sh AND and OR in one command

Question

Trying to check for 3 conditions in one line of code, but I'm stuck.

Essentially, I need to code for the following:

IF

string1 is not equal to string2

AND

string3 is not equal to string4

OR

bool1 = true

THEN

display "conditions met - running code ...".

As requested in the comments, I've updated my example to try to make the problem clearer.

#!/bin/sh

string1="a"
string2="b"
string3="c"
string4="d"
bool1=true

# the easy-to-read way ....
if [ "$string1" != "$string2" ] && [ "$string3" != "$string4" ] ; then
    echo "conditions met - running code ..."
fi

if $bool1 ; then
    echo "conditions met - running code ..."
fi

# or the shorter way ...
[ "$string1" != "$string2" ] && [ "$string3" != "$string4" ] && echo "conditions met - running code ..."

$bool1 && echo "conditions met - running code ..."

The code above will potentially run twice: if the first 2 conditions are met, and then again if the 3rd condition is met. This is not what I need.

The issue with this example is that it involves 2 distinct calls to 'echo' - (note: in the real code, it's not an echo, but you get the idea). I'm trying to reduce the code duplication by combining the 3 condition check into a single command.

I'm sure there's a few people now shaking their heads and shouting at the screen "That's NOT how you do it!"

And there's probably others waiting to mark this as a duplicate ... well, I looked but I'm damned if I could figure out how to do this from the answers I've read.

Can someone please enlighten me ? :)

Answer 1

This will work:

if    [ "$string1" != "$string2" ] \
   && [ "$string3" != "$string4" ] \
   || [ "$bool1" = true ]; then
    echo "conditions met - running code ...";
fi;

Or surround with { ;} for readability and easy to maintain in future.

if { [ "$string1" != "$string2" ] && [ "$string3" != "$string4" ] ;} \
   || [ "$bool1" = true ] ; then
    echo "conditions met - running code ...";
fi;

Points to note:

1. There is no such thing as a boolean variable..
2. Braces need the final semicolon ({ somecmd;}).
3. && and || evaluate left-to-right in the above — && has higher precedence than || only within (( )) and [[..]]

&& higher precedence only happen in [[ ]] is proven as follows. Assume bool1=true.

With [[ ]] :

bool1=true
if [[ "$bool1" == true || "$bool1" == true && "$bool1" != true ]]; then echo 7; fi #1 # Print 7, due to && higher precedence than ||
if [[ "$bool1" == true ]] || { [[ "$bool1" == true && "$bool1" != true ]] ;}; then echo 7; fi # Same as #1
if { [[ "$bool1" == true || "$bool1" == true ]] ;} && [[ "$bool1" != true ]] ; then echo 7; fi # NOT same as #1
if [[ "$bool1" != true && "$bool1" == true || "$bool1" == true ]]; then echo 7; fi # Same result as #1, proved that #1 not caused by right-to-left factor, or else will not print 7 here

With [ ] :

bool1=true
if [ "$bool1" == true ] || [ "$bool1" == true ] && [ "$bool1" != true ]; then echo 7; fi #1, no output, due to && IS NOT higher precedence than ||
if [ "$bool1" == true ] || { [ "$bool1" == true ] && [ "$bool1" != true ] ;}; then echo 7; fi # NOT same as #1
if { [ "$bool1" == true ] || [ "$bool1" == true ] ;} && [ "$bool1" != true ]; then echo 7; fi # Same as #1
if [ "$bool1" != true ] && [ "$bool1" == true ] || [ "$bool1" == true ]; then echo 7; fi # Proved that #1 not caused by || higher precedence than &&, or else will not print 7 here, instead #1 is only left-to-right evaluation

Answer 2

The && that you are using at the end of your conditional expression is meant for running a subsequent command (its second argument) if the preceding command (its first argument) returns a status of zero, which is traditionally considered a success code for programs. Also, the $bool1 that you have put there is standing alone, which makes Bash interpret its value as the name of a command or program to run. What you are trying to achieve seems to be a conditional expression evaluation, which is done using the if command as follows:-

if [ "$string1" != "$string2" ]  && [ "$string3" != "$string4" ] || [ $bool1 -ne 0 ]; then echo "conditions met - running code ..."; fi

Another way is to use the built-in "and", -a, and "or", -o, operators:-

if [ "$string1" != "$string2" -a "$string3" != "$string4" -o $bool1 -ne 0 ]; then echo "conditions met - running code ..."; fi

Again, there is yet another way that does not use the if command:-

[[ "$string1" != "$string2" && "$string3" != "$string4" || $bool1 -ne 0 ]] && echo "conditions met - running code ..."

Note that the Boolean value is expressed as an integer, so the integer expression test for "not equal to", -ne, can be used here. The following is a script that demonstrates the working:-

#!/bin/bash

string1="Hello"
string2="Hello"
string3="Hello"
string4="Hello"
bool1=1

if [ "$string1" != "$string2" ]  && [ "$string3" != "$string4" ] || [ $bool1 -ne 0 ]; then echo "conditions met - running code ..."; fi

exit 0