如何评估/常量化字符串到方法，然后将参数传递给它

发布于 2024-12-07 15:47:20 字数 226 浏览 4 评论 0原文

给出：

s = "foo_bar_path"

我如何评估或常量化 s，并将参数传递给它，例如我的最终结果相当于：

foo_bar_path(@myvar, @foobar)

我正在尝试 eval(s).send 但这似乎并不工作。而且 Constantize 似乎只适用于类？

原文

Given:

s = "foo_bar_path"

How can I eval or constantize s, and pass arguments to it, such as my final result would be the equivalent of:

foo_bar_path(@myvar, @foobar)

I was trying eval(s).send but that doesn't seem to work. And constantize seems to only work on Classes?

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南薇 2024-12-14 15:47:20

您只需使用 send 方法（或 public_send根据您的需要）在适当的对象上：

some_object.send(s, @myvar, @foobar)

或者如果您想对自己调用方法：

self.send(s, @myvar, @foobar)

文档中说“符号”，但 send 也很乐意获取字符串中的方法名称。

You would just use the send method (or public_send depending on your needs) on the appropriate object:

some_object.send(s, @myvar, @foobar)

or if you want to call a method on yourself:

self.send(s, @myvar, @foobar)

The documentation says "symbol" but send is just as happy to get the method name in a string.

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临风闻羌笛 2024-12-14 15:47:20

另一种处理此类事情的方法（对于 OpenStruct 更好）：

#for more in this domain, see the Ruby core API for the Object class
class Demo
  def initialize(n)
    @iv = n
  end
  def hello()
    "Hello, @iv = #{@iv}"
  end
end

k = Demo.new(99)
m = k.method(:hello)
m.call   #=> "Hello, @iv = 99"

Another way to go about this sort of this sort of thing (better for OpenStruct):

#for more in this domain, see the Ruby core API for the Object class
class Demo
  def initialize(n)
    @iv = n
  end
  def hello()
    "Hello, @iv = #{@iv}"
  end
end

k = Demo.new(99)
m = k.method(:hello)
m.call   #=> "Hello, @iv = 99"

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