linux - 如何保存来自 telnet 脚本的响应？

Question

我的脚本可以连接、登录和运行命令。但是我不知道如何将命令的响应保存到文件，而不保存整个 session 。

#!/bin/sh
Var=1
while [ $Var -lt 20 ]
do
HOST='IPa.ddr.ess.'$Var
USER='MyUser'
PASSWD='MyPassword'
CMD='MyCommand'

(
    echo open "$HOST"
    sleep 1
    echo "$USER"
    sleep 1
    echo "$PASSWD"
    sleep 1
    echo "$CMD"
    #I want to save the output from my $cmd to an varaible $Output
    #Then I want to write "$HOST - $Output" to a file named "output.txt"
    sleep 2
    echo "exit"
    ) | telnet
Var=$((Var + 1))
done

我将不胜感激任何帮助或正确方向的指示

Answer 1

好的，这看起来比我最初想象的更具挑战性。我喜欢它:-)

#!/bin/sh
Var=1
while [ $Var -lt 20 ]
do
HOST='IPa.ddr.ess.'$Var
USER='MyUser'
PASSWD='MyPassword'
CMD='MyCommand'
MARKER='XXXX1234:AUIE'

(echo "$HOST - " ; (
    echo unset echo
    echo open "$HOST"
    sleep 1
    echo "$USER"
    sleep 1
    echo "$PASSWD"
    sleep 1
    echo echo "$MARKER"
    echo "$CMD"
    #I want to save the output from my $cmd to an varaible $Output
    #Then I want to write "$HOST - $Output" to a file named "output.txt"
    sleep 2
    echo "exit"
    ) | telnet | sed -e "1,/$MARKER/d" ) >> output.txt
Var=$((Var + 1))
done

它的作用是:

• 它在 telnet 中禁用回显
• 登录 session 后，打印一个标记
• 标记保存到 output.txt 之后的任何内容
• 我嵌入了另一个 shell，它将打印“$HOST -”部分