I tried to write a program to see if the count divisible by 2 without a remainder Here is my program
count=$((count+0))
while read line; do
if [ $count%2==0 ]; then
printf "%x\n" "$line" >> file2.txt
else
printf "%x\n" "$line" >> file1.txt
fi
count=$((count+1))
done < merge.bmp
This program doesnt work its every time enter to the true
In the shell, the [ command does different things depending on how many arguments you give it. See https://www.gnu.org/software/bash/manual/bashref.html#index-test
With this:
[ $count%2==0 ]
you give [ a single argument (not counting the trailing ]), and in that case, if the argument is not empty then the exit status is success (i.e. "true"). This is equivalent to [ -n "${count}%2==0" ]
You want
if [ "$(( $count % 2 ))" -eq 0 ]; then
or, if you're using bash
if (( count % 2 == 0 )); then
Some more "exotic" way to do this:
count=0
files=(file1 file2 file3)
num=${#files[@]}
while IFS= read -r line; do
printf '%s\n' "$line" >> "${files[count++ % num]}"
done < input_file
This will put 1st line to file1, 2nd line to file2, 3rd line to file3, 4th line to file1 and so on.
awk to the rescue!
what you're trying to do is a one-liner
$ seq 10 | awk '{print > (NR%2?"file1":"file2")}'
==> file1 <==
1
3
5
7
9
==> file2 <==
2
4
6
8
10
try
count=$((count+0))
while read line; do
if [ $(($count % 2)) == 0 ]; then
printf "%x\n" "$line" >> file2.txt
else
printf "%x\n" "$line" >> file1.txt
fi
count=$((count+1))
done < merge.bmp
You have to use the $(( )) around a mod operator as well.
How to use mod operator in bash?
This will print "even number":
count=2;
if [ $(($count % 2)) == 0 ]; then
printf "even number";
else
printf "odd number";
fi
This will print "odd number":
count=3;
if [ $(($count % 2)) == 0 ]; then
printf "even number";
else
printf "odd number";
fi
