如何使用 Java HttpClient 库与 PHP 一起上传文件

Question

我想编写 Java 应用程序，使用 PHP 将文件上传到 Apache 服务器。 Java 代码使用 Jakarta HttpClient 库版本 4.0 beta2：

import java.io.File;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.HttpVersion;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.FileEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.params.CoreProtocolPNames;
import org.apache.http.util.EntityUtils;


public class PostFile {
  public static void main(String[] args) throws Exception {
    HttpClient httpclient = new DefaultHttpClient();
    httpclient.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1);

    HttpPost httppost = new HttpPost("http://localhost:9002/upload.php");
    File file = new File("c:/TRASH/zaba_1.jpg");

    FileEntity reqEntity = new FileEntity(file, "binary/octet-stream");

    httppost.setEntity(reqEntity);
    reqEntity.setContentType("binary/octet-stream");
    System.out.println("executing request " + httppost.getRequestLine());
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity resEntity = response.getEntity();

    System.out.println(response.getStatusLine());
    if (resEntity != null) {
      System.out.println(EntityUtils.toString(resEntity));
    }
    if (resEntity != null) {
      resEntity.consumeContent();
    }

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

PHP 文件 upload.php 非常简单：

<?php
if (is_uploaded_file($_FILES['userfile']['tmp_name'])) {
  echo "File ". $_FILES['userfile']['name'] ." uploaded successfully.\n";
  move_uploaded_file ($_FILES['userfile'] ['tmp_name'], $_FILES['userfile'] ['name']);
} else {
  echo "Possible file upload attack: ";
  echo "filename '". $_FILES['userfile']['tmp_name'] . "'.";
  print_r($_FILES);
}
?>

读取响应我得到以下结果：

executing request POST http://localhost:9002/upload.php HTTP/1.1

HTTP/1.1 200 OK
Possible file upload attack: filename ''.
Array
(
)

所以请求成功，我能够与服务器通信，但是 PHP 没有注意到该文件 - 方法 is_uploaded_file 返回 false 并且 $_FILES 变量为空。 我不知道为什么会发生这种情况。 我已经跟踪了 HTTP 响应和请求，它们看起来没问题：
请求是：

POST /upload.php HTTP/1.1
Content-Length: 13091
Content-Type: binary/octet-stream
Host: localhost:9002
Connection: Keep-Alive
User-Agent: Apache-HttpClient/4.0-beta2 (java 1.5)
Expect: 100-Continue

˙Ř˙ŕ..... the rest of the binary file...

响应：

HTTP/1.1 100 Continue

HTTP/1.1 200 OK
Date: Wed, 01 Jul 2009 06:51:57 GMT
Server: Apache/2.2.8 (Win32) DAV/2 mod_ssl/2.2.8 OpenSSL/0.9.8g mod_autoindex_color PHP/5.2.5 mod_jk/1.2.26
X-Powered-By: PHP/5.2.5
Content-Length: 51
Keep-Alive: timeout=5, max=100
Connection: Keep-Alive
Content-Type: text/html

Possible file upload attack: filename ''.Array
(
)

我正在使用 xampp 的本地 Windows XP 和远程 Linux 服务器上对此进行测试。 我还尝试使用以前版本的 HttpClient - 版本 3.1 - 结果更不清楚，is_uploaded_file 返回 false，但是 $_FILES数组已填充正确的数据。

Answer 1

我知道我迟到了，但下面是处理这个问题的正确方法，关键是使用 InputStreamBody 代替 FileBody 上传多部分文件。

   try {
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost postRequest = new HttpPost("https://someserver.com/api/path/");
        postRequest.addHeader("Authorization",authHeader);
        //don't set the content type here            
        //postRequest.addHeader("Content-Type","multipart/form-data");
        MultipartEntity reqEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);


        File file = new File(filePath);
        FileInputStream fileInputStream = new FileInputStream(file);
        reqEntity.addPart("parm-name", new InputStreamBody(fileInputStream,"image/jpeg","file_name.jpg"));

        postRequest.setEntity(reqEntity);
        HttpResponse response = httpclient.execute(postRequest);

        }catch(Exception e) {
            Log.e("URISyntaxException", e.toString());
   }

Answer 2

如果您在本地 WAMP 上对此进行测试，您可能需要设置用于文件上传的临时文件夹。 您可以在 PHP.ini 文件中执行此操作：

upload_tmp_dir = "c:\mypath\mytempfolder\"

您需要授予文件夹权限以允许上传 - 您需要授予的权限因操作系统而异。

Answer 3

对于那些很难实现接受的答案（需要 org.apache.http.entity.mime.MultipartEntity）的人，您可能正在使用 org.apache.httpcomponents 4.2.*
在这种情况下，您必须显式安装 httpmime 依赖项，在我的例子中：

<dependency>
    <groupId>org.apache.httpcomponents</groupId>
    <artifactId>httpmime</artifactId>
    <version>4.2.5</version>
</dependency>

Answer 4

好吧，我使用的 Java 代码是错误的，这里是正确的 Java 类：

import java.io.File;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.HttpVersion;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.mime.MultipartEntity;
import org.apache.http.entity.mime.content.ContentBody;
import org.apache.http.entity.mime.content.FileBody;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.params.CoreProtocolPNames;
import org.apache.http.util.EntityUtils;


public class PostFile {
  public static void main(String[] args) throws Exception {
    HttpClient httpclient = new DefaultHttpClient();
    httpclient.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1);

    HttpPost httppost = new HttpPost("http://localhost:9001/upload.php");
    File file = new File("c:/TRASH/zaba_1.jpg");

    MultipartEntity mpEntity = new MultipartEntity();
    ContentBody cbFile = new FileBody(file, "image/jpeg");
    mpEntity.addPart("userfile", cbFile);


    httppost.setEntity(mpEntity);
    System.out.println("executing request " + httppost.getRequestLine());
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity resEntity = response.getEntity();

    System.out.println(response.getStatusLine());
    if (resEntity != null) {
      System.out.println(EntityUtils.toString(resEntity));
    }
    if (resEntity != null) {
      resEntity.consumeContent();
    }

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

注意使用 MultipartEntity。

Answer 5

对于那些尝试使用 MultipartEntity 的人来说，更新...

org.apache.http.entity.mime.MultipartEntity 在 4.3.1 中已弃用。

您可以使用 MultipartEntityBuilder 创建 HttpEntity 对象。

File file = new File();

HttpEntity httpEntity = MultipartEntityBuilder.create()
    .addBinaryBody("file", file, ContentType.create("image/jpeg"), file.getName())
    .build();

对于 Maven 用户，该类可在以下依赖项中使用（与 fervisa 的答案几乎相同，只是版本更高）。

<dependency>
  <groupId>org.apache.httpcomponents</groupId>
  <artifactId>httpmime</artifactId>
  <version>4.3.1</version>
</dependency>

Answer 6

正确的方法是使用多部分 POST 方法。 请参阅 这里是客户端的示例代码。

对于 PHP，有许多可用的教程。 这是我发现的第一个。 我建议您首先使用 html 客户端测试 PHP 代码，然后尝试 java 客户端。

Answer 7

我遇到了同样的问题，发现 httpclient 4.x 需要文件名才能与 PHP 后端一起使用。 httpclient 3.x 的情况并非如此。

所以我的解决方案是在FileBody构造函数中添加一个name参数。
ContentBody cbFile = new FileBody(文件, "image/jpeg", "FILE_NAME");

希望能帮助到你。

Answer 8

A新版本示例在这里。

下面是原始代码的副本：

/*
 * ====================================================================
 * Licensed to the Apache Software Foundation (ASF) under one
 * or more contributor license agreements.  See the NOTICE file
 * distributed with this work for additional information
 * regarding copyright ownership.  The ASF licenses this file
 * to you under the Apache License, Version 2.0 (the
 * "License"); you may not use this file except in compliance
 * with the License.  You may obtain a copy of the License at
 *
 *   http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing,
 * software distributed under the License is distributed on an
 * "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
 * KIND, either express or implied.  See the License for the
 * specific language governing permissions and limitations
 * under the License.
 * ====================================================================
 *
 * This software consists of voluntary contributions made by many
 * individuals on behalf of the Apache Software Foundation.  For more
 * information on the Apache Software Foundation, please see
 * <http://www.apache.org/>.
 *
 */
package org.apache.http.examples.entity.mime;

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;

/**
 * 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();
        }
    }

}

Answer 9

我有一个使用 apache http 库发送图像的工作解决方案（这里非常重要的是边界添加，如果没有它在我的连接中它将无法工作）：

            ByteArrayOutputStream baos = new ByteArrayOutputStream();
            bitmap.compress(Bitmap.CompressFormat.PNG, 100, baos);
            byte[] imageBytes = baos.toByteArray();

            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(StaticData.AMBAJE_SERVER_URL + StaticData.AMBAJE_ADD_AMBAJ_TO_GROUP);

            String boundary = "-------------" + System.currentTimeMillis();

            httpPost.setHeader("Content-type", "multipart/form-data; boundary="+boundary);

            ByteArrayBody bab = new ByteArrayBody(imageBytes, "pic.png");
            StringBody sbOwner = new StringBody(StaticData.loggedUserId, ContentType.TEXT_PLAIN);
            StringBody sbGroup = new StringBody("group", ContentType.TEXT_PLAIN);

            HttpEntity entity = MultipartEntityBuilder.create()
                    .setMode(HttpMultipartMode.BROWSER_COMPATIBLE)
                    .setBoundary(boundary)
                    .addPart("group", sbGroup)
                    .addPart("owner", sbOwner)
                    .addPart("image", bab)
                    .build();

            httpPost.setEntity(entity);

            try {
                HttpResponse response = httpclient.execute(httpPost);
                ...then reading response

Answer 10

地方添加一个名称参数即可。

FileBody constructor. ContentBody cbFile = new FileBody(file, "image/jpeg", "FILE_NAME");

啊，您只需要在希望有帮助的