“未知转义序列” Go中的错误 [英] "Unknown escape sequence" error in Go

问题描述

我用Go编写了以下函数。这个想法是函数有一个字符串传递给它并返回找到的第一个IPv4 IP地址。

  func parseIp（checkIpBody string）string {
 reg ，err：= regexp.Compile（[0-9] + \。[0-9] + \。[0-9] + \。[0-9] +）
 if err == nil {
 return
} 
 return reg.FindString（checkIpBody）
} 
  

我得到的编译时错误是

未知的转义序列：。 p>

我该如何告诉Go '。'是实际的字符I'米寻找？我认为转义它会做到这一点，但显然我错了。 反斜杠不被正则表达式解析器解释，它在字符串文字中被解释。你应该再次避开反斜线：

  regexp.Compile（ [0-9] + \\ [0-9] + \\。[0-9] + \\。[0-9] +）
  

用双引号字符引用的字符串称为解释的字符串字面，在大多数语言中，解释字符串文字就像字符串文字： \ 反斜线字符不是字面意义上的，它们用于给下一个字符赋予特殊含义字符，源代码必须在一行中包含 \\ 两个反斜杠，以获得解析值中的单个反斜杠字符。

由于 Evan Shaw在评论中指出，Go有另一种选择这在为正则表达式编写字符串文字时很有用，原始字符串文字由<引用code>`反引号字符在原始字符串文本中没有特殊字符，所以只要你的模式不包含反引号，你可以使用这个语法而不会转义任何东西：

  regexp.Compile（`[0-9] + \。[0-9] + \\ \\。[0-9] + \。[0-9] +`）
  

这是如 Go spec 的字符串文字部分所述。

I have the following function written in Go. The idea is the function has a string passed to it and returns the first IPv4 IP address found. If no IP address is found, an empty string is returned.

func parseIp(checkIpBody string) string {
    reg, err := regexp.Compile("[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+")
    if err == nil {
        return ""
    }   
    return reg.FindString(checkIpBody)
}

The compile-time error I'm getting is

unknown escape sequence: .

How can I tell Go that the '.' is the actual character I'm looking for? I thought escaping it would do the trick, but apparently I'm wrong.

解决方案

The \ backslash isn't being interpreted by the regex parser, it's being interpreted in the string literal. You should escape the backslash again:

regexp.Compile("[0-9]+\\.[0-9]+\\.[0-9]+\\.[0-9]+")

A string quoted with " double-quote characters is known as an "interpreted string literal" in Go. Interpreted string literals are like string literals in most languages: \ backslash characters aren't included literally, they're used to give special meaning to the next character. The source must included \\ two backslashes in a row to obtain an a single backslash character in the parsed value.

As Evan Shaw pointed out in the comments, Go has another alternative which can be useful when writing string literals for regular expressions. A "raw string literal" is quoted by ` backtick characters. There are no special characters in a raw string literal, so as long as your pattern doesn't include a backtick you can use this syntax without escaping anything:

regexp.Compile(`[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+`)

This is described in the "String literals" section of the Go spec.

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