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
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.
}
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();
}
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
调用HttpURLConnection.setRequestMethod("POST")和HttpURLConnection.setDoOutput(true);实际上只需要后者，因为POST将成为默认方法。
使用okhttp:
okhttp的源代码可以在这里找到https://github.com/square/okhttp。
如果您正在编写一个pom项目，请添加此依赖项
<dependency>
<groupId>com.squareup.okhttp3</groupId>
<artifactId>okhttp</artifactId>
<version>4.2.2</version>
</dependency>
如果不是，可以在网上搜索“下载okhttp”。在下载jar的地方会出现几个结果。
你的代码:
import okhttp3.*;
import java.io.IOException;
public class ClassName{
private void sendPost() throws IOException {
// form parameters
RequestBody formBody = new FormBody.Builder()
.add("id", 10)
.build();
Request request = new Request.Builder()
.url("http://www.example.com/page.php")
.post(formBody)
.build();
OkHttpClient httpClient = new OkHttpClient();
try (Response response = httpClient.newCall(request).execute()) {
if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);
// Get response body
System.out.println(response.body().string());
}
}
}
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.
}
