mjhjmtu

This thing here

n=0
x=1
while [ $n -lt 6 ]
do
 n=$(( n+1 ))
 echo "sasadgsad gsda $n" >> /home/test/rptest

if [ $n -eq 5 ]
 then
 while [ $x -le 5 ]
 do
 echo "end of line$x" >> /home/test/rptest
 x=$(( x+1 ))
 done
 fi
done

Output this thing

sasadgsad gsda 1
sasadgsad gsda 2
sasadgsad gsda 3
sasadgsad gsda 4
sasadgsad gsda 5
end of line1
end of line2
end of line3
end of line4
end of line5
sasadgsad gsda 6

That 11th line shouldn't be there... Shouldn't the first while thing finish when n = 5? Why does it create that last line?

Thank you for the help :)

I even tried it with n=1 and while [ $n -le 5 ]

Your script with proper indentation:

n=0
x=1
while [ $n -lt 6 ]; do
 n=$(( n+1 ))
 echo "sasadgsad gsda $n" >> /home/test/rptest

 if [ $n -eq 5 ]; then
 while [ $x -le 5 ]; do
 echo "end of line$x" >> /home/test/rptest
 x=$(( x+1 ))
 done
 fi
done

Your outer loop runs from 0 to 5, which is six times. Since you update n at the start of the outer loop, the value of n will go from 1 to 6 in the body of the loop. When n is 5, you run another loop from 1 to 5 outputting end of line.... When that's done, you still have one iteration of the outer loop to do.

Another way to write the script in bash:

for (( n=1; n<=5; ++n )); do
 printf 'sasadgsad gsda %sn' "$n"

 if (( n == 5 )); then
 for (( x=1; x<=5; ++x )); do
 printf 'end of line%sn' "$x"
 done
 fi
done >>/home/test/rptest

This would not have the same issue because the outer loop stops when n reaches 6.

However, if you just want to append the output of the inner loop after the that of the outer, you might just as well run the after each other:

for (( n=1; n<=5; ++n )); do
 printf 'sasadgsad gsda %sn' "$n"
done >>/home/test/rptest

for (( n=1; n<=5; ++x )); do
 printf 'end of line%sn' "$n"
done >>/home/test/rptest

or, for this simple example only,

printf 'sasadgsad gsda %sn' 1..5 >>/home/test/rptest
printf 'end of line%sn' 1..5 >>/home/test/rptest