CODEWITHSUNDEEP
bash
Elliott B
Elliott B

Reputation: 1199

Concatenate multiple variables isn't working

I have these variables:

$reel = XF_1

$base = AA000201

$output_dir = directory

This command isn't working. It produces a file called directory/AA000201.mp4

ffmpeg -i $1 $output_dir"/$reel_$base.mp4

This command works. It produces a file called directory/XF_1_AA000201.mp4. 

ffmpeg -i $1 $output_dir"/"$reel"_"$base".mp4"

But it gives this error: /Volumes/RAID/LIGHTS/RawFootage/Day01/B_Camera/XF_1/CONTENTS/CLIPS001/AA0002/AA000201.MXF: line 5: continue: only meaningful in a for',while', or `until' loop

Note: I have shortened directory names here for simplicity.

Upvotes: 1

Views: 100

Answers (2)

Bertrand Caron
Bertrand Caron

Reputation: 2657

Polbrelkey gave you the answer, but i'm still gonna add this.

Lets see why ffmpeg -i $1 $output_dir"/$reel_$base.mp4 is un-interpretable unambiguously :

Look at those three variables : 

reel="A"
reel_="B"   #Nothing prevents you from ending a variable with an underscore
base="C"

Now, let's interpret ffmpeg -i $1 $output_dir"/$reel_$base.mp4 : Depending on how your decompose the string, you get either :

#ffmpeg -i $1 $output_dir"/${reel_}${base}.mp4
ffmpeg -i $1 $output_dir"/BC.mp4    

#ffmpeg -i $1 $output_dir"/${reel}_${base}.mp4
ffmpeg -i $1 $output_dir"/A_C.mp4

In your case, Bash is actually interpreting the first one (actually, he looks for the longest possible variable name) and since reel_="" is not assigned, outputs directory/AA000201.mp4. You don't have these kind of problems with base.mp4 since a variable name can't have a dot.

Source : Parameter Expansion on Bash-Hackers

Upvotes: 1

pobrelkey
pobrelkey

Reputation: 5973

ffmpeg -i "${1}" "${output_dir}/${reel}_${base}.mp4"

Upvotes: 1

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