Comments (1)
你好,我之前没考虑到这个问题,合并分片时直接使用框架的task_id拼接文件名,所以是个乱码。真实文件名可以通过框架的“file.source['name']”来获取。
谢谢你的反馈,我已经更新了源码和README。
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