如何在 bash 脚本中将参数传递给内联期望？

Question

我正在创建一个 shell 脚本来自动安装具有自己的安装程序脚本的软件，但我不想向用户提示。所以我正在使用 expect 来回答所述脚本。现在我有以下内容：

expect $package -c 'set timeout -1
set package [lindex $argv 0]
spawn sh install_pipeline
expect "Where should I install the software packages ? *"
send "/usr/local/$package\r"
expect "Where should I install the pipeline calibration files ? *"
send "/usr/local/$package\r"
expect eof'

其中 package 是一个包含要安装的模块名称的变量。问题是，此脚本不起作用并打印一条消息："Couldnt read file "包的名称”：没有这样的文件或目录。

如何将变量传递给 expect 脚本？我不想为 expect 创建一个单独的脚本来保持简单。

Answer 1

您的 expect 命令用单引号括起来。里面的变量不能被shell.

展开

试试这个：

expect "$package" -c "set timeout -1
set package [lindex $argv 0]
spawn sh install_pipeline
expect \"Where should I install the software packages ? *\"
send \"/usr/local/$package\r\"
expect \"Where should I install the pipeline calibration files ? *\"
send \"/usr/local/$package\r\"
expect eof"

或者如果您想保留单引号：

expect "$package" -c 'set timeout -1
set package [lindex $argv 0]
spawn sh install_pipeline
expect "Where should I install the software packages ? *"
send "/usr/local/'"$package"'\r"
expect "Where should I install the pipeline calibration files ? *"
send "/usr/local/'"$package"'\r"
expect eof'

Answer 2

根据 Kenavoz 的回答，我找到了解决方案。正如他所说，我的代码被简单的引号括起来，所以 expect 无法从主脚本中获取变量。因此，只用普通引号解决了问题，而不需要将变量作为参数传递：

expect -c "set timeout -1
spawn sh install_pipeline
expect \"Where should I install the software packages ? *\"
send \"/usr/local/$package\r\"
expect \"Where should I install the pipeline calibration files ? *\"
send \"/usr/local/$package\r\"
expect eof"

Answer 3

可以通过环境传入：

env pkg="$package" expect -c '
    set timeout -1
    spawn sh install_pipeline
    expect "Where should I install the software packages ? *"
    send "/usr/local/$env(pkg)\r"
    expect "Where should I install the pipeline calibration files ? *"
    send "/usr/local/$env(pkg)\r"
    expect eof
'

如何在 bash 脚本中将参数传递给内联期望？

How to pass arguments to inline expect in a bash script?

bash

expect