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

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());

其他回答

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());
                }
        }
    }
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/

在普通Java中发送POST请求很容易。从一个URL开始,我们需要使用URL . openconnection();将其转换为URLConnection。之后,我们需要将它转换为一个HttpURLConnection,这样我们就可以访问它的setRequestMethod()方法来设置我们的方法。我们最后说,我们将通过连接发送数据。

URL url = new URL("https://www.example.com/login");
URLConnection con = url.openConnection();
HttpURLConnection http = (HttpURLConnection)con;
http.setRequestMethod("POST"); // PUT is another valid option
http.setDoOutput(true);

然后我们需要声明我们要发送的内容:

发送一个简单的表单

来自http表单的普通POST具有定义良好的格式。我们需要将输入转换为以下格式:

Map<String,String> arguments = new HashMap<>();
arguments.put("username", "root");
arguments.put("password", "sjh76HSn!"); // This is a fake password obviously
StringJoiner sj = new StringJoiner("&");
for(Map.Entry<String,String> entry : arguments.entrySet())
    sj.add(URLEncoder.encode(entry.getKey(), "UTF-8") + "=" 
         + URLEncoder.encode(entry.getValue(), "UTF-8"));
byte[] out = sj.toString().getBytes(StandardCharsets.UTF_8);
int length = out.length;

然后,我们可以将表单内容附加到http请求,并发送它。

http.setFixedLengthStreamingMode(length);
http.setRequestProperty("Content-Type", "application/x-www-form-urlencoded; charset=UTF-8");
http.connect();
try(OutputStream os = http.getOutputStream()) {
    os.write(out);
}
// Do something with http.getInputStream()

发送JSON

我们也可以使用java发送json,这也很简单:

byte[] out = "{\"username\":\"root\",\"password\":\"password\"}" .getBytes(StandardCharsets.UTF_8);
int length = out.length;

http.setFixedLengthStreamingMode(length);
http.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
http.connect();
try(OutputStream os = http.getOutputStream()) {
    os.write(out);
}
// Do something with http.getInputStream()

记住,不同的服务器接受不同的json内容类型,请看这个问题。


用java post发送文件

发送文件的难度更大,因为格式更复杂。我们还将添加以字符串形式发送文件的支持,因为我们不想将文件完全缓冲到内存中。

为此,我们定义了一些helper方法:

private void sendFile(OutputStream out, String name, InputStream in, String fileName) {
    String o = "Content-Disposition: form-data; name=\"" + URLEncoder.encode(name,"UTF-8") 
             + "\"; filename=\"" + URLEncoder.encode(filename,"UTF-8") + "\"\r\n\r\n";
    out.write(o.getBytes(StandardCharsets.UTF_8));
    byte[] buffer = new byte[2048];
    for (int n = 0; n >= 0; n = in.read(buffer))
        out.write(buffer, 0, n);
    out.write("\r\n".getBytes(StandardCharsets.UTF_8));
}

private void sendField(OutputStream out, String name, String field) {
    String o = "Content-Disposition: form-data; name=\"" 
             + URLEncoder.encode(name,"UTF-8") + "\"\r\n\r\n";
    out.write(o.getBytes(StandardCharsets.UTF_8));
    out.write(URLEncoder.encode(field,"UTF-8").getBytes(StandardCharsets.UTF_8));
    out.write("\r\n".getBytes(StandardCharsets.UTF_8));
}

然后我们可以使用这些方法创建一个多部分的post请求,如下所示:

String boundary = UUID.randomUUID().toString();
byte[] boundaryBytes = 
           ("--" + boundary + "\r\n").getBytes(StandardCharsets.UTF_8);
byte[] finishBoundaryBytes = 
           ("--" + boundary + "--").getBytes(StandardCharsets.UTF_8);
http.setRequestProperty("Content-Type", 
           "multipart/form-data; charset=UTF-8; boundary=" + boundary);

// Enable streaming mode with default settings
http.setChunkedStreamingMode(0); 

// Send our fields:
try(OutputStream out = http.getOutputStream()) {
    // Send our header (thx Algoman)
    out.write(boundaryBytes);

    // Send our first field
    sendField(out, "username", "root");

    // Send a seperator
    out.write(boundaryBytes);

    // Send our second field
    sendField(out, "password", "toor");

    // Send another seperator
    out.write(boundaryBytes);

    // Send our file
    try(InputStream file = new FileInputStream("test.txt")) {
        sendFile(out, "identification", file, "text.txt");
    }

    // Finish the request
    out.write(finishBoundaryBytes);
}


// Do something with http.getInputStream()

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: