将文件从 Java 客户端上传到 HTTP 服务器

Question

我想将一些文件上传到 HTTP 服务器。基本上我需要的是带有一些参数和文件的对服务器的某种 POST 请求。我见过仅上传文件的示例，但没有找到如何传递其他参数。

这样做最简单且免费的解决方案是什么？有人有我可以学习的文件上传示例吗？我已经在谷歌上搜索了几个小时，但是（也许这只是那些日子之一）无法找到我需要的东西。最好的解决方案是不涉及任何第三方类或库的解决方案。

Answer 1

您通常会使用 java.net.URLConnection 来触发 HTTP 请求。您通常还可以使用 multipart/form- data 混合 POST 内容（二进制和字符数据）的编码。单击该链接，它包含如何组成 multipart/form-data 请求正文的信息和示例。 RFC2388 中更详细地描述了该规范。

下面是一个启动示例：

String url = "http://example.com/upload";
String charset = "UTF-8";
String param = "value";
File textFile = new File("/path/to/file.txt");
File binaryFile = new File("/path/to/file.bin");
String boundary = Long.toHexString(System.currentTimeMillis()); // Just generate some unique random value.
String CRLF = "\r\n"; // Line separator required by multipart/form-data.

URLConnection connection = new URL(url).openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);

try (
    OutputStream output = connection.getOutputStream();
    PrintWriter writer = new PrintWriter(new OutputStreamWriter(output, charset), true);
) {
    // Send normal param.
    writer.append("--" + boundary).append(CRLF);
    writer.append("Content-Disposition: form-data; name=\"param\"").append(CRLF);
    writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF);
    writer.append(CRLF).append(param).append(CRLF).flush();

    // Send text file.
    writer.append("--" + boundary).append(CRLF);
    writer.append("Content-Disposition: form-data; name=\"textFile\"; filename=\"" + textFile.getName() + "\"").append(CRLF);
    writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF); // Text file itself must be saved in this charset!
    writer.append(CRLF).flush();
    Files.copy(textFile.toPath(), output);
    output.flush(); // Important before continuing with writer!
    writer.append(CRLF).flush(); // CRLF is important! It indicates end of boundary.

    // Send binary file.
    writer.append("--" + boundary).append(CRLF);
    writer.append("Content-Disposition: form-data; name=\"binaryFile\"; filename=\"" + binaryFile.getName() + "\"").append(CRLF);
    writer.append("Content-Type: " + URLConnection.guessContentTypeFromName(binaryFile.getName())).append(CRLF);
    writer.append("Content-Transfer-Encoding: binary").append(CRLF);
    writer.append(CRLF).flush();
    Files.copy(binaryFile.toPath(), output);
    output.flush(); // Important before continuing with writer!
    writer.append(CRLF).flush(); // CRLF is important! It indicates end of boundary.

    // End of multipart/form-data.
    writer.append("--" + boundary + "--").append(CRLF).flush();
}

// Request is lazily fired whenever you need to obtain information about response.
int responseCode = ((HttpURLConnection) connection).getResponseCode();
System.out.println(responseCode); // Should be 200

当您使用像 Apache Commons HttpComponents Client 这样的第 3 方库时，此代码会更简洁。

Apache Commons FileUpload 正如一些人错误地建议的那样，仅对服务器端感兴趣。您不能在客户端使用也不需要它。

另请参阅

• 使用 java.net.URLConnection 来触发和处理HTTP 请求

Answer 2

以下是使用 Apache HttpClient 的方法（此解决方案适用于那些不介意使用第 3 方的人）图书馆）：

    HttpEntity entity = MultipartEntityBuilder.create()
                       .addPart("file", new FileBody(file))
                       .build();

    HttpPost request = new HttpPost(url);
    request.setEntity(entity);

    HttpClient client = HttpClientBuilder.create().build();
    HttpResponse response = client.execute(request);

Answer 3

单击链接获取示例文件上传 clint java 与 apache HttpComponents

http://hc.apache.org/httpcomponents-client-ga/httpmime/examples/org/apache/http/examples/entity/mime/ClientMultipartFormPost.java

和库 downalod链接

https://hc.apache.org/downloads.cgi

使用 4.5.3.zip 它是 中工作正常..

在我的代码和工作代码

import java.io.File;
import org.apache.http.HttpEntity;
import org.apache.http.client.methods.CloseableHttpResponse;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.ContentType;
import org.apache.http.entity.mime.MultipartEntityBuilder;
import org.apache.http.entity.mime.content.FileBody;
import org.apache.http.entity.mime.content.StringBody;
import org.apache.http.impl.client.CloseableHttpClient;
import org.apache.http.impl.client.HttpClients;
import org.apache.http.util.EntityUtils;

public class ClientMultipartFormPost {

     public static void main(String[] args) throws Exception {

          CloseableHttpClient httpclient = HttpClients.createDefault();
          try {
             HttpPost httppost = new HttpPost("http://localhost:8080/MyWebSite1/UploadDownloadFileServlet");

             FileBody bin = new FileBody(new File("E:\\meter.jpg"));
             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();
      }
   }

}

Answer 4

以下是使用 Java 11 的 java.net.http 包执行此操作的方法：

    var fileA = new File("a.pdf");
    var fileB = new File("b.pdf");

    var mimeMultipartData = MimeMultipartData.newBuilder()
            .withCharset(StandardCharsets.UTF_8)
            .addFile("file1", fileA.toPath(), Files.probeContentType(fileA.toPath()))
            .addFile("file2", fileB.toPath(), Files.probeContentType(fileB.toPath()))
            .build();

    var request = HttpRequest.newBuilder()
            .header("Content-Type", mimeMultipartData.getContentType())
            .POST(mimeMultipartData.getBodyPublisher())
            .uri(URI.create("http://somehost/upload"))
            .build();

    var httpClient = HttpClient.newBuilder().build();
    var response = httpClient.send(request, BodyHandlers.ofString());

使用以下 MimeMultipartData：

public class MimeMultipartData {

    public static class Builder {

        private String boundary;
        private Charset charset = StandardCharsets.UTF_8;
        private List<MimedFile> files = new ArrayList<MimedFile>();
        private Map<String, String> texts = new LinkedHashMap<>();

        private Builder() {
            this.boundary = new BigInteger(128, new Random()).toString();
        }

        public Builder withCharset(Charset charset) {
            this.charset = charset;
            return this;
        }

        public Builder withBoundary(String boundary) {
            this.boundary = boundary;
            return this;
        }

        public Builder addFile(String name, Path path, String mimeType) {
            this.files.add(new MimedFile(name, path, mimeType));
            return this;
        }

        public Builder addText(String name, String text) {
            texts.put(name, text);
            return this;
        }

        public MimeMultipartData build() throws IOException {
            MimeMultipartData mimeMultipartData = new MimeMultipartData();
            mimeMultipartData.boundary = boundary;

            var newline = "\r\n".getBytes(charset);
            var byteArrayOutputStream = new ByteArrayOutputStream();
            for (var f : files) {
                byteArrayOutputStream.write(("--" + boundary).getBytes(charset)); 
                byteArrayOutputStream.write(newline);
                byteArrayOutputStream.write(("Content-Disposition: form-data; name=\"" + f.name + "\"; filename=\"" + f.path.getFileName() + "\"").getBytes(charset));
                byteArrayOutputStream.write(newline);
                byteArrayOutputStream.write(("Content-Type: " + f.mimeType).getBytes(charset));
                byteArrayOutputStream.write(newline);
                byteArrayOutputStream.write(newline);
                byteArrayOutputStream.write(Files.readAllBytes(f.path));
                byteArrayOutputStream.write(newline);
            }
            for (var entry: texts.entrySet()) {
                byteArrayOutputStream.write(("--" + boundary).getBytes(charset));
                byteArrayOutputStream.write(newline);
                byteArrayOutputStream.write(("Content-Disposition: form-data; name=\"" + entry.getKey() + "\"").getBytes(charset));
                byteArrayOutputStream.write(newline);
                byteArrayOutputStream.write(newline);
                byteArrayOutputStream.write(entry.getValue().getBytes(charset));
                byteArrayOutputStream.write(newline);
            }
            byteArrayOutputStream.write(("--" + boundary + "--").getBytes(charset));

            mimeMultipartData.bodyPublisher = BodyPublishers.ofByteArray(byteArrayOutputStream.toByteArray());
            return mimeMultipartData;
        }

        public class MimedFile {

            public final String name;
            public final Path path;
            public final String mimeType;

            public MimedFile(String name, Path path, String mimeType) {
                this.name = name;
                this.path = path;
                this.mimeType = mimeType;
            }
        }
    }

    private String boundary;
    private BodyPublisher bodyPublisher;

    private MimeMultipartData() {
    }

    public static Builder newBuilder() {
        return new Builder();
    }

    public BodyPublisher getBodyPublisher() throws IOException {
        return bodyPublisher;
    }

    public String getContentType() {
        return "multipart/form-data; boundary=" + boundary;
    }

}

Answer 5

public static String simSearchByImgURL(int  catid ,String imgurl) throws IOException{
    CloseableHttpClient httpClient = HttpClients.createDefault();
    CloseableHttpResponse response = null;
    String result =null;
    try {
        HttpPost httppost = new HttpPost("http://api0.visualsearchapi.com:8084/vsearchtech/api/v1.0/apisim_search");
        StringBody catidBody = new StringBody(catid+"" , ContentType.TEXT_PLAIN);
        StringBody keyBody = new StringBody(APPKEY , ContentType.TEXT_PLAIN);
        StringBody langBody = new StringBody(LANG , ContentType.TEXT_PLAIN);
        StringBody fmtBody = new StringBody(FMT , ContentType.TEXT_PLAIN);
        StringBody imgurlBody = new StringBody(imgurl , ContentType.TEXT_PLAIN);
        MultipartEntityBuilder builder = MultipartEntityBuilder.create();
        builder.addPart("apikey", keyBody).addPart("catid", catidBody)
        .addPart("lang", langBody)
        .addPart("fmt", fmtBody)
        .addPart("imgurl", imgurlBody);
        HttpEntity reqEntity =  builder.build();
        httppost.setEntity(reqEntity);
        response = httpClient.execute(httppost);
        HttpEntity resEntity = response.getEntity();
        if (resEntity != null) {
           // result = ConvertStreamToString(resEntity.getContent(), "UTF-8");
            String charset = "UTF-8";   
          String content=EntityUtils.toString(response.getEntity(), charset);   
            System.out.println(content);
        }
        EntityUtils.consume(resEntity);
    }catch(Exception e){
        e.printStackTrace();
    }finally {
        response.close();
        httpClient.close();
    }
    return result;
}

Answer 6

我建议使用 Apache http 类而不是 Vanilla Java。这是一个兼容 Java 8 的简单解决方案：

// Create http client.

CloseableHttpClient httpClient = HttpClients.createDefault();
final File file = new File("<FILE PATH TO POST>");

// Specify the content type of the attached file.

FileBody filebody = new FileBody(file, ContentType.MULTIPART_FORM_DATA);
MultipartEntityBuilder entitybuilder = MultipartEntityBuilder.create();
entitybuilder.setMode(HttpMultipartMode.BROWSER_COMPATIBLE);

// Add the binary file to the entity we send later on.

entitybuilder.addBinaryBody("file", file);
HttpEntity mutiPartHttpEntity = entitybuilder.build();
RequestBuilder reqbuilder = RequestBuilder.post("URL TO POST DATA WITH ALL THE PARAMETERS");
reqbuilder.setEntity(mutiPartHttpEntity);
HttpUriRequest multipartRequest = reqbuilder.build();

// Using the http client, execute http post. Variable httpresponse would contain the reply back explaining http response code, and ...

HttpResponse httpresponse = httpClient.execute(multipartRequest);
httpClient.close();

1. 为了将参数嵌入到 post 消息中，您可以简单地将它们嵌入到用于发布数据的 URL 中。使用这个规则：?=&=。一个例子如下：
   https://company.com/jobs/status?jobId=876&apiKey= 123
2. 要了解哪种数据类型适合您通过 Http post 发布的文件，请参阅此链接：https://developer.mozilla.org/en-US/docs/Web/HTTP/Basics_of_HTTP/MIME_types
3. 导入org.apache.http的包。*
4. 以防万一导入不起作用，您可能需要更改 pom.xml 文件（如果您使用的是 Maven）并将 Apache 包添加到此 xml 文件中。 查看此内容。

Answer 7

protected void doPost(HttpServletRequest request,
        HttpServletResponse response) throws ServletException, IOException {

    boolean isMultipart = ServletFileUpload.isMultipartContent(request);

    if (!isMultipart) {
        return;
    }

    DiskFileItemFactory factory = new DiskFileItemFactory();

    factory.setSizeThreshold(MAX_MEMORY_SIZE);

    factory.setRepository(new File(System.getProperty("java.io.tmpdir")));

    String uploadFolder = getServletContext().getRealPath("")
            + File.separator + DATA_DIRECTORY;//DATA_DIRECTORY is directory where you upload this file on the server

    ServletFileUpload upload = new ServletFileUpload(factory);

    upload.setSizeMax(MAX_REQUEST_SIZE);//MAX_REQUEST_SIZE is the size which size you prefer

并使用

并在 html 中使用

Answer 8

这可能取决于您的框架。 （对于他们每个人都可以存在一个更简单的解决方案）。

但要回答你的问题：有很多外部库这个功能。请查看此处如何使用 apache commons fileupload。