在Java中发送HTTP POST请求-Java 学习之路

238

我们假设这个URL ...

http://www.example.com/page.php?id=10

（这里需要在POST请求中发送id）

我想将 id = 10 发送到服务器的 page.php ，它在POST方法中接受它 .

我怎样才能从Java中做到这一点？

我试过这个：

URL aaa = new URL("http://www.example.com/page.php");
URLConnection ccc = aaa.openConnection();

但我仍然无法弄清楚如何通过POST发送它

8 回答

我建议使用基于apache http api构建的http-request .

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();
}

回复于 2024-04-23T18:03:28+08:00

144

使用Apache HTTP组件的一种简单方法是

Request.Post("http://www.example.com/page.php")
            .bodyForm(Form.form().add("id", "10").build())
            .execute()
            .returnContent();

看看Fluent API

回复于 2024-04-23T18:03:28+08:00

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());

回复于 2024-04-23T18:03:28+08:00

95

调用 HttpURLConnection.setRequestMethod("POST") 和 HttpURLConnection.setDoOutput(true); 实际上只需要后者作为POST然后成为默认方法 .

回复于 2024-04-23T18:03:28+08:00

在vanilla Java中发送POST请求很容易 . 从 URL 开始，我们需要使用 url.openConnection(); 将其转换为 URLConnection . 之后，我们需要将其转换为 HttpURLConnection ，因此我们可以访问其 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);

然后我们需要说明我们要发送的内容：

发送简单表格

来自http表单的普通POST具有well defined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，我们可以使用正确的标头将表单内容附加到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()

请记住，不同的服务器接受json的不同内容类型，请参阅this问题 .

使用java post发送文件

由于格式更复杂，因此可以认为发送文件更具挑战性 . 我们还将添加支持将文件作为字符串发送，因为我们不希望将文件完全缓冲到内存中 .

为此，我们定义了一些辅助方法：

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发布请求，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()

回复于 2024-04-23T18:03:28+08:00

第一个答案很棒，但我必须添加try / catch以避免Java编译器错误 .
另外，我在如何阅读使用Java库的 HttpResponse 时遇到了麻烦 .

这是更完整的代码：

/*
 * 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();
}

回复于 2024-04-23T18:03:28+08:00

276

更新答案：

由于原始答案中的某些类在较新版本的Apache HTTP Components中已弃用，因此我发布此更新 .

顺便说一句，您可以访问完整文档以获取更多示例here .

HttpClient httpclient = HttpClients.createDefault();
HttpPost httppost = new HttpPost("http://www.a-domain.com/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
    }
}

原文答案：

我建议使用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.

有关更多信息，请查看此网址：http://hc.apache.org/

回复于 2024-04-23T18:03:28+08:00

使用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);

你做完了 . 现在你可以使用 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.
}

回复于 2024-04-23T18:03:28+08:00

在Java中发送HTTP POST请求

8 回答

发送简单表格

发送JSON

使用java post发送文件

更新答案：

原文答案：

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