Can this function be achieved

[rookiefront]

Can this function be achieved，If you have enough time
jquery - How to calculate the XPath position of an element using Javascript?

Javascript get XPath of a node

[rookiefront]

function getXPathForElement(element) {
    const idx = (sib, name) => sib
        ? idx(sib.previousElementSibling, name||sib.localName) + (sib.localName == name)
        : 1;
    const segs = elm => !elm || elm.nodeType !== 1
        ? ['']
        : elm.id && document.getElementById(elm.id) === elm
            ? [`id("${elm.id}")`]
            : [...segs(elm.parentNode), `${elm.localName.toLowerCase()}[${idx(elm)}]`];
    return segs(element).join('/');
}

[rookiefront]

my problem is solved

func  GetCssSelector(elm *goquery.Selection) string {

	path := []string{}
	for {
		selector := strings.ToLower(goquery.NodeName(elm))
		if selector =="" {
			break
		}
		//如果是ID选择器直接退出
		id_val, id_exists := elm.Attr("id")
		if id_exists && id_val != "" {
			selector+="#"+id_val
			path = append(path,selector)
			break
		}else{
			sib := elm
			nth := 1
			sib.PrevAll().Each(func(i int, selection *goquery.Selection) {
				//fmt.Println("234",strings.ToLower(goquery.NodeName(selection)) , selector)
				if strings.ToLower(goquery.NodeName(selection)) == selector {
					nth++
				}

			})
			if nth != 1 {
				selector += fmt.Sprintf(`:nth-of-type(%d)`,nth)
			}
		}
		path = append(path,selector)
		elm = elm.Parent()
	}
	path = Reverse(path)
	join := strings.Join(path, " > ")

	return join
}

[mna]

For future reference, see #198 . Closing since you seem to be happy with the solution you have.