Go 中的“未知转义序列”错误

我用 Go 编写了以下函数。其思想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。如果没有找到 IP 地址，则返回一个空字符串。

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。

我怎样才能告诉 Go，'.'就是我正在寻找的字符？我以为逃出去就能解决问题，但显然我错了。

61657

小开

最佳答案

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 include \\ two backslashes in a row to obtain an a single backslash character in the parsed value.

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]+`)

These are described in the "String literals" section of the Go spec.