problem with $( printf "%03d" $i in script

[zimbot]

friends,

I am attempting to use jpg2yuv on many jpgs
I have a dir with 300 jpegs ( 000.jpg - 299.jpg )

here is my script
#!/bin/bash
# n=300
n=300
for (( i=000; i < n; i++ ))
do
jpeg2yuv -n 30 -I p -L 0 -f 30 -j "$( printf "%03d" $i).jpg | mpeg2enc -a 2 -f 8 -b 5800 -o "$( printf "%03d" $i).mpg
printf " OK\n"
done

it fails - complaining that there is no such file----------
<snip>
INFO: [jpeg2yuv] Parsing & checking input files.
**ERROR: [jpeg2yuv] System error while opening: "000.jpg | mpeg2enc -a 2 -f 8 -b 5800 -o 000.mpg": No such file or directory
</snip>

but - of course - there IS a 000.jpg and this works
jpeg2yuv -n 30 -I p -L 0 -f 30 -j 000.jpg | mpeg2enc -a 2 -f 8 -b 5800 -o 000.mpg

i get a good 000.mpg - so I know all is ok but for my script.
( i am not a real good scripter....)

I am certain that it is something about my " or ' or lack of.....
Thanks!

[radoulov]

At first glance, you should change this:

jpeg2yuv -n 30 -I p -L 0 -f 30 -j "$( printf "%03d" $i).jpg | mpeg2enc -a 2 -f 8 -b 5800 -o "$( printf "%03d" $i).mpg

to this:

jpeg2yuv -n 30 -I p -L 0 -f 30 -j "$( printf "%03d" $i).jpg" | mpeg2enc -a 2 -f 8 -b 5800 -o "$( printf "%03d" $i).mpg"

At second, you could rewrite the script as:

for f in *.jpg; do
  printf "processing %s ...\n" "$f"
  jpeg2yuv -n 30 -I p -L 0 -f 30 -j "$f" |
    mpeg2enc -a 2 -f 8 -b 5800 -o "${f%.jpg}.mpg"
  printf "%s done\n" "$f"	
done