我如何期待’echo -e“？c”’？

我被要求編寫一個與另一個互動的腳本 - 但我被卡住了。我與之互動的腳本回顯以下內容，我需要發回“1”。但我不能完全到達那裡……

echo -e "Select option 1"
echo -e "? \c"

到目前為止，我已經嘗試過 - 據我所知：

expect "?"
send "1" 

expect "? "
send "1"

expect "? \n"
send "1"

expect "? \c"
send "1"

這一切似乎都不起作用。有人可以在正確的方向上輕推我嗎…？:)

PS：我想一旦我跨過第一個障礙，我就需要添加\r一個1…

在echo -e "? \c"中，該\c部分不是列印出來的任何東西，它是命令的指令，在echo作為參數傳遞的字元串之後不列印換行符¹。所以在期望中，你需要期望字元串"? "（問號，空格）。由於expect命令的參數?是萬用字元的模式，因此您需要按字面解釋問號：

expect -ex "? "
send "1\r"

¹的其他一些實現echo，例如內置的 bash，使用此語法echo -n "?"。

您快到了。考慮使用read內置（來自TDLP：Catching User Input）：

閱讀範例

cat leaptest.sh

#!/bin/bash
# This script will test if you have given a leap year or not.

echo "Type the year that you want to check (4 digits), followed by [ENTER]:"

read year

if (( ("$year" % 400) == "0" )) || (( ("$year" % 4 == "0") && ("$year" % 100 !=
"0") )); then
 echo "$year is a leap year."
else
 echo "This is not a leap year."
fi

注意第 6 行。變數 year 是由 BASH 動態創建的，用於保存 echo 語句的輸出。

測試使用者輸入範例

cat friends.sh

#!/bin/bash

# This is a program that keeps your address book up to date.

friends="/var/tmp/michel/friends"

echo "Hello, "$USER".  This script will register you in Michel's friends database."

echo -n "Enter your name and press [ENTER]: "
read name
echo -n "Enter your gender and press [ENTER]: "
read -n 1 gender
echo

grep -i "$name" "$friends"

if  [ $? == 0 ]; then
 echo "You are already registered, quitting."
 exit 1
elif [ "$gender" == "m" ]; then
 echo "You are added to Michel's friends list."
 exit 1
else
 echo -n "How old are you? "
 read age
 if [ $age -lt 25 ]; then
   echo -n "Which colour of hair do you have? "
   read colour
   echo "$name $age $colour" >> "$friends" 
   echo "You are added to Michel's friends list.  Thank you so much!"
 else
   echo "You are added to Michel's friends list."
   exit 1
 fi
fi

在您的特定情況下，您將使用腳本中的選項列表替換第 5行，並從第 17 行開始修改 if 以匹配您作為ANS. 如果選項與 匹配if，則執行您的腳本，如sh myscript.sh --option ANS

引用自：https://unix.stackexchange.com/questions/174724