Go: Retrieve a string from between two characters or other strings - string

Question

Let's say for example that I have one string, like this:
<h1>Hello World!</h1>
What Go code would be able to extract Hello World! from that string? I'm still relatively new to Go. Any help is greatly appreciated!

Answer 1

If the string looks like whatever;START;extract;END;whatever you can use this which will get the string in between:
// GetStringInBetween Returns empty string if no start string found
func GetStringInBetween(str string, start string, end string) (result string) {
s := strings.Index(str, start)
if s == -1 {
return
}
s += len(start)
e := strings.Index(str[s:], end)
if e == -1 {
return
}
e += s + e - 1
return str[s:e]
}
What happens here is it will find first index of START, adds length of START string and returns all that exists from there until first index of END.

Answer 2

There are lots of ways to split strings in all programming languages.
Since I don't know what you are especially asking for I provide a sample way to get the output
you want from your sample.
package main
import "strings"
import "fmt"
func main() {
initial := "<h1>Hello World!</h1>"
out := strings.TrimLeft(strings.TrimRight(initial,"</h1>"),"<h1>")
fmt.Println(out)
}
In the above code you trim <h1> from the left of the string and </h1> from the right.
As I said there are hundreds of ways to split specific strings and this is only a sample to get you started.
Hope it helps, Good luck with Golang :)
DB

Answer 3

I improved the Jan Kardaš`s answer.
now you can find string with more than 1 character at the start and end.
func GetStringInBetweenTwoString(str string, startS string, endS string) (result string,found bool) {
s := strings.Index(str, startS)
if s == -1 {
return result,false
}
newS := str[s+len(startS):]
e := strings.Index(newS, endS)
if e == -1 {
return result,false
}
result = newS[:e]
return result,true
}

Answer 4

Here is my answer using regex. Not sure why no one suggested this safest approach
package main
import (
"fmt"
"regexp"
)
func main() {
content := "<h1>Hello World!</h1>"
re := regexp.MustCompile(`<h1>(.*)</h1>`)
match := re.FindStringSubmatch(content)
if len(match) > 1 {
fmt.Println("match found -", match[1])
} else {
fmt.Println("match not found")
}
}
Playground - https://play.golang.org/p/Yc61x1cbZOJ

Answer 5

In the strings pkg you can use the Replacer to great affect.
r := strings.NewReplacer("<h1>", "", "</h1>", "")
fmt.Println(r.Replace("<h1>Hello World!</h1>"))
Go play!

Answer 6

func findInString(str, start, end string) ([]byte, error) {
var match []byte
index := strings.Index(str, start)
if index == -1 {
return match, errors.New("Not found")
}
index += len(start)
for {
char := str[index]
if strings.HasPrefix(str[index:index+len(match)], end) {
break
}
match = append(match, char)
index++
}
return match, nil
}

Answer 7

Read up on the strings package. Have a look into the SplitAfter function which can do something like this:
var sample = "[this][is my][string]"
t := strings.SplitAfter(sample, "[")
That should produce a slice something like: "[", "this][", "is my][", "string]". Using further functions for Trimming you should get your solution. Best of luck.

Answer 8

func Split(str, before, after string) string {
a := strings.SplitAfterN(str, before, 2)
b := strings.SplitAfterN(a[len(a)-1], after, 2)
if 1 == len(b) {
return b[0]
}
return b[0][0:len(b[0])-len(after)]
}
the first call of SplitAfterN will split the original string into array of 2 parts divided by the first found after string, or it will produce array containing 1 part equal to the original string.
second call of SplitAfterN uses a[len(a)-1] as input, as it is "the last item of array a". so either string after after or the original string str. the input will be split into array of 2 parts divided by the first found before string, or it will produce array containing 1 part equal to the input.
if after was not found than we can simply return b[0] as it is equal to a[len(a)-1]
if after is found, it will be included at the end of b[0] string, therefore you have to trim it via b[0][0:len(b[0])-len(after)]
all strings are case sensitive