如何在java中发送带有post参数的简单http post请求

我需要一个简单的代码示例，发送带有从表单输入获得的post参数的http post请求。 我找到了Apache HTTPClient，它有很多API和许多复杂的例子，但我找不到一个简单的例子，用输入参数发送http post请求并获得文本响应。

更新：我对Apache HTTPClient v.4.x感兴趣，因为不推荐使用3.x.

Answer 1

以下是使用Apache HTTPClient API的Http POST示例代码。

import java.io.InputStream; import org.apache.commons.httpclient.HttpClient; import org.apache.commons.httpclient.methods.PostMethod; public class PostExample { public static void main(String[] args){ String url = "http://www.google.com"; InputStream in = null; try { HttpClient client = new HttpClient(); PostMethod method = new PostMethod(url); //Add any parameter if u want to send it with Post req. method.addParameter("p", "apple"); int statusCode = client.executeMethod(method); if (statusCode != -1) { in = method.getResponseBodyAsStream(); } System.out.println(in); } catch (Exception e) { e.printStackTrace(); } } } 

Answer 2

我从我在应用程序中使用的Andrew Gertig的Android项目中提取了此代码。 它允许您执行HTTPost。 如果我有时间，我会创建一个POJO示例，但希望您可以剖析代码并找到所需内容。

Arshak

https://github.com/AndrewGertig/RubyDroid/blob/master/src/com/gertig/rubydroid/AddEventView.java

 private void postEvents() { DefaultHttpClient client = new DefaultHttpClient(); /** FOR LOCAL DEV HttpPost post = new HttpPost("http://192.168.0.186:3000/events"); //works with and without "/create" on the end */ HttpPost post = new HttpPost("http://cold-leaf-59.heroku.com/myevents"); JSONObject holder = new JSONObject(); JSONObject eventObj = new JSONObject(); Double budgetVal = 99.9; budgetVal = Double.parseDouble(eventBudgetView.getText().toString()); try { eventObj.put("budget", budgetVal); eventObj.put("name", eventNameView.getText().toString()); holder.put("myevent", eventObj); Log.e("Event JSON", "Event JSON = "+ holder.toString()); StringEntity se = new StringEntity(holder.toString()); post.setEntity(se); post.setHeader("Content-Type","application/json"); } catch (UnsupportedEncodingException e) { Log.e("Error",""+e); e.printStackTrace(); } catch (JSONException js) { js.printStackTrace(); } HttpResponse response = null; try { response = client.execute(post); } catch (ClientProtocolException e) { e.printStackTrace(); Log.e("ClientProtocol",""+e); } catch (IOException e) { e.printStackTrace(); Log.e("IO",""+e); } HttpEntity entity = response.getEntity(); if (entity != null) { try { entity.consumeContent(); } catch (IOException e) { Log.e("IO E",""+e); e.printStackTrace(); } } Toast.makeText(this, "Your post was successfully uploaded", Toast.LENGTH_LONG).show(); } 

Answer 3

使用Apache HttpClient v.4.x的 HTTP POST请求示例

 HttpClient httpClient = HttpClients.createDefault(); HttpPost httpPost = new HttpPost(url); MultipartEntityBuilder builder = MultipartEntityBuilder.create(); builder.addTextBody("param1", param1Value, ContentType.TEXT_PLAIN); builder.addTextBody("param2", param2Value, ContentType.TEXT_PLAIN); HttpEntity multipart = builder.build(); httpPost.setEntity(multipart); HttpResponse response = httpClient.execute(httpMethod); 

Answer 4

http://httpunit.sourceforge.net/doc/cookbook.html使用PostMethodWebRequest和setParameter方法

Answer 5

显示了一个非常简单的exapmle，你从Html页面发布，servlet处理它并发送文本响应..

http://java.sun.com/developer/onlineTraining/Programming/BasicJava1/servlet.html