linux - 使用输入参数执行 shell 脚本

Question

下面的 shell 脚本询问用户小时和分钟。当您输入这两个输入时，它会安排成像。当我尝试以 root 身份运行此脚本时: bash -x /run/opt/corp/cmtg/2.0.0/sesm/admin/scripts/configure_gwcautoimaging 13 56 .我收到以下错误:typset:'13': not a valid identifier and typset:'56': not a valid identifier.

我的问题是如何在一行中执行带有参数的脚本。我也试过./run/opt/corp/cmtg/2.0.0/sesm/admin/scripts/configure_gwcautoimaging << "16 16"和 sh /run/opt/corp/cmtg/2.0.0/sesm/admin/scripts/configure_gwcautoimaging 16 50但他们没有工作我得到了同样的错误。

#!/bin/bash
    DEBUG="";
    #DEBUG="set -x";
    eval $DEBUG;
    typeset inputHour=2;
    typeset inputMinute=0;
    typeset crontabsFile="/etc/crontab";
    typeset tmp_crontabsFile="/tmp/crontab.tmp";
    typeset imagingsh=$BIN_DIR/autoimaging.sh;
    function validateNumber
    {
      eval $DEBUG;
      typeset -i vInput=$1;
      if [[ $? != 0 ]]
      then
        return 1;
      fi
      typeset -i vBase=$2;
      if [[ $? != 0 ]]
      then
        return 1;
      fi

      if (( vInput < 0 || vInput >= vBase ))
      then
        return 1;
      fi
      return 0;
    }
    function updatecrontabs
    {
      eval $DEBUG;
      typeset hour=$1;
      typeset minute=$2;

      #remove any entries that already have the autoimaging.sh
      /bin/grep -v "${imagingsh}" $crontabsFile > ${tmp_crontabsFile};   

      cat <<- EOF >> $tmp_crontabsFile
    $inputMinute $inputHour * * * $imagingsh
    EOF
      crontab ${tmp_crontabsFile}  
      if [[ $? != 0 ]]
      then
        return 1;
      fi
      cp ${tmp_crontabsFile} ${crontabsFile}; 
    }
    # If this script is invoked with two parameters, parse the two parameters as Hour and Minute
    # Otherwise, get input from user.
    if [[ $# != 2 ]]
    then
      echo " Configure Auto Imaging ";
      while ( true )
      do
        echo -n "Please input the hour when the Auto Imaging is to be executed: ";
        read inputHour
        if [[ -n $inputHour ]]
        then
          validateNumber $inputHour 24;
          if [[ $? == 0 ]]
          then
            break;
          else
            echo "You must input the number between 0 and 24";
        echo " ";
          fi
        else
          echo "You must input the number between 0 and 24";
          echo " ";
         fi 
      done
      while ( true )
      do
        echo -n "Please input the minute when the Auto Imaging is to be executed: ";
        read inputMinute
        if [[ -n $inputMinute ]]
        then
          validateNumber $inputMinute 60;
          if [[ $? == 0 ]]
          then
            break;
          else
            echo "You must input the number between 0 and 60";
        echo " ";
          fi
        else
          echo "You must input the number between 0 and 60";
          echo " ";
        fi  
      done
    else
      inputHour=$1;
      inputMinute=$2;
      typeset vResult=validateNumber $inputHour 24;
      if [[ $vResult != 0 ]]
      then
        echo "The number for hour must be between 0 and 24";
        exit 1;
      fi
      vResult= validateNumber $inputMinute 60;
      if [[ $vResult != 0 ]]
      then
        echo "The number for minute must be between 0 and 60";
        exit 1;
      fi 
    fi
    # Begin update the entry: auto imaging in crontabs
    updatecrontabs $inputHour $inputMinute;
    if [[ $? == 0 ]]
    then
      echo "update crontabs successfully! The auto imaging will be executed at: " $inputHour ":" $inputMinute;
      echo " ";
    else
      echo "Errors happen during update crontabs";
      echo " "
      exit 1;
    fi
    rm ${tmp_crontabsFile} > /dev/null;
    exit 0;

Answer 1

这看起来很可疑:

typeset vResult=validateNumber $inputHour 24;

它不调用该函数，您需要使用 $(validateNumber $inputHour 24) 或反引号。

同样，下面这行可能是错误的:

vResult= validateNumber $inputMinute 60;

它清除 $vResult 并以 $inputMinute 和 60 作为参数调用 validateNumber。

如果我调用 typeset 并将数字作为第一个参数，我会得到同样的错误，但我在您的代码中没有看到这样的用法。您确定粘贴的是导致问题的版本吗？

a=12 ; typeset -i $a=b
bash: typeset: `12=b': not a valid identifier