Tat*_*ta2 56 java xml request apache-httpclient-4.x
我想用XML内容类型编写请求的主体,但我不知道如何使用HttpClient对象(http://hc.apache.org/httpclient-3.x/apidocs/index.html)
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpRequest = new HttpPost(this.url);
httpRequest.setHeader("Content-Type", "application/xml");
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而且我不知道如何继续使用我的XML编写正文...
Lar*_*y.Z 109
如果您的xml是由java.lang.String您编写的,您可以HttpClient这样使用
public void post() throws Exception{
HttpClient client = new DefaultHttpClient();
HttpPost post = new HttpPost("http://www.baidu.com");
String xml = "<xml>xxxx</xml>";
HttpEntity entity = new ByteArrayEntity(xml.getBytes("UTF-8"));
post.setEntity(entity);
HttpResponse response = client.execute(post);
String result = EntityUtils.toString(response.getEntity());
}
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注意例外情况.
顺便说一下,这个例子是由httpclient版本4.x编写的
San*_*osh 24
扩展代码(假设您要发送的XML是xmlString):
String xmlString = "</xml>";
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpRequest = new HttpPost(this.url);
httpRequest.setHeader("Content-Type", "application/xml");
StringEntity xmlEntity = new StringEntity(xmlString);
httpRequest.setEntity(xmlEntity );
HttpResponse httpresponse = httpclient.execute(httppost);
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