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I am getting a syntax error at line 17 : unexpected token `else'

declare -i A

echo "enter any numeric value"
read value


if [$value > 0]; 
    if [[ "$value" =~ ^[0-9]+$ ]]; then
        A=$value
    else
        if ! [[$value =~ ^[0-9]+$ ] || $value !=0];then
        A=$[RANDOM%20+1]
        fi
    fi
else 
    A=$((RANDOM%25+16))
fi
echo"the value of |A| is $A"
codeholic24
  • 307
Wissi
  • 1
  • 5
    The first if needs a then. Download from shellcheck.net and save yourself much trouble in the future. – Paul_Pedant Nov 20 '20 at 13:32
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    Also see Why are bash tests so picky about whitespace? – steeldriver Nov 20 '20 at 13:32
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    shellcheck will also tell you where you started a test with [[ and ended it with ]. – Paul_Pedant Nov 20 '20 at 13:34
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    Also, > is not a "greater than" test. It tests for sort order. You probably meant [ "$value" -gt 0 ]. You also seem to mix the arcane and obsolete $[ ... ] syntax for arithmetic expansions with the modern $(( ... )) syntax. There is a general issue with whitespaces in your code, consider re-writing and testing while doing so. As others have pointed out, the shellcheck.net website will be of use to you. – Kusalananda Nov 20 '20 at 13:36
  • It worked and the link was very useful. Thank you. – Wissi Nov 20 '20 at 13:45
  • leaving aside the syntax, what's the logic in if ! [[$value =~ ^[0-9]+$ ] || $value !=0];then meant to be? It looks to me the condition will end up falsy (after the outer negation) regardless of the value of $value, so the main body of the if will never run. – ilkkachu Nov 20 '20 at 14:30
  • @ilkkachu In that test, the value of $value will never be zero (due to the earlier test in the code), so $value != 0 can be removed, leaving if ! [[ $value =~ ^[0-9]+$ ]]. This would be true for any string that evaluates to an integer for the $value -gt 0 test, but that contains characters other than digits, e.g. 12aa. What was actually meant from the start is unknown, but the whole script could possible have wanted to test for negative, zero, and positive values of $value, to then set A depending on this. – Kusalananda Nov 20 '20 at 14:51
  • @Kusalananda, right, because of the outer test, it'll never be zero, making $value !=0 true. The whole [[ ... || $value != 0 ]] will be true, and the ! [[ ... ]] will be false. Note that it's || there, not &&. – ilkkachu Nov 20 '20 at 16:23

1 Answers1

0

Used shellcheck.net to fix syntax issue

Updated code :

#!/bin/bash

declare -i A


echo "enter any numeric value"
read -r value


if [ "$value" -gt 0 ] ; then
    if [[ "$value" =~ ^[0-9]+$ ]]; then
        A=$value
    else
        if ! [[ "$value" =~ ^[0-9]+$ || "$value" != 0 ]] 
        then
            A=$((RANDOM%20+1))
        fi
    fi
else 
    A=$((RANDOM%25+16))
fi
echo "The value of |A| is $A"
codeholic24
  • 307
  • 1
    Not that it matters syntactically, but that innermost if statement basically says "if $value is a zero and at the same time not an integer". – Kusalananda Nov 20 '20 at 15:42
  • @Kusalananda yes i too saw that it's contradict but his issue was to solve unexpected error so i considered that only – codeholic24 Nov 20 '20 at 16:42