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
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());
其他回答
我建议使用http-request建立在apache http api上。
HttpRequest<String> httpRequest = HttpRequestBuilder.createPost("http://www.example.com/page.php", String.class)
.responseDeserializer(ResponseDeserializer.ignorableDeserializer()).build();
public void send(){
String response = httpRequest.execute("id", "10").get();
}
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());
从java 11开始，HTTP请求可以通过使用java.net.http.HttpClient进行，代码更少。
var values = new HashMap<String, Integer>() {{
put("id", 10);
}};
var objectMapper = new ObjectMapper();
String requestBody = objectMapper
.writeValueAsString(values);
HttpClient client = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder()
.uri(URI.create("http://www.example.com/abc"))
.POST(HttpRequest.BodyPublishers.ofString(requestBody))
.build();
HttpResponse<String> response = client.send(request,
HttpResponse.BodyHandlers.ofString());
System.out.println(response.body());
第一个答案很好，但我必须添加try/catch以避免Java编译器错误。 另外，我在弄清楚如何使用Java库读取HttpResponse时遇到了麻烦。
以下是更完整的代码:
/*
* Create the POST request
*/
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://example.com/");
// Request parameters and other properties.
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("user", "Bob"));
try {
httpPost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));
} catch (UnsupportedEncodingException e) {
// writing error to Log
e.printStackTrace();
}
/*
* Execute the HTTP Request
*/
try {
HttpResponse response = httpClient.execute(httpPost);
HttpEntity respEntity = response.getEntity();
if (respEntity != null) {
// EntityUtils to get the response content
String content = EntityUtils.toString(respEntity);
}
} catch (ClientProtocolException e) {
// writing exception to log
e.printStackTrace();
} catch (IOException e) {
// writing exception to log
e.printStackTrace();
}
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
