What does the variable ${i%.*} mean in terminal?

Question

This shell command converts all .avi to .mp4 as per this answer by @llogan

Can someone explain how ${i%.*} works? In particular, what does % do?

for i in *.avi; do ffmpeg -i "$i" "${i%.*}.mp4"; done

Answer 1

From BashGuide/Parameters/Parameter Expansion:

${parameter%pattern}
The pattern is matched against the end of parameter. The result is the expanded value of parameter with the shortest match deleted.

From How do I do string manipulations in bash?:

The % means "remove the shortest possible match from the end of the variable's contents"

In other words, if the expanded value of parameter (i in your case) is myvideo.avi, then %.* will result in myvideo. This is so the output file does not end up being named myvideo.avi.mp4, and instead is named myvideo.mp4.