Problem with the arithmetics in the bash scripting

I'm trying to cut a video into 2-minute clips using FFMpeg. I am using Ubuntu 10.10.

Her开发者_开发问答e is my code:

#!/bin/sh
COUNTER=0
BEG=0
MIN=`ffmpeg -i ${1} 2>&1 | grep "Duration" | cut -d ' ' -f 4 | sed s/,// | cut -d ":" -f 2`
echo $MIN
((MIN=MIN-2))
before_last_dot=${1%.*};
while [ $COUNTER -lt $MIN ]; do
    ((BEG=COUNTER*60))
    echo "MIN:${MIN}"
    echo "ffmpeg -sameq -i ${1} -ss ${BEG} -t 120 ${before_last_dot}.${COUNTER}.wmv"
    ((COUNTER=COUNTER+2))
done

echo "ffmpeg -sameq -i ${1} -ss ${BEG} -t 120 ${before_last_dot}.${COUNTER}.wmv" should be ffmpeg -sameq -i ${1} -ss ${BEG} -t 120 ${before_last_dot}.${COUNTER}.wmv. I print it to check it. ${1} is the video name.

But the problem is, ((COUNTER=COUNTER+2)) or ((COUNTER+=2))never works! COUNTER is always 0, BEG is always 0 too. ((MIN=MIN-2)) never works too.

I tried to replace ((MIN=MIN-2)) with let "MIN-=2" I get an error: let: not found

I+ve double checked but still don't know why. I'm getting gray hair on this.

---

The ((MIN=MIN-2)) syntax that you're using is a bash-specific feature.

I don't have Ubuntu 10.10 to hand to test with, but I'd guess that your /bin/sh is not bash, but a smaller and simpler shell with only the basic features required by POSIX. (In which case, ((MIN=MIN-2)) probably launches a sub-shell, which launches a sub-shell, which does nothing but set a variable MIN to the string MIN-2 and then exit.)

Try #!/bin/bash on the first line instead.