在 bash 脚本中期望

Question

我正在尝试将 Expect 脚本实现到 bash 脚本中。请耐心等待，因为我对 bash/expect 还很陌生。

这是按预期工作的期望脚本：

log_user 0

file delete foo.txt

set fh [open foo.txt a]

set servers {xxx@server1 xxx@server2}

foreach s $servers {
spawn ssh $s
expect "password: "
send "PASSWORD\r"
expect "$ "
send "grep "something" /some/log/file.log"
expect "$ " { puts $fh "$expect_out(buffer)"}
send "exit\r"
}

close $fh

现在，我希望将此 Expect 脚本包含在 bash 脚本中，但它没有按预期工作。

这是我到目前为止所拥有的：

#!/bin/bash

XYZ=$(expect -c "
file delete foo.txt

set fh [open foo.txt a]

set servers {xxx@server1 xxx@server2}

foreach s $servers {
spawn ssh $s
expect "password: "
send "PASSWORD\r"
expect "$ "
send "grep "something" /some/log/file.log"
expect "$ " { puts $fh "$expect_out(buffer)"}
send "exit\r"
}

close $fh
")

echo "$XYZ"

我收到的错误是：

command substitution: line 42: syntax error near unexpected token `('
command substitution: line 42: `expect "$ " { puts $fh "$expect_out(buffer)"}'

我愿意接受任何其他方式来实现这一点！:)

Answer 1

您可以使用/usr/bin/expect -c来执行expect命令：

#!/bin/bash

/usr/bin/expect -c ' 

file delete foo.txt

set fh [open foo.txt a]

set servers {xxx@server1 xxx@server2}

foreach s $servers {
spawn ssh $s
expect "password: "
send "PASSWORD\r"
expect "$ "
send "grep "something" /some/log/file.log"
expect "$ " { puts $fh "$expect_out(buffer)"}
send "exit\r"
}

close $fh
'