Bash if statement with multiple conditions throws an error

BashIf StatementFlags

Bash Problem Overview


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?

Bash Solutions


Solution 1 - Bash

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

Solution 2 - Bash

You can use either [[ or (( keyword. When you use [[ keyword, you have to use string operators such as -eq, -lt. I think, (( is most preferred for arithmetic, because you can directly use operators such as ==, < and >.

Using [[ operator

a=$1
b=$2
if [[ a -eq 1 || b -eq 2 ]] || [[ a -eq 3 && b -eq 4 ]]
then
     echo "Error"
else
     echo "No Error"
fi

Using (( operator

a=$1
b=$2
if (( a == 1 || b == 2 )) || (( a == 3 && b == 4 ))
then
     echo "Error"
else
     echo "No Error"
fi

Do not use -a or -o operators Since it is not Portable.

Solution 3 - Bash

Please try following

if ([ $dateR -ge 234 ] && [ $dateR -lt 238 ]) || ([ $dateR -ge 834 ] && [ $dateR -lt 838 ]) || ([ $dateR -ge 1434 ] && [ $dateR -lt 1438 ]) || ([ $dateR -ge 2034 ] && [ $dateR -lt 2038 ]) ;
then
	echo "WORKING"
else
	echo "Out of range!"

Solution 4 - Bash

You can get some inspiration by reading an entrypoint.sh script written by the contributors from MySQL that checks whether the specified variables were set.

As the script shows, you can pipe them with -a, e.g.:

if [ -z "$MYSQL_ROOT_PASSWORD" -a -z "$MYSQL_ALLOW_EMPTY_PASSWORD" -a -z "$MYSQL_RANDOM_ROOT_PASSWORD" ]; then
    ...
fi

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionSimply_meView Question on Stackoverflow
Solution 1 - BashmkhatibView Answer on Stackoverflow
Solution 2 - BashFizer KhanView Answer on Stackoverflow
Solution 3 - BashManash NathView Answer on Stackoverflow
Solution 4 - BashDaniel Andrei MincăView Answer on Stackoverflow