Bash“not”：反转命令的退出状态

Question

我知道我可以这样做......

if diff -q $f1 $f2
then
    echo "they're the same"
else
    echo "they're different"
fi

但是，如果我想否定我正在检查的条件怎么办？即这样的事情（显然不起作用）

if not diff -q $f1 $f2
then
    echo "they're different"
else
    echo "they're the same"
fi

我可以做这样的事......

diff -q $f1 $f2
if [[ $? > 0 ]]
then
    echo "they're different"
else
    echo "they're the same"
fi

我检查上一个命令的退出状态是否大于0.但这感​​觉有点尴尬。有没有更惯用的方法呢？

Answer 1

if ! diff -q $f1 $f2; then ...

Answer 2

如果你想否定，你正在寻找!：

if ! diff -q $f1 $f2; then
    echo "they're different"
else
    echo "they're the same"
fi

或（简化if / else动作的反转）：

if diff -q $f1 $f2; then
    echo "they're the same"
else
    echo "they're different"
fi

或者，也可以尝试使用cmp：

执行此操作

if cmp &>/dev/null $f1 $f2; then
    echo "$f1 $f2 are the same"
else
    echo >&2 "$f1 $f2 are NOT the same"
fi

Answer 3

否定使用if ! diff -q $f1 $f2;。记录在man test：

! EXPRESSION
      EXPRESSION is false

不完全确定为什么需要否定，因为你处理这两种情况......如果你只需要处理它们不匹配的情况：

diff -q $f1 $f2 || echo "they're different"