver to convert.wav,.mp3,.flac, and.ogginto the appropriate .wav file for whisper (16-bit 16000Hz) without any dependency on ffmpeg.The ffmpeg support still remains under the
--convertflag.The main change here is giving
read_wava file instead of a buffer. Before it was given a buffer when run through the server, and a filename when run through the cli. Now it is unified to always use a filename.In addition
is_wav_bufferwas removed, as the codepath is dead with the changes to use a filename throughout. This function was always expecting a buffer, but was receiving both filenames and buffers.