Bash scripting, multiple conditions in while loop

Question

I'm trying to get a simple while loop working in bash that uses two conditions, but after trying many different syntax from various forums, I can't stop throwing an error. Here is what I have:

while [ $stats -gt 300 ] -o [ $stats -eq 0 ] 

I have also tried:

while [[ $stats -gt 300 ] || [ $stats -eq 0 ]] 

... as well as several others constructs. I want this loop to continue while $stats is > 300 or if $stats = 0.

Answer 1

The correct options are (in increasing order of recommendation):

# Single POSIX test command with -o operator (not recommended anymore). # Quotes strongly recommended to guard against empty or undefined variables. while [ "$stats" -gt 300 -o "$stats" -eq 0 ]  # Two POSIX test commands joined in a list with ||. # Quotes strongly recommended to guard against empty or undefined variables. while [ "$stats" -gt 300 ] || [ "$stats" -eq 0 ]  # Two bash conditional expressions joined in a list with ||. while [[ $stats -gt 300 ]] || [[ $stats -eq 0 ]]  # A single bash conditional expression with the || operator. while [[ $stats -gt 300 || $stats -eq 0 ]]  # Two bash arithmetic expressions joined in a list with ||. # $ optional, as a string can only be interpreted as a variable while (( stats > 300 )) || (( stats == 0 ))  # And finally, a single bash arithmetic expression with the || operator. # $ optional, as a string can only be interpreted as a variable while (( stats > 300 || stats == 0 )) 

Some notes:

1. Quoting the parameter expansions inside [[ ... ]] and ((...)) is optional; if the variable is not set, -gt and -eq will assume a value of 0.

2. Using $ is optional inside (( ... )), but using it can help avoid unintentional errors. If stats isn't set, then (( stats > 300 )) will assume stats == 0, but (( $stats > 300 )) will produce a syntax error.