如何将“空格”和“或”条件与bash脚本匹配？

Question

我想在bash中匹配一个包含字符串中的“空格”和"OR“条件的条件。我无法执行新的shell脚本编写，请帮助它转到其他循环中，它将匹配${myarrlin1}。我得到了"Centrify是残疾人“，但我希望条件是真实的这里。这是我的代码：

FILE_CENT="/etc/nsswitch.conf" 
OS=`uname`    
if [[ $OS = 'Linux' ]]; then
 if [[ -e $FILE_CENT ]]; then
echo "nsswitch.conf found, Proceeding further..."
  while read -r LINE
  do
  if [[ $LINE =~ ^passwd ]]; then
  myarrlin=($LINE)
    if [[ ${myarrlin[1]} =~ ^(centrify)|(centrifydc)[[:space:]]* || ${myarrlin[1]} =~ [[:space:]]+(centrify)|(centrifydc)[[:space:]]* ]];  then
     echo "Centrify is enabled"
     else
     echo "Centrify is disabled"
    fi
  fi
  done < $FILE_CENT
else
echo "nsswitch.conf does not exist in $OS, cannot fetch CENTRIFY information!"
 fi 
fi

nsswitch.conf >>>

passwd:     centrify files
  or
passwd:     centrifydc files
  or
passwd:     files centrify
  or
passwd:     files centrifydc

Answer 1

你为什么要这么做：myarrlin=($LINE)？

如果您只想知道这条线是否包含centrify：

  while read -r LINE
  do
  if [[ ${LINE} =~ ^passwd ]]; then
    if [[ ${LINE} == *"centrify"* ]];  then
     echo "Centrify is enabled"
    else
     echo "Centrify is disabled"
    fi
  fi
  done < $FILE_CENT