参数检查控制逻辑和语法混乱

Question

我有以下几点：

if [ ! $# == 2 ] || [ $1 != "mount" -o $1 != "unmount" ]; then

例如，对于当前的||运算符，没有任何东西可以通过验证器，但是如果我使用&&运算符，所有的东西都可以通过验证器。

if [ ! $# == 2 ] && [ $1 != "mount" -o $1 != "unmount" ]; then

有人能帮我解决这个问题吗？

下面是整个代码块和预期用途

if [ ! $# == 2 ] || [ $1 != "mount" -o $1 != "unmount" ]; then
  echo "Usage:"
  echo "encmount.sh mount remotepoint       # mount the remote file system"
  echo "encmount.sh unmount remotepoint   # unmount the remote file system"
  exit
fi

Answer 1

你可以这样做：

if [ "$#" -ne 2 ] || [ "$1" != "mount" -a "$1" != "unmount" ]; then
    echo "Usage:"
    echo "encmount.sh mount remotepoint       # mount the remote file system"
    echo "encmount.sh unmount remotepoint   # unmount the remote file system"
    exit -1
fi
echo "OK" 

您的测试中有一个小的逻辑错误，因为如果$1不等于"mount"和"unmount"，您应该进入usage分支。此外，您还应该使用-eq和-ne运算符(see here)或使用(( ))来比较数字。

请注意，您应该在test ([ ])中引用变量

您也可以像这样组合这两个表达式：

if [ "$#" -ne 2 -o \( "$1" != "mount" -a "$1" != "unmount" \) ]; then

如果您有]]，还可以使用[[ bash语法：

if [[ $# -ne 2 || ( $1 != "mount" && $1 != "unmount" ) ]]; then