Question

我正在UNIX中编写一个程序加倍编号

if [ $# == l ]; then
  let "twice=$l*2

if [ $? == 0 ]; then
   clear
   echo "twice Program"
   echo "-------------"
   echo "l * 2 = $twice"
   exit 0
 else
   clear
   echo "The argument must be an integer."
   exit 1
 fi

 exit 0
else
 clear
 echo "Only one argument is acceptable with twice."
 echo :Usage: twice argument"
 exit 1
fi

Answer 1

您在第二行打开双引号，但未正确关闭。

  let "twice=$l*2

应该是

  let "twice=$1*2"

shell解释器在处理不相关的双引号时会出现阻塞。

请注意，您也会在多个地方将l（ell）字母与数字1（一个）混淆，因为它不会让shell了解您的脚本。

Answer 2

您正在使用l（小L）而不是1

if [ $# ==  1 ]
then
  let "twice=$1*2"
  if [ $? == 0 ]; then
     echo "twice Program"
     clear
     echo "-------------"
     echo "$1 * 2 = $twice"
     exit 0
 else
     clear
     echo "The argument must be an integer."
     exit 1
 fi

 exit 0
 else
   clear
   echo "Only one argument is acceptable with twice."
   echo ":Usage: twice argument"
   exit 1
fi

在第7行获得一元运算符期望的错误消息

2 个答案: