Include string into another string using regex - regex

Question

I have some set of strings. Strings might have items listed between square brackets. I'd like to include into strings with brackets a constant number of extra items. Brackets might be empty, or absent. For example:
string1 --> string1 # added nothing
string2[] --> string2[extra1="1",extra2="2"] # added two items
string3[item="1"] --> string3[item="1",extra1="1",extra2="2"] # added two items
Currently I achieve this with the following code (Golang):
str1 := "test"
str2 := `test[]`
str3 := `test[item1="1"]`
re := regexp.MustCompile(`\[(.+)?\]`)
for _, s := range []string{str1, str2, str3} {
s = re.ReplaceAllString(s, fmt.Sprintf(`[item1="a",item2="b",$1]`))
fmt.Println(s)
}
But in the output, in case of empty brackets I also got an unwanted comma "," in the end:
test
test[item1="a",item2="b",]
test[item1="a",item2="b",item1="1"]
Is it possible to avoid paste comma in case of empty brackets?
Of course it's possible to parse string again and trim the comma, but it seems suboptimal.
Code example on Go playground

Answer 1

You can have two regexes, where one matches for empty [] and other
matches for string with text inside []. Below is the tested code -
https://play.golang.org/p/_DOOGDMUOCm
Second way is just look back in the string after replacing it. If the
last two characters are ,] and you can substring till , and add ]. I
guess you already know this approach.
package main
import (
"fmt"
"regexp"
)
func main() {
str1 := "test"
str2 := `test[]`
str3 := `test[item1="1"]`
re := regexp.MustCompile(`\[(.*)\]`)
for _, s := range []string{str1, str2, str3} {
matched,err := regexp.Match(`\[(.+)\]`, []byte(s));
_ = err;
if(matched==true){
s = re.ReplaceAllString(s, fmt.Sprintf(`[item1="a",item2="b",$1]`));
}else {
s = re.ReplaceAllString(s, fmt.Sprintf(`[item1="a",item2="b"]`));
}
fmt.Println(s)
}
}