如何使用 Java 发出多部分/表单数据 POST 请求？

How can I make a multipart/form-data POST request using Java?

提问by

In the days of version 3.x of Apache Commons HttpClient, making a multipart/form-data POST request was possible (an example from 2004). Unfortunately this is no longer possible in version 4.0 of HttpClient.

在 Apache Commons HttpClient 3.x 版的时代，可以进行 multipart/form-data POST 请求（2004 年的示例）。不幸的是，这在HttpClient 的 4.0 版中不再可能。

For our core activity "HTTP", multipart is somewhat out of scope. We'd love to use multipart code maintained by some other project for which it is in scope, but I'm not aware of any. We tried to move the multipart code to commons-codec a few years ago, but I didn't take off there. Oleg recently mentioned another project that has multipart parsing code and might be interested in our multipart formatting code. I don't know the current status on that. (http://www.nabble.com/multipart-form-data-in-4.0-td14224819.html)

对于我们的核心活动“HTTP”，multipart 有点超出范围。我们很想使用由它在范围内的其他项目维护的多部分代码，但我不知道任何。几年前，我们尝试将多部分代码移至 commons-codec，但我并没有在那里起飞。Oleg 最近提到了另一个具有多部分解析代码的项目，并且可能对我们的多部分格式化代码感兴趣。我不知道目前的状况。( http://www.nabble.com/multipart-form-data-in-4.0-td14224819.html)

Is anybody aware of any Java library that allows me to write an HTTP client that can make a multipart/form-data POST request?

有人知道任何允许我编写可以发出多部分/表单数据 POST 请求的 HTTP 客户端的 Java 库吗？

Background: I want to use the Remote API of Zoho Writer.

背景：我想使用Zoho Writer的远程 API。

采纳答案by ZZ Coder

We use HttpClient 4.x to make multipart file post.

我们使用 HttpClient 4.x 进行多部分文件发布。

UPDATE: As of HttpClient 4.3, some classes have been deprecated. Here is the code with new API:

更新：从HttpClient 4.3 开始，一些类已被弃用。这是带有新 API 的代码：

CloseableHttpClient httpClient = HttpClients.createDefault();
HttpPost uploadFile = new HttpPost("...");
MultipartEntityBuilder builder = MultipartEntityBuilder.create();
builder.addTextBody("field1", "yes", ContentType.TEXT_PLAIN);

// This attaches the file to the POST:
File f = new File("[/path/to/upload]");
builder.addBinaryBody(
    "file",
    new FileInputStream(f),
    ContentType.APPLICATION_OCTET_STREAM,
    f.getName()
);

HttpEntity multipart = builder.build();
uploadFile.setEntity(multipart);
CloseableHttpResponse response = httpClient.execute(uploadFile);
HttpEntity responseEntity = response.getEntity();

Below is the original snippet of code with deprecated HttpClient 4.0 API:

以下是不推荐使用的 HttpClient 4.0 API的原始代码片段：

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);

FileBody bin = new FileBody(new File(fileName));
StringBody comment = new StringBody("Filename: " + fileName);

MultipartEntity reqEntity = new MultipartEntity();
reqEntity.addPart("bin", bin);
reqEntity.addPart("comment", comment);
httppost.setEntity(reqEntity);

HttpResponse response = httpclient.execute(httppost);
HttpEntity resEntity = response.getEntity();

回答by Bob Yoplait

httpcomponents-client-4.0.1worked for me. However, I had to add the external jar apache-mime4j-0.6.jar(org.apache.james.mime4j) otherwise reqEntity.addPart("bin", bin);would not compile. Now it's working like charm.

httpcomponents-client-4.0.1为我工作。但是，我必须添加外部 jar apache-mime4j-0.6.jar( org.apache.james.mime4j) 否则 reqEntity.addPart("bin", bin);将无法编译。现在它就像魅力一样工作。

回答by Jaco van Niekerk

These are the Maven dependencies I have.

这些是我拥有的 Maven 依赖项。

Java Code:

Java代码：

HttpClient httpclient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);

FileBody uploadFilePart = new FileBody(uploadFile);
MultipartEntity reqEntity = new MultipartEntity();
reqEntity.addPart("upload-file", uploadFilePart);
httpPost.setEntity(reqEntity);

HttpResponse response = httpclient.execute(httpPost);

Maven Dependencies in pom.xml:

pom.xml 中的 Maven 依赖项：

<dependency>
  <groupId>org.apache.httpcomponents</groupId>
  <artifactId>httpclient</artifactId>
  <version>4.0.1</version>
  <scope>compile</scope>
</dependency>
<dependency>
  <groupId>org.apache.httpcomponents</groupId>
  <artifactId>httpmime</artifactId>
  <version>4.0.1</version>
  <scope>compile</scope>
</dependency>

回答by Johan

You can also use REST Assuredwhich builds on HTTP Client. It's very simple:

您还可以使用基于 HTTP 客户端的REST Assured。这很简单：

given().multiPart(new File("/somedir/file.bin")).when().post("/fileUpload");

回答by anre

If size of the JARs matters (e.g. in case of applet), one can also directly use httpmime with java.net.HttpURLConnection instead of HttpClient.

如果 JAR 的大小很重要（例如在小程序的情况下），您还可以直接使用 httpmime 和 java.net.HttpURLConnection 而不是 HttpClient。

httpclient-4.2.4:      423KB
httpmime-4.2.4:         26KB
httpcore-4.2.4:        222KB
commons-codec-1.6:     228KB
commons-logging-1.1.1:  60KB
Sum:                   959KB

httpmime-4.2.4:         26KB
httpcore-4.2.4:        222KB
Sum:                   248KB

Code:

代码：

HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setRequestMethod("POST");

FileBody fileBody = new FileBody(new File(fileName));
MultipartEntity multipartEntity = new MultipartEntity(HttpMultipartMode.STRICT);
multipartEntity.addPart("file", fileBody);

connection.setRequestProperty("Content-Type", multipartEntity.getContentType().getValue());
OutputStream out = connection.getOutputStream();
try {
    multipartEntity.writeTo(out);
} finally {
    out.close();
}
int status = connection.getResponseCode();
...

Dependency in pom.xml:

pom.xml 中的依赖：

<dependency>
    <groupId>org.apache.httpcomponents</groupId>
    <artifactId>httpmime</artifactId>
    <version>4.2.4</version>
</dependency>

回答by Java Man

Use this code to upload images or any other files to the server using post in multipart.

使用此代码使用多部分发布将图像或任何其他文件上传到服务器。

import java.io.File;
import java.io.IOException;
import java.io.UnsupportedEncodingException;

import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.ResponseHandler;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.mime.MultipartEntity;
import org.apache.http.entity.mime.content.FileBody;
import org.apache.http.entity.mime.content.StringBody;
import org.apache.http.impl.client.BasicResponseHandler;
import org.apache.http.impl.client.DefaultHttpClient;

public class SimplePostRequestTest {

    public static void main(String[] args) throws UnsupportedEncodingException, IOException {
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://192.168.0.102/uploadtest/upload_photo");

        try {
            FileBody bin = new FileBody(new File("/home/ubuntu/cd.png"));
            StringBody id = new StringBody("3");
            MultipartEntity reqEntity = new MultipartEntity();
            reqEntity.addPart("upload_image", bin);
            reqEntity.addPart("id", id);
            reqEntity.addPart("image_title", new StringBody("CoolPic"));

            httppost.setEntity(reqEntity);
            System.out.println("Requesting : " + httppost.getRequestLine());
            ResponseHandler<String> responseHandler = new BasicResponseHandler();
            String responseBody = httpclient.execute(httppost, responseHandler);
            System.out.println("responseBody : " + responseBody);

        } catch (ClientProtocolException e) {

        } finally {
            httpclient.getConnectionManager().shutdown();
        }
    }

}

it requires below files to upload.

它需要以下文件才能上传。

libraries are httpclient-4.1.2.jar,httpcore-4.1.2.jar,httpmime-4.1.2.jar,httpclient-cache-4.1.2.jar,commons-codec.jarand commons-logging-1.1.1.jarto be in classpath.

库现在 httpclient-4.1.2.jar,httpcore-4.1.2.jar,httpmime-4.1.2.jar,httpclient-cache-4.1.2.jar,commons-codec.jar和 commons-logging-1.1.1.jar将在类路径中。

回答by Atul Soman

We have a pure java implementation of multipart-form submit without using any external dependencies or libraries outside jdk. Refer https://github.com/atulsm/https-multipart-purejava/blob/master/src/main/java/com/atul/MultipartPure.java

我们有一个多部分表单提交的纯 java 实现，不使用任何外部依赖项或 jdk 之外的库。参考https://github.com/atulsm/https-multipart-purejava/blob/master/src/main/java/com/atul/MultipartPure.java

private static String body = "{\"key1\":\"val1\", \"key2\":\"val2\"}";
private static String subdata1 = "@@ -2,3 +2,4 @@\r\n";
private static String subdata2 = "<data>subdata2</data>";

public static void main(String[] args) throws Exception{        
    String url = "https://" + ip + ":" + port + "/dataupload";
    String token = "Basic "+ Base64.getEncoder().encodeToString((userName+":"+password).getBytes());

    MultipartBuilder multipart = new MultipartBuilder(url,token);       
    multipart.addFormField("entity", "main", "application/json",body);
    multipart.addFormField("attachment", "subdata1", "application/octet-stream",subdata1);
    multipart.addFormField("attachment", "subdata2", "application/octet-stream",subdata2);        
    List<String> response = multipart.finish();         
    for (String line : response) {
        System.out.println(line);
    }
}

回答by Amber

I found this samplein Apache's Quickstart Guide. It's for version 4.5:

我在 Apache 的快速入门指南中找到了这个示例。它适用于 4.5 版：

/**
 * Example how to use multipart/form encoded POST request.
 */
public class ClientMultipartFormPost {

    public static void main(String[] args) throws Exception {
        if (args.length != 1)  {
            System.out.println("File path not given");
            System.exit(1);
        }
        CloseableHttpClient httpclient = HttpClients.createDefault();
        try {
            HttpPost httppost = new HttpPost("http://localhost:8080" +
                    "/servlets-examples/servlet/RequestInfoExample");

            FileBody bin = new FileBody(new File(args[0]));
            StringBody comment = new StringBody("A binary file of some kind", ContentType.TEXT_PLAIN);

            HttpEntity reqEntity = MultipartEntityBuilder.create()
                    .addPart("bin", bin)
                    .addPart("comment", comment)
                    .build();


            httppost.setEntity(reqEntity);

            System.out.println("executing request " + httppost.getRequestLine());
            CloseableHttpResponse response = httpclient.execute(httppost);
            try {
                System.out.println("----------------------------------------");
                System.out.println(response.getStatusLine());
                HttpEntity resEntity = response.getEntity();
                if (resEntity != null) {
                    System.out.println("Response content length: " + resEntity.getContentLength());
                }
                EntityUtils.consume(resEntity);
            } finally {
                response.close();
            }
        } finally {
            httpclient.close();
        }
    }
}

回答by user1005939

Here's a solution that does not require any libraries.

这是一个不需要任何库的解决方案。

This routine transmits every file in the directory d:/data/mpf10to urlToConnect

此例程将目录中的每个文件传输d:/data/mpf10到urlToConnect

String boundary = Long.toHexString(System.currentTimeMillis());
URLConnection connection = new URL(urlToConnect).openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
PrintWriter writer = null;
try {
    writer = new PrintWriter(new OutputStreamWriter(connection.getOutputStream(), "UTF-8"));
    File dir = new File("d:/data/mpf10");
    for (File file : dir.listFiles()) {
        if (file.isDirectory()) {
            continue;
        }
        writer.println("--" + boundary);
        writer.println("Content-Disposition: form-data; name=\"" + file.getName() + "\"; filename=\"" + file.getName() + "\"");
        writer.println("Content-Type: text/plain; charset=UTF-8");
        writer.println();
        BufferedReader reader = null;
        try {
            reader = new BufferedReader(new InputStreamReader(new FileInputStream(file), "UTF-8"));
            for (String line; (line = reader.readLine()) != null;) {
                writer.println(line);
            }
        } finally {
            if (reader != null) {
                reader.close();
            }
        }
    }
    writer.println("--" + boundary + "--");
} finally {
    if (writer != null) writer.close();
}
// Connection is lazily executed whenever you request any status.
int responseCode = ((HttpURLConnection) connection).getResponseCode();
// Handle response

回答by Martin521Wang

My code post multipartFile to server.

我的代码将 multipartFile 发布到服务器。

  public static HttpResponse doPost(
    String host,
    String path,
    String method,
    MultipartFile multipartFile
  ) throws IOException
  {

    HttpClient httpClient = wrapClient(host);
    HttpPost httpPost = new HttpPost(buildUrl(host, path));

    if (multipartFile != null) {

      HttpEntity httpEntity;

      ContentBody contentBody;
      contentBody = new ByteArrayBody(multipartFile.getBytes(), multipartFile.getOriginalFilename());
      httpEntity = MultipartEntityBuilder.create()
                                         .addPart("nameOfMultipartFile", contentBody)
                                         .build();

      httpPost.setEntity(httpEntity);

    }
    return httpClient.execute(httpPost);
  }