为什么不能链接String.replace？ [英] Why can't I chain String.replace?

问题描述

我正在开发一种价格格式功能，该功能需要一个浮点数并正确表示它。

I'm working on a price format function, which takes a float, and represent it properly.

例如。 190.5，应该是190,50

ex. 190.5, should be 190,50

这就是我想出的内容

  def format_price(price) do
    price
    |> to_string
    |> String.replace ".", ","
    |> String.replace ~r/,(\d)$/, ",\\1 0"
    |> String.replace " ", ""
  end

如果我运行以下命令。

format_price(299.0)
# -> 299,0

看起来它只通过了第一次替换。现在，如果我将其更改为以下内容。

It looks like it only ran through the first replace. Now if i change this to the following.

  def format_price(price) do
    formatted = price
    |> to_string
    |> String.replace ".", ","

    formatted = formatted
    |> String.replace ~r/,(\d)$/, ",\\1 0"

    formatted = formatted
    |> String.replace " ", ""
  end

然后一切似乎都正常。

Then everything seems to work just fine.

format_price(299.0)
# -> 299,00

为什么？

推荐答案

编辑在Elixir的master分支上，如果有参数，则将函数插入没有括号的编译器将发出警告。

EDIT On the master branch of Elixir, the compiler will warn if a function is piped into without parentheses if there are arguments.

这是一个优先级问题，可以使用明确的括号来解决：

This is an issue of precedence that can be fixed with explicit brackets:

price
|> to_string
|> String.replace(".", ",")
|> String.replace(~r/,(\d)$/, ",\\1 0")
|> String.replace(" ", "")

因为函数调用的优先级高于 |> 运算符，您的代码与以下代码相同：

Because function calls have a higher precedence than the |> operator your code is the same as:

price
|> to_string
|> String.replace(".",
  ("," |> String.replace ~r/,(\d)$/,
    (",\\1 0" |> String.replace " ", "")))

如果我们替换最后一个子句，则： / p>

Which if we substitute the last clause:

price
|> to_string
|> String.replace(".",
  ("," |> String.replace ~r/,(\d)$/, ".\\10"))

再次：

price
|> to_string
|> String.replace(".", ",")

应解释为什么会得到该结果。

Should explain why you get that result.

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