尝试使ajax php后退按钮正常工作

我尝试使用pushState（）和popstate（）使php和ajax返回按钮工作。我让pushState在某种程度上起作用了，但popState却不起作用

            for ($i=1; $i<=$lastPage; $i++) {  
            //$pageLink[] = '<a href="ajax.php?action=merchantlistpage&merchant_id=' . $merchant_id . '&p='.$i.'" onclick="showmerchantitems(' . $merchant_id . ', ' . $i . ');">'.$i.'</a>';  
            $pageLink[] = '<a href="#p' . $i . '" onclick="showmerchantitems(' . $merchant_id . ', ' . $i . ');">'.$i.'</a>';  

        };  

用于（$i=1；$i）
<script>

function showmerchantitems(str, str2) {

    var elements = document.getElementsByClassName("itemwrapper");

    if (str.length == 0) { 

        for (var i = 0; i < elements.length; i++) {
            elements[i].innerHTML = "";
        }

        return;
    } else {
        var xmlhttp = new XMLHttpRequest();
        xmlhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {

                for (var i = 0; i < elements.length; i++) {
                    elements[i].innerHTML = this.responseText;
                }

            }
        };
        xmlhttp.open("GET", "ajax.php?action=merchantlistpage&merchant_id=" + str + "&p=" + str2, true);
        xmlhttp.send();

        history.pushState(null, "", "ajax.php?action=merchantlistpage&merchant_id=" + str + "&p=" + str2);

        $(window).bind("popstate", function(){
            $.getScript(location.href);
        });
    }
}

</script>