【发布时间】:2012-10-14 02:26:08
【问题描述】:
所以我有以下代码:
public void SendToApplication(HttpServletRequest request) throws IOException, TransformerException {
BufferedReader br = new BufferedReader(new FileReader(new File("CreatePoll.xml")));
String line;
StringBuilder sb = new StringBuilder();
while((line=br.readLine())!= null) sb.append(line.trim());
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost postRequest = new HttpPost("http://localhost:8080/cs9322.simple.rest.doodle/rest/polls/comment");
StringEntity input = new StringEntity(sb.toString());
input.setContentType("text/xml");
postRequest.setEntity(input);
HttpResponse response = httpClient.execute(postRequest);
HttpEntity entity = response.getEntity();
}
读取一个XML文件(CreatePoll.xml),
<Comment xmlns:xs="http://localhost:8080/cs9322.simple.rest.doodle/CommentSchema">
<Poll_ID>2</Poll_ID>
<Name>Bob</Name>
<Text>testing junk</Text>
<Timestamp>2012-10-14T12:37:04</Timestamp>
</Comment>
并将其发布到 Web 服务,我现在遇到的问题是在发送后尝试从 Web 服务接收 XML 响应。我打算接收的 XML 是:
<comment>
<address>
</address>
</comment>
谁能帮帮我,不胜感激!
【问题讨论】:
-
您永远不应该使用 Reader 读取 xml(或将其转换为字符串)。您应该使用 InputStream 并将 xml 数据视为字节,以免意外破坏字符编码。处理响应也是如此。
标签: java xml apache restful-url