Why is Bash not recognising my variable as a string? -


i've made bash script. let's call "runner.sh". inside i'm running omxplayer flags.

omxplayer $targetfolder/*.mp4

this works fine. let's resize , position video.

omxplayer —win "0 0 640 480" $targetfolder/*.mp4

this works fine. let's try put "win flag" in variable. i'm escaping quote.

sizeandposition="—win \"0 0 640 480\""

allright, i'm trying out.

omxplayer $sizeandposition $targetfolder/*.mp4

nope, doesn't work. error file "0" not found. sure, print whole command screen.

echo "omxplayer $sizeandposition $targetfolder/*.mp4"

… , output…

omxplayer —win "0 0 640 480" /homefolder/*.mp4

why doesn't recognise variable other string in command? can it?

edit: sorry confusion. question sizeandposition, why not treated argument?

instead of storing command arguments in variable, use array:

sizeandposition=(-win '0 0 640 480') omxplayer "${sizeandposition[@]}" "$targetfolder"/*.mp4

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