我想在 Android Activity 中发出 HTTP POST 请求。我(认为我)知道如何执行此操作,但我的问题是我不知道如何创建 XML 文件。我尝试过之前帖子中描述的不同方法,但没有成功。
我的xml格式如下:
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<IAM version="1.0">
<ServiceRequest>
<RequestTimestamp>2012-07-20T11:10:12Z</RequestTimestamp
<RequestorRef>username</RequestorRef>
<StopMonitoringRequest version="1.0">
<RequestTimestamp>2012-07-20T11:10:12Z</RequestTimestamp>
<MessageIdentifier>12345</MessageIdentifier>
<MonitoringRef>112345</MonitoringRef>
</StopMonitoringRequest>
</ServiceRequest>
</IAM>
我编写了以下 Java 代码行:
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
//What to write here to add the above XML lines?
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
xml = EntityUtils.toString(httpEntity);
编辑
虽然我设法使用以下几行以某种方式创建 xml,但我得到的结果不正确。
StringBuilder sb = new StringBuilder();
sb.append("<?xml version='1.0' encoding='UTF-8' standalone='yes'?>");
sb.append("<IAM version'1.0'>");
sb.append("<ServiceRequest>");
sb.append("<RequestTimestamp>2012-07-20T12:33:00Z</RequestTimestamp");
sb.append("<RequestorRef>username</RequestorRef>");
sb.append("<StopMonitoringRequest version='1.0'>");
sb.append("<RequestTimestamp>2012-07-20T12:33:00Z</RequestTimestamp>");
sb.append("<MessageIdentifier>12345</MessageIdentifier>");
sb.append("<MonitoringRef>32900109</MonitoringRef>");
sb.append("</StopMonitoringRequest>");
sb.append("</ServiceRequest>");
sb.append("</IAM>");
String xmlContentTosend = sb.toString();
StringEntity entity = new StringEntity(xmlContentTosend, "UTF-8");
httpPost.setEntity(entity);
httpPost.addHeader(BasicScheme.authenticate(new UsernamePasswordCredentials("username", "password"), "UTF-8", false));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
xml = EntityUtils.toString(httpEntity);
我得到一个字符串文件(xml),它不是我应该得到的完整答案。如果我使用 Firefox 的 HTTP 资源测试,我会得到正确的答案,而使用我的解决方案,我会得到部分答案。当我删除
时,我设法在 HTTP 资源测试中收到相同的部分答案<IAM version="1.0">
行或其他一些行(通常在“破坏”xml 结构时)。不过不知道有没有关系。
编辑(找到解决方案) 你能发现它吗? xml 结构中的第一个 RequestTimestamp 处缺少一个“>”。我整天都在复制粘贴,所以我没有提到它。噗……
最佳答案
你可以使用 Dom 解析器来做到这一点
这是一些代码
public class WriteXMLFile {
public static void main(String argv[]) {
try {
DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
// root elements
Document doc = docBuilder.newDocument();
Element rootElement = doc.createElement("company");
doc.appendChild(rootElement);
// staff elements
Element staff = doc.createElement("Staff");
rootElement.appendChild(staff);
// set attribute to staff element
Attr attr = doc.createAttribute("id");
attr.setValue("1");
staff.setAttributeNode(attr);
// shorten way
// staff.setAttribute("id", "1");
// firstname elements
Element firstname = doc.createElement("firstname");
firstname.appendChild(doc.createTextNode("yong"));
staff.appendChild(firstname);
// lastname elements
Element lastname = doc.createElement("lastname");
lastname.appendChild(doc.createTextNode("mook kim"));
staff.appendChild(lastname);
// nickname elements
Element nickname = doc.createElement("nickname");
nickname.appendChild(doc.createTextNode("mkyong"));
staff.appendChild(nickname);
// salary elements
Element salary = doc.createElement("salary");
salary.appendChild(doc.createTextNode("100000"));
staff.appendChild(salary);
// write the content into xml file
TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
DOMSource source = new DOMSource(doc);
StreamResult result = new StreamResult(new File("C:\\file.xml"));
// Output to console for testing
// StreamResult result = new StreamResult(System.out);
transformer.transform(source, result);
System.out.println("File saved!");
} catch (ParserConfigurationException pce) {
pce.printStackTrace();
} catch (TransformerException tfe) {
tfe.printStackTrace();
}
}
}
这会创建一个类似的 xml
<?xml version="1.0" encoding="UTF-8" standalone="no" ?>
<company>
<staff id="1">
<firstname>yong</firstname>
<lastname>mook kim</lastname>
<nickname>mkyong</nickname>
<salary>100000</salary>
</staff>
</company>
通过 http post 发送:
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://192.168.192.131/");
try {
StringEntity se = new StringEntity( "<aaaLogin inName=\"admin\" inPassword=\"admin123\"/>", HTTP.UTF_8);
se.setContentType("text/xml");
httppost.setEntity(se);
HttpResponse httpresponse = httpclient.execute(httppost);
HttpEntity resEntity = httpresponse.getEntity();
tvData.setText(EntityUtils.toString(resEntity));
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
顺便说一句,请考虑使用 JSON而不是 XML。它更高效且更易于使用。
关于java - HTTP POST 请求 XML 创建,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11578793/