tcl - 如何在expect脚本中清理之前的expect_buf

Question

我写了一个期望函数来获取命令的输出，我的代码如下

proc do_cmd {cmd id} {
  set spawn_id $id
  send "$cmd\r"
  expect "$cmd\r"
  expect {
    -re "\n(.*)\r\n" {return $expect_out(1,string)}
    default {exit 1}
  }
}

如果我只调用一次函数，它会工作正常并返回我想要的东西，但如果我不停地调用它，它会返回不需要的东西。

# test case 1
set ret [do_cmd $mycmd $spawn_id]
puts "$mycmd returns $ret"  # the return value is ok

# test case 2
set ret [do_cmd $mycmd $spawn_id]
set ret [do_cmd $mycmd $spawn_id]
puts "$mycmd returns $ret"  # the return value is not something I want

我使用'exp_internal 1'来调试它，发现第二个调用命令中的expect_out仍然保存着之前的输出信息并导致了匹配的问题，那么我该如何清理expect_out缓冲区(我试图将其设置为空字符串但它不起作用，)或者我还能做些什么来避免这个问题？提前致谢。

Answer 1

Don Libes 对您的方案的建议如下，

Sometimes it is even useful to say:

expect *

Here the * matches anything. This is like saying, "I don't care what's in the input buffer. Throw it away." This pattern always matches, even if nothing is there. Remember that * matches anything, and the empty string is anything! As a corollary of this behavior, this command always returns immediately. It never waits for new data to arrive. It does not have to since it matches everything.

引用: Exploring Expect

在这种情况下，在您所需的匹配之后，最好尝试将匹配保存到某个变量，然后简单地添加代码 expect *最后。这将清空缓冲区。您的代码可以更改如下。

proc do_cmd {cmd id} {
  set spawn_id $id
  send "$cmd\r"
  #Looks like you are looking for a particular command to arrive
  expect "$cmd\r"
  #Then you have one more expect here which is you want to get it
  expect {
    #Saving the value sub match to the variable 'result'
    -re "\n(.*)\r\n" {set result $expect_out(1,string)}}
    }
  #Causing the buffer to clear and it will return quickly
  expect *
  return $result
}

除此之外，还有一种方法可以取消设置 expect_out(buffer)内容本身将从 expect_out 中删除“缓冲区”索引可以表示为的数组

unset expect_out(buffer)

当下一场比赛发生时， expect_out数组将更新索引“缓冲区”，我们可以获得新的 expect_out(buffer)值(value)。更换 expect *如果您更喜欢使用这种方式，请使用上面的代码。

这是获得我们真正想要的东西的一种解决方法。您可以继续使用任何方法。选择权在你。 :)