在Golang中提取部分字符串？

Question

I'm learning Golang so I can rewrite some of my shell scripts.

I have URL's that look like this:

https://example-1.example.com/a/c482dfad3573acff324c/list.txt?parm1=value,parm2=value,parm3=https://example.com/a?parm1=value,parm2=value

I want to extract the following part:

https://example-1.example.com/a/c482dfad3573acff324c/list.txt

In a shell script I would do something like this:

echo "$myString" | grep -o 'http://.*.txt'

What is the best way to do the same thing in Golang, only by using the standard library?

Answer 1

There are a few options:

// match regexp as in question
pat := regexp.MustCompile(`https?://.*\.txt`)
s := pat.FindString(myString)

// everything before the query 
s := strings.Split(myString, "?")[0] string

// same as previous, but avoids []string allocation
s := myString
if i := strings.IndexByte(s, '?'); i >= 0 {
    s = s[:i]
}

// parse and clear query string
u, err := url.Parse(myString)
u.RawQuery = ""
s := u.String()

The last option is the best because it will handle all possible corner cases.

<kbd>try it on the playground</kbd>