怎么用java读取一个http：//。。。。。 .xml并返回json数据啊

如题所述

可以用httpClient 发起一个 get或者post请求然后得到返回的结果再做json的解析即可
httpClient 用法:
1. GET 方式传递参数
//先将参数放入List，再对参数进行URL编码
List<BasicNameValuePair> params = new LinkedList<BasicNameValuePair>();
params.add(new BasicNameValuePair("param1", "数据")); //增加参数1
params.add(new BasicNameValuePair("param2", "value2"));//增加参数2
String param = URLEncodedUtils.format(params, "UTF-8");//对参数编码
String baseUrl = "服务器接口完整URL";
HttpGet getMethod = new HttpGet(baseUrl + "?" + param);//将URL与参数拼接
HttpClient httpClient = new DefaultHttpClient();
try {
HttpResponse response = httpClient.execute(getMethod); //发起GET请求
Log.i(TAG, "resCode = " + response.getStatusLine().getStatusCode()); //获取响应码
Log.i(TAG, "result = " + EntityUtils.toString(response.getEntity(), "utf-8"));//获取服务器响应内容
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}

2. POST方式 方式传递参数
//和GET方式一样，先将参数放入List
params = new LinkedList<BasicNameValuePair>();
params.add(new BasicNameValuePair("param1", "Post方法"));//增加参数1
params.add(new BasicNameValuePair("param2", "第二个参数"));//增加参数2
try {
HttpPost postMethod = new HttpPost(baseUrl);//创建一个post请求
postMethod.setEntity(new UrlEncodedFormEntity(params, "utf-8")); //将参数填入POST Entity中
HttpResponse response = httpClient.execute(postMethod); //执行POST方法
Log.i(TAG, "resCode = " + response.getStatusLine().getStatusCode()); //获取响应码
Log.i(TAG, "result = " + EntityUtils.toString(response.getEntity(), "utf-8")); //获取响应内容
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}

温馨提示：内容为网友见解，仅供参考

当前网址：https://11.t2y.org/zz/424svf2s7q2vvpp4ps.html