如何使用Java HttpClient库上传文件

Pio*_*ski 51 php java java-http-client

我想编写Java应用程序,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();
  }
}
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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);
}
?>
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阅读响应我得到以下结果:

executing request POST http://localhost:9002/upload.php HTTP/1.1
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HTTP/1.1 200 OK
Possible file upload attack: filename ''.
Array
(
)

因此请求成功,我能够与服务器通信,但PHP没有注意到该文件 - is_uploaded_file返回的方法false$_FILES变量为空.我不知道为什么会这样.我已跟踪HTTP响应和请求,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

????..... the rest of the binary file...

和回应:

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
(
)

我在使用xampp的本地windows xp和远程Linux服务器上测试了这个.我也尝试使用以前版本的HttpClient - 版本3.1 - 结果更加不清楚,is_uploaded_file返回false,但$_FILES阵列充满了适当的数据.

Pio*_*ski 66

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();
  }
}
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请注意使用MultipartEntity.

  • 答案:来自http://hc.apache.org/downloads.cgi的HttpClient 4.1-alpha1和HttpCore 4.1-alpha1 - 支持的Apache HttpComponents Java代码.有了这些,该错误消息消失了:) (2认同)
  • Apache HttpComponents MIME功能可以在组中找到:artifact`org.apache.httpcomponents:httpmime`. (2认同)

Bre*_*son 29

那些试图使用的人的更新MultipartEntity......

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

您可以使用它MultipartEntityBuilder来创建HttpEntity对象.

File file = new File();

HttpEntity httpEntity = MultipartEntityBuilder.create()
    .addBinaryBody("file", file, ContentType.create("image/jpeg"), file.getName())
    .build();
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对于Maven用户,该类具有以下依赖关系(几乎与fervisa的答案相同,只是在更高版本中).

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