Question

我是ajax初学者。我试图写一个简单的搜索建议。当我使用firefox测试它工作，但当我使用chrome我得到了：未捕获TypeError：无法读取属性＆＃39; getElementsByTagName＆＃39;为null

我在回调函数

上写了这个

xhr.onreadystatechange = function () {
    if (xhr.readyState === 4 && xhr.status === 200) {
        var data = xhr.responseXML;
        var x = data.getElementsByTagName("suggestion");
        var div = document.getElementById("suggest")
        div.innerHTML = "";
        showSuggest();
        for (var i = 0; i < x.length; i++) {
            var result= x[i].firstChild.nodeValue;
            div.innerHTML += "<div id='sResult' onmouseover='over(this)' onmouseout='out(this)' onclick='replace(this)' >"
                + result+ "</div>";
        }
    }
}

和Servlet doGet方法：

 protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        response.setContentType("test/xml;charset=UTF-8");
        response.setHeader("Cache-Control", "no-cache");
        response.setHeader("Pragma", "no-cache");
        String keyword = request.getParameter("keyword");

        //get suggestion from  db
        List<KeyWord> kw = SearchService.suggest(keyword);
        PrintWriter pw = response.getWriter();
        pw.println("<?xml version='1.0' encoding='UTF-8'?>");
        pw.write("<suggestions>");
        for(KeyWord k:kw){
            pw.write("<suggestion>");
            pw.write(k.getContent());
            pw.write("</suggestion>");
        }
        pw.write("</suggestions>");
        pw.flush();
        pw.close();
    }

Answer 1

如果您的帖子中response.setContentType("test/xml;charset=UTF-8");不是拼写错误，请将其更正为response.setContentType("text/xml;charset=UTF-8");，并希望IE和Chrome填充responseXML。

Answer 2

你试过这个吗？

var x = data.documentElement.getElementsByTagName("suggestion");

而不是

var x = data.getElementsByTagName("suggestion");

Ajax，xhr.responseXML.getElementsByTagName（）不是在chrome和IE上工作但是在firefox上工作

2 个答案: