11

If want to express the following test in shell (sh) :

if ( a == 1 && ( b == 1 || b == 2 )) { ... }

So far, the best I have been able to write is this :

if [[ $a -eq 1 ]]; then
  if [[ $b -eq 1 || $b -eq 2 ]]; then
     ...
  fi
fi

I don't know how to compound && and || with correct precedence. Googling has not given me any answer (tutorials only give basic examples, if any)

What is the syntax to combine those two if into one ?

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1
  • 1
    Any reason to not use ksh? Then your initial code would require only one more pair of parenthesis: if (( a == 1 && ( b == 1 || b == 2 ))); then :; fi – manatwork Jan 10 '12 at 17:18
15

Note that [[ ]] is not in either Bourne or POSIX sh. For true sh syntax, there are several ways to do this.

Using only one [ ] pair

if [ 1 -eq "$a" -a \( 1 -eq "$b" -o 2 -eq "$b" \) ]; then
    # ...
fi

or Avoiding the POSIX -a and -o options1

if [ 1 -eq "$a" ] && { [ 1 -eq "$b" ] || [ 2 -eq "$b" ]; }; then
    # ...
fi

1 One reason for avoiding -a and -o is maximum portability - not all test or [ implementations can handle more than 4 arguments, which is precisely what you get if you chain expressions with -a and -o and \( \).

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8
  • My scripts have the #! /bin/sh shebang, and the default shell is ksh. Why would sh not be used ? – Offirmo Jan 10 '12 at 16:45
  • 2
    @Offirmo If you declare your script as /bin/sh, it is incorrect to write the script using features not in sh. You should be aware that ksh is NOT sh, but rather a superset of sh. If you take your ksh-but-declared-as-sh script to a system where sh is not a symlink to ksh, it may break. You can avoid that problem by making sure your script matches the declaration. – jw013 Jan 10 '12 at 17:03
  • I thought that setting /bin/sh as shebang would make my scripts invoked via /bin/sh. Do you mean that my default shell (ksh) ignore the shebang and interpret it directly ? – Offirmo Jan 10 '12 at 18:04
  • @Offirmo You are correct: a shebang of /bin/sh will invoke /bin/sh. The problem is /bin/sh does not accept [[. You can either not use [[ or use a shell that does accept [[, e.g. ksh, as your shebang. – jw013 Jan 10 '12 at 22:25
  • 2
    The problem is not that much that -a/-o are not supported by some implementations but that it's not reliable for arbitrary arguments. Like [ "$a" = "$b" -o "$b" = "$c" ] would fail for values of $a like ! with many [ implementations. – Stéphane Chazelas Mar 16 '15 at 16:17
2

I tried some syntax and found that

if [[ $a -eq 1 ]] && [[ $b -eq 1 || $b -eq 2 ]]; then
 ...
fi

is working.

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1
  • 2
    this works for bash version above 4. – Nikhil Mulley Jan 10 '12 at 17:18
2

Another POSIX approach:

if [ "$((a == 1 && ( b == 1 || b == 2 )))" -ne 0 ]; then
  ...
fi
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0

The right way to do it with a mere shell:

if test 1 -eq "$a" && test 1 -eq "$b" -o 2 -eq "$b"; then 
    ...
fi

Explanation:

test here serves as parenthesis and -o as logical or. If you a logical and inside the "parenthesis" you would use -a.

More on that can be found here

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