bash - Bash "not"：反转命令的退出状态

Question

我知道我可以做到这一点...

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？即类似这样的东西（这显然不起作用）

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

我可以做这样的事情......

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

在这里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来做到这一点？

score 13 · Accepted Answer

13

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

于 2013-02-25T17:45:38.343 回答

score 2 · Accepted Answer

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（只需反转 if/else 操作）：

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

或者，尝试使用cmp：

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

score 0 · Accepted Answer

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

! EXPRESSION
      EXPRESSION is false

不太清楚为什么你需要否定，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"

3 回答 3