I'm trying to write a script that will check two error flags, and in case one flag (or both) are changed it'll echo-- error happened. My script:
my_error_flag=0
my_error_flag_o=0
do something.....
if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then
echo "$my_error_flag"
else
echo "no flag"
fi
Basically, it should be, something along:
if ((a=1 or b=2) or (a=1 and b=2))
then
display error
else
no error
fi
The error I get is:
line 26: conditional binary operator expected
line 26: syntax error near `]'
line 26: `if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then'
Are my brackets messed up?
Use -a
(for and) and -o
(for or) operations.
tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html
Update
Actually you could still use &&
and ||
with the -eq
operation. So your script would be like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ] || ([ $my_error_flag -eq 1 ] && [ $my_error_flag_o -eq 2 ]); then
echo "$my_error_flag"
else
echo "no flag"
fi
Although in your case you can discard the last two expressions and just stick with one or operation like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ]; then
echo "$my_error_flag"
else
echo "no flag"
fi