Bash if statement produces different result with exclamation mark in different positions

Question

My first question is why putting ! in the front of if statement fails to produce syntax error when status is not double quoted. That is how is [ ! $x = "string" ] different from [ $x != "string" ]?

My script is as follows.

#!/bin/bash status=" " # Comparison 1: without error echo "-----Comparison 1-----" if [ ! $status = "success" ] then echo 'Error: status was not success but: ' $status else echo "The status is success." fi echo "-----Comparison 2-----" # Comparison 2: with error message but still shows success if [ $status != "success" ] then echo 'Error: status was not success but: ' $status else echo "The status is success." fi echo "-----Comparison 3-----" # Comparison 3: Correct result after quoting status if [ ! "$status" == "success" ] then echo 'Error: status was not success but: ' $status else echo "The status is success." fi echo "-----Comparison 4-----" # Comparison 4: Correct result after quoting status if [ "$status" != "success" ] then echo 'Error: status was not success but: ' $status else echo "The status is success." fi

The output is

-----Comparison 1----- The status is success. -----Comparison 2----- ./test2.sh: line 14: [: !=: unary operator expected The status is success. -----Comparison 3----- Error: status was not success but: -----Comparison 4----- Error: status was not success but:

Additional questions

Why does it produce "The status is success." in Comparison 2 after a syntax error? How would a syntax error in an if statement affects the evaluated result of such if statement?

p.s. I know we need "" around $status to make output right.

Use [[ .. ]] instead of [ .. ]

anubhava
– anubhava

2018-10-01 14:29:40 +00:00
Commented Oct 1, 2018 at 14:29 — anubhava
– anubhava, Commented Oct 1, 2018 at 14:29
@anubhava thanks for the suggestion.

Tai
– Tai

2018-10-01 14:34:07 +00:00
Commented Oct 1, 2018 at 14:34 — Tai
– Tai, Commented Oct 1, 2018 at 14:34

chepner · Accepted Answer · 2018-10-01 14:30:52Z

You need to quote $status. Without the quotes, your first comparison is

if [ ! = "success" ]

which will fail because ! and success are not equal strings.

Your second one results in a syntactically invalid expression for [:

if [ != "success" ]

The syntax error you see in condition 2 isn't a shell syntax error, but a condition syntax error raised by the [ command. The command itself runs fine, but exits with a non-zero exit status.

Because success means the condition did not hold, and ! = success is indeed false. ! isn't an operator in this context; it's just the left-hand operand of the = operator.

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