How to use arithmetic in an if statement for Bash

Question

Hi Guys I am trying to perform a check in my bash script that needs to meet three conditions, I was able to make the first 2 conditions work in the if statement as I wanted, but when I wanted to implement a third check with some arithmetic using the && operator, the script does not even launch.

DOUBLE_CHECK=0

 if [[ -z "$avail" &&  "$WIFI_ID" == "some_str"  && 'expr $DOUBLE_CHECK % 2' -eq "0"]]; then
        sudo caffeinate xterm -geometry 70x20+0+0 -fa monospace -fs 8 -e './script1.sh'  & disown

fi
(($DOUBLE_CHECK++))

The Idea is I want the third check to have a number that increments over time inside my while loop and then checked, whenever it is divisible by 2 it passes the 3rd condition of the if statement

if should be ((DOUBLE_CHECK++))

user12007102
– user12007102

2020-02-27 08:59:05 +00:00
Commented Feb 27, 2020 at 8:59 — user12007102
– user12007102, Commented Feb 27, 2020 at 8:59

Fred · Accepted Answer · 2017-04-16 23:39:44Z

As noted in a separate answer, you could use backticks or preferably $() to expand the expression to the output of the inner command, like this :

if [[ -z "$avail" && $WIFI_ID == "some_str" && $(expr $DOUBLE_CHECK % 2) -eq "0" ]]

Please note I have added a (required) space before the final ]].

Another possibility is to use an arithmetic expression :

if [[ -z $avail && $WIFI_ID == "some_str" ]] && ((DOUBLE_CHECK % 2 == 0))

Please note I have removed harmless but unnecessary quotes : the double-bracketed conditional expression is not a normal command, it is special shell syntax, and it does not perform word splitting. I have left quotes around some_str, because equality/inequality comparisons will perform pattern matching on the right-hand expression if it is not quoted. There is also no $ before the variable name in the arithmetic expression : it works, but is not required inside (()) or $(()).

The expression could also be expressed as :

if [[ -z $avail && $WIFI_ID == "some_str" ]] && ! ((DOUBLE_CHECK % 2))

The reason for this is that (( )) has a return code of 0 if its numerical result is non-zero, and a non-zero return code otherwise.

Michael Gamble · Accepted Answer · 2017-04-17 00:03:33Z

0

There are a couple small issues

Your third condition should be evaluated using back-ticks ('`'), instead of quotes. Single-quotes in bash mean string literal. So, 'expr $DOUBLE_CHECK %2' -eq "0" will always be false
The check -eq is used to compare numbers. Your statement is using strings (to compare strings, use the normal == syntax)

Please try,

DOUBLE_CHECK=0

if [[ -z "$avail" ]] &&  [[ "$WIFI_ID" == "some_str" ]] && [[ `expr $DOUBLE_CHECK % 2` -eq 0 ]]; then
        sudo caffeinate xterm -geometry 70x20+0+0 -fa monospace -fs 8 -e './script1.sh'  & disown

fi
(($DOUBLE_CHECK++))

edited Apr 17, 2017 at 0:03

answered Apr 16, 2017 at 23:24

Michael Gamble

13 bronze badges

5 Comments

Fred Over a year ago

I do not think your #3 recommendation is valid. Bash conditionals support logical operators, and I do not know of a reason to avoid them (contrary to -a and -o in standard [ tests).

Michael Gamble Over a year ago

[[ "A" == "A" ]] && [[ "B" == "B" ]] && echo 'foo' works for me

Fred Over a year ago

Yes, it works. I am not saying the alternative syntax you suggest is wrong, my point was saying that I do not think that recommending to put the logical operators outside the conditionals is good advice. If it is (it could be), it is something new to me.

Fred Over a year ago

I am not saying the way you do it is bad at all. It may be more readable to you, and in that case there is nothing to prevent you from doing it. In most cases, I would prefer shorter syntax, but I would not say either way is better in an absolute sense. My intervention was more because you recommended this as being better without giving a clear reason why. It is always helpful telling why one recommends X over Y, especially when both solutions work.

Michael Gamble Over a year ago

Thanks for feedback. I appreciate it. I have removed the point from my post to avoid future confusion

codeforester · Accepted Answer · 2017-04-17 02:34:16Z

0

You can use [[ .. ]] for string logic (where = and == are equivalent) and (( ... )) for numeric comparison, and do away with expr:

if [[ -z "$avail" && "$WIFI_ID" = "some_str" ]] && (( DOUBLE_CHECK % 2 == 0 )); then
  # your logic here
fi

Note that prefixing variables with $ is optional inside (( ... )) and quoting variables is optional inside [[ ... ]] since doesn't do word splitting and globbing there.

edited Apr 17, 2017 at 2:34

answered Apr 16, 2017 at 23:37

codeforester

43.8k21 gold badges122 silver badges159 bronze badges

Collectives™ on Stack Overflow

How to use arithmetic in an if statement for Bash

3 Answers 3

Comments

5 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

5 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related