Bash while loop syntax error in do

问题

I´m trying do do a simple counter:

max=100
count=1
while [[ $count -le $max]]
  do
    echo "$count"
    ((count++))
  done

This gives me a syntax error in conditional expression near do.

What´s my issue? (probably something obvious)

The idea is then to raise the max from 100 to 200 and so forth in a superior loop so I will get a new file to manipulate with a python program 100 lines each time, but that´s irrelevant here.

回答1:

Your mistake is that it need one more space in [[ $count -le 100]]

max=100
count=1
while [[ $count -le $max ]]
do
    echo "$count"
    ((count++))
done

Another solution :

while ((count < max+1)); do echo $((count++)); done

or

for ((i=count; i<max; i++)) { echo $i; }

or

for ((i=count; i<max; i++)); do echo $i; done

or

for i in {1..100}; do echo $i; done

回答2:

Change the line:

while [[ $count -le 100]]

to:

while [[ $count -le 100 ]];

Notice the space after 100.

来源：https://stackoverflow.com/questions/12840769/bash-while-loop-syntax-error-in-do