Question

我需要使用脚本将xxx#yyy@zzz转换为ZZZ xxx-yyy。我已经在脚本中使用了case语句，所以我应该在case语句之前使用if语句还是应该继续case语句？这是我目前的剧本

function usage
{
  echo "usage: $0 arguments ..."
  if [ $# -eq 1 ]
  then echo "ERROR: $1"
  fi
}

# Script begins here
if [ $# -gt 0 ]
then
  echo "Joe Fosteson"
  date
  echo
   for arg
  do
    case $arg in
      rootjobs)
        echo "Root is running $(ps -U root -u root u | wc -l) processes on $(una
me)."
         ;;
      student_accounts)
        echo "There are $(ls -1 /home/student/ | wc -l) student accounts on $(ho
stname)."
         ;;
      *)
        usage  "Do not know what to with $arg"
         ;;
    esac

    echo "*****"
  done
else
  usage

fi

这是我使用的sed命令，但格式不正确........

if [ $# -eq 0 ]
then echo "Please enter a file name"
else echo
 if test -f $1
  then
    #s1.sh
    starting_data=$1
    sed 's/ //' $starting_data > raw_dataz
    sed "/^id/ d" raw_dataz > new
    cut -f1 -d, new > id1
    sed 's/ //' id1 > id2
    cut -f2 -d, new > lastname1
    sed 's/ //' lastname1 > lastname2
    cut -f3 -d, new > firstname1
    sed 's/ //' firstname1 > firstname2
    awk '{print $0":"}' lastname2 > lastname3
    sed -e 's/-//g' id2 > id3
    paste -d\  firstname2 lastname3 id3 > final
    cat final
  else echo "$1 cannot be found"
 fi
fi

Answer 1

要在bash中将xxx#yyy@zzz之类的字符串转换为ZZZ xxx-yyy：

str='xxx#yyy@zzz'
if [[ "$str" =~ (.+)#(.+)@(.+) ]]; then
  str2="${BASH_REMATCH[3]^^} ${BASH_REMATCH[1]}-${BASH_REMATCH[2]}"
  echo "$str2"
else
  echo "Bad string format"
fi

在sed

sed -r 's/(.+)#(.+)@(.+)/\U\3\E \1-\2' <<<"$str"

使用sed命令转换数据

1 个答案: