让我们假设这个URL…
http://www.example.com/page.php?id=10
(这里id需要在POST请求中发送)
我想将id = 10发送到服务器的page.php，该服务器在POST方法中接受它。
我如何从Java中做到这一点?
我试了一下:
URL aaa = new URL("http://www.example.com/page.php");
URLConnection ccc = aaa.openConnection();
但是我仍然不知道如何通过POST发送它
当前回答
使用Apache HTTP Components的一种简单方法是
Request.Post("http://www.example.com/page.php")
.bodyForm(Form.form().add("id", "10").build())
.execute()
.returnContent();
看一看Fluent API
其他回答
String rawData = "id=10";
String type = "application/x-www-form-urlencoded";
String encodedData = URLEncoder.encode( rawData, "UTF-8" );
URL u = new URL("http://www.example.com/page.php");
HttpURLConnection conn = (HttpURLConnection) u.openConnection();
conn.setDoOutput(true);
conn.setRequestMethod("POST");
conn.setRequestProperty( "Content-Type", type );
conn.setRequestProperty( "Content-Length", String.valueOf(encodedData.length()));
OutputStream os = conn.getOutputStream();
os.write(encodedData.getBytes());
更新后的答案
由于原始答案中的一些类在Apache HTTP组件的新版本中已弃用，所以我发布了此更新。
顺便说一下，您可以在这里访问更多示例的完整文档。
HttpClient httpclient = HttpClients.createDefault();
HttpPost httppost = new HttpPost("http://www.a-domain.example/foo/");
// Request parameters and other properties.
List<NameValuePair> params = new ArrayList<NameValuePair>(2);
params.add(new BasicNameValuePair("param-1", "12345"));
params.add(new BasicNameValuePair("param-2", "Hello!"));
httppost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));
//Execute and get the response.
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
if (entity != null) {
try (InputStream instream = entity.getContent()) {
// do something useful
}
}
原来的答案
我推荐使用Apache HttpClient。它更快更容易实现。
HttpPost post = new HttpPost("http://jakarata.apache.org/");
NameValuePair[] data = {
new NameValuePair("user", "joe"),
new NameValuePair("password", "bloggs")
};
post.setRequestBody(data);
// execute method and handle any error responses.
...
InputStream in = post.getResponseBodyAsStream();
// handle response.
欲了解更多信息，请查看此URL: http://hc.apache.org/
用post请求发送参数的最简单方法:
String postURL = "http://www.example.com/page.php";
HttpPost post = new HttpPost(postURL);
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("id", "10"));
UrlEncodedFormEntity ent = new UrlEncodedFormEntity(params, "UTF-8");
post.setEntity(ent);
HttpClient client = new DefaultHttpClient();
HttpResponse responsePOST = client.execute(post);
你做到了。现在您可以使用responsePOST。 获取响应内容为字符串:
BufferedReader reader = new BufferedReader(new InputStreamReader(responsePOST.getEntity().getContent()), 2048);
if (responsePOST != null) {
StringBuilder sb = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
System.out.println(" line : " + line);
sb.append(line);
}
String getResponseString = "";
getResponseString = sb.toString();
//use server output getResponseString as string value.
}
使用Apache HTTP Components的一种简单方法是
Request.Post("http://www.example.com/page.php")
.bodyForm(Form.form().add("id", "10").build())
.execute()
.returnContent();
看一看Fluent API
我建议使用Postman来生成请求代码。简单地使用Postman进行请求，然后点击code选项卡:
然后你会看到下面的窗口，选择你想要的请求代码的语言:
