Question

我对如何使用AJAX从.xml文件中读取信息感到困惑。我希望程序执行的只是显示用户可以按下的按钮，然后它将在.xml文件中显示电影列表。现在我的网页显示按钮，但当用户按下它时没有任何反应。我将发布我的3个文件，希望有人可以帮助我。

的index.html

<!DOCTYPE html>
<html>
<head>
    <title>AJAX - responseXML</title>
    <script src="ajax.js"></script>
    <script>
    function getMovies() {
            var xmlHttp = xmlHttpObjCreate();
            if (!xmlHttp) {
                    alert("The browser doesn't support this action.");
                    return;
            }

            xmlHttp.onload = function() {
                    if (xmlHttp.status == 200) {
                            // Get XML Document
                            var xmlDoc = xmlHttp.responseXML;

                            // Variable for our output
                            var output = '';

                            // Build output by parsing XML
                            movieTitles = xmlDoc.getElementsByTagName('title');

                            for (i = 0; i < movieTitles.length; i++) {
                                    output += movieTitles[i].childNodes[0].nodeValue + "<br>";
                            }

                            // Get div object
                            var divObj = document.getElementById('movieBox');

                            // Set the div's innerHTML
                            divObj.innerHTML = output;
                    }
            }

            xmlHttp.open("GET", "movies.xml", true);
            xmlHttp.send();
    }
    </script>
</head>
<body>
    <h3>My Movies</h3>
    <div id="movieBox"></div>
    <button type="button" onclick="getMovies()">Get Movie Titles</button>
</body>
</html>

ajax.js

function xmlHttpObjCreate() {
var xmlHttp; 
try
 {xmlHttp=new XMLHttpRequest();
 }
catch (e)
{try
  {
 xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
 }
catch (e)
  {
 try
   {
   xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
   }
 catch (e)
   {
   return false;
   }
 }
}
 return xmlHttp;
}

movies.xml

 <?xml version= "1.0"?>
 <collection>
    <movies>
            <title> Gone with the Wind
                        <year> 1939 </year>
                                  </title>

            <title> 2001: A Space Odyssey
                            <year> 1968 </year>
                                            </title>

            <title> Jurassic Park
                            <year> 1993 </year>
                                            </title>

            <title> The Shawshank Redmption
                            <year> 1994 </year>
                                            </title>

            <title> Balto
                            <year> 1995 </year>
                                            </title>

    </movies>
</collection>

这是我得到的错误

Uncaught TypeError: Cannot read property 'getElementsByTagName' of null

如何使用AJAX显示xml信息

0 个答案: