Question

我在自己的主机上有一个外部mysql数据库，我还在服务器端创建了php文件，它将从数据库中提取记录并将它们放入xml格式。（如下所示）

<?php
mysql_connect("localhost","username","password") or die("Could not      connect to host.");
mysql_select_db("footieDB") or die("Could not find database.");
$resultID=mysql_query("select * from league"); //$resultID was $sql


$xml_output = "<?xml version=\"1.0\"?>\n"; 
$xml_output .= "<leagues>\n"; 

for($x = 0 ; $x < mysql_num_rows($resultID) ; $x++){ 
    $row = mysql_fetch_assoc($resultID); 
    $xml_output .= "\t<league>\n"; 
    $xml_output .= "\t\t<leagueID>" . $row['leagueID'] . "</leagueID>\n"; 
    $xml_output .= "\t\t<leagueName>" . $row['leagueName'] . "</leagueName>\n"; 
    $xml_output .= "\t</league>\n"; 
} 

$xml_output .= "</leagues>"; 

echo $xml_output;

mysql_close();
?>

我想知道如何解析这个xml以便在android模拟器中显示信息。任何例子都将不胜感激。

提前致谢。

Answer 1

老实说，您应该将数据放在JSON格式而不是XML中。 XML解析并不好玩，JSON快速而简单。如果你确定了：

String example1;
String example2;
// establish a connection with the server
            HttpClient httpclient = new DefaultHttpClient();
            HttpGet httpget = new HttpGet("http://www........./your_stuff.xml");

            // execute the post and get the response
            HttpResponse response = httpclient.execute(httpget);

            byte[] responsebody = EntityUtils.toByteArray(response.getEntity());
            String result = new String(responsebody, 0, responsebody.length, "UTF-8");

            // Go through the XML with an XML pull parser, identifying items
            // by their tag names
            final XmlPullParser xpp = Xml.newPullParser();
            xpp.setInput(new StringReader(result));
            int eventType = xpp.getEventType();
            while (eventType != XmlPullParser.END_DOCUMENT) {
                if(eventType == XmlPullParser.START_TAG) {
                    if (xpp.getName().equalsIgnoreCase("example1")) {
                        eventType = xpp.next();
                        example1 = xpp.getText();
                    } else if (xpp.getName().equalsIgnoreCase("example2")) {
                        eventType = xpp.next();
                        example2 = xpp.getText();
                    }
                } 
            eventType = xpp.next();
            }
           ....... etc

我如何解析xml来显示android模拟器中的内容？

1 个答案: