英文:
"Unknown escape sequence" error in Go
问题
我在Go中编写了以下函数。这个函数的想法是将一个字符串传递给它，并返回找到的第一个IPv4 IP地址。如果找不到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: .
我该如何告诉Go . 是我要寻找的实际字符？我以为转义它会起作用，但显然我错了。
英文:
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.
答案1
得分: 187
The \ backslash isn't being interpreted by the regex parser, it's being interpreted in the string literal. You should escape the backslash again:
<!-- language-all: go -->
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 <code> 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.
英文:
The \ backslash isn't being interpreted by the regex parser, it's being interpreted in the string literal. You should escape the backslash again:
<!-- language-all: go -->
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 <code>`</code> 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.
答案2
得分: 1
"(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])"
英文:
IPv4 address (accurate capture)
Matches 0.0.0.0 through 255.255.255.255
Use this regex to match IP numbers with accurracy.
Each of the 4 numbers is stored into a capturing group, so you can access them for further processing.
"(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])"
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