A Linux administrator wants a loop in a Bash script to ignore any file whose name does not end with the extension .log and writes the following check:
if [[ $file != *.log ]]; then
echo "Skipping $file"
fi
Which statement correctly describes what the != operator does inside the [[ … ]] test used here?
When used inside [[ … ]], != applies an extended regular expression to the left-hand string, so a glob like *.log would be taken literally and never match.
Inside [[ … ]], != interprets the right-hand operand as an unquoted glob pattern and is true when the left-hand string does not match that pattern.
!= performs numeric inequality inside [[ … ]]; to compare strings you should use -ne instead.
!= compares two strings while ignoring case differences, so DEBUG.LOG and debug.log are considered identical unless a special option is set.
Within the double-bracket conditional [[ … ]], the operators == and != compare the left-hand string with a shell glob pattern on the right. Because the pattern side is unquoted, metacharacters such as * and ? are treated according to the normal filename-matching rules. The expression therefore returns a zero exit status ("true") when the left-hand string does not match the pattern and a non-zero status when it does. Numeric comparison would require -ne, and regular-expression matching would require =~. Case is significant unless the nocasematch shell option is enabled, so log and LOG are not considered the same by default.
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What is the difference between `!=` and `==` in a `[[ … ]]` test?
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What is the difference between a glob pattern and a regular expression?
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How does case sensitivity work in string comparisons within `[[ … ]]`?
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