使用java进行类似于ajax的调用
使用java进行类似于ajax的调用
提问于 2015-06-05 09:05:56
我已经研究过这个Previous post，如果请求返回json，如何在java.How中使用ajax等效调用来检索响应。
final URL url = new URL("http://localhost:8080/SearchPerson.aspx/PersonSearch");
final URLConnection urlConnection = url.openConnection();
urlConnection.setDoOutput(true);
urlConnection.setRequestProperty("Content-Type", "application/json; charset=utf-8");
urlConnection.connect();
final OutputStream outputStream = urlConnection.getOutputStream();
outputStream.write(("{\"fNamn\": \"" + stringData + "\"}").getBytes("UTF-8"));
outputStream.flush();
final InputStream inputStream = urlConnection.getInputStream();`Stack Overflow用户
发布于 2015-06-05 10:04:03
尝试处理getInputStream()对象的HttpURLConnection方法。它将给出来自目标URI的响应。
尝尝这个,
public JSONObject getResult() {
String uri = "http://example.com"
JSONObject jsonResponse = null;
try {
URL url = new URL(uri);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setDoOutput(true);
conn.setRequestMethod("GET");
conn.setRequestProperty("Content-Type", "application/json");
if (conn.getResponseCode() != HttpURLConnection.OK) {
throw new RuntimeException("Failed : HTTP error code : "
+ conn.getResponseCode());
}
BufferedReader br = new BufferedReader(new InputStreamReader(
(conn.getInputStream())));
String jsonResponseString = br.readLine();
/**
* jsonResponse will be having data from target, convert it to JSONObject
*/
jsonResponse = new JSONObject(jsonResponseString);
conn.disconnect();
} catch (MalformedURLException e) {
} catch (IOException e) {
} catch (JSONException ex) {
}
return jsonResponse;
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/30662533
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