Go implementation Impolde method

This is a creation in Article, where the information may have evolved or changed.

Impolde method is commonly used in PHP string stitching method, in Golang also has the function of string concatenation:

strings.Join(a []string, sep string) string

This function is similar to implode, but has one drawback: the first argument must be []string, and if you want to stitch []int into a string, you need to convert.
The following uses reflection to implement a common "implode" approach:

func Implode(list interface{}, seq string) string {    listValue := reflect.Indirect(reflect.ValueOf(list))    if listValue.Kind() != reflect.Slice {        return ""    }    count := listValue.Len()    listStr := make([]string, 0, count)    for i := 0; i < count; i++ {        v := listValue.Index(i)        if str, err := getValue(v); err == nil {            listStr = append(listStr, str)        }    }    return strings.Join(listStr, seq)}func getValue(value reflect.Value) (res string, err error) {    switch value.Kind() {    case reflect.Ptr:        res, err = GetValue(value.Elem())    default:        res = fmt.Sprint(value.Interface())    }    return}

Then make a simple test:

func TestImplode(t *testing.T) {    list := []int{1, 2, 3, 4, 5, 6, 7}    res := Implode(list, ",")    fmt.Println(res)    list1 := []int16{1, 2, 3, 4, 5, 6, 7}    res = Implode(list1, ",")    fmt.Println(res)    list2 := []float64{1.5, 2.1, 3.0, 4, 5.9, 6.7, 7.7}    res = Implode(list2, ",")    fmt.Println(res)    var list3  []float64    res = Implode(list3, ",")    fmt.Println("res: ", res)    list4 := make([]interface{}, 4, 4)    list4[0] = "str"    list4[1] = 19    list4[2] = 19.999    list4[3] = true    res = Implode(list4, ",")    fmt.Println(res)}

Output:

1,2,3,4,5,6,71,2,3,4,5,6,71.5,2.1,3,4,5.9,6.7,7.7res:  str,19,19.999,true

is not convenient many:)