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.