Java – How to upload a file using Java HttpClient library working with PHP

javajava-http-clientPHP

I want to write Java application that will upload a file to the Apache server with PHP. The Java code uses Jakarta HttpClient library version 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();
  }
}

The PHP file upload.php is very simple:

<?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);
}
?>

Reading the response I get the following result:

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

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

So the request was successful, I was able to communicate with server, however PHP didn't notice the file – the method is_uploaded_file returned false and $_FILES variable is empty. I have no idea why this might happend. I have tracked HTTP response and request and they look ok:
request is:

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...

and response:

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
(
)

I was testing this both on the local windows xp with xampp and remote Linux server. I have also tried to use previous version of HttpClient – version 3.1 – and the result was even more unclear, is_uploaded_file returned false, however $_FILES array was filled with proper data.

Best Answer

Ok, the Java code I used was wrong, here comes the right Java class:

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();
  }
}

note using MultipartEntity.

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It's a common misconception that user input can be filtered. PHP even has a (now deprecated) "feature", called magic-quotes, that builds on this idea. It's nonsense. Forget about filtering (or cleaning, or whatever people call it).

What you should do, to avoid problems, is quite simple: whenever you embed a a piece of data within a foreign code, you must treat it according to the formatting rules of that code. But you must understand that such rules could be too complicated to try to follow them all manually. For example, in SQL, rules for strings, numbers and identifiers are all different. For your convenience, in most cases there is a dedicated tool for such an embedding. For example, when you need to use a PHP variable in the SQL query, you have to use a prepared statement, that will take care of all the proper formatting/treatment.

Another example is HTML: If you embed strings within HTML markup, you must escape it with htmlspecialchars. This means that every single echo or print statement should use htmlspecialchars.

A third example could be shell commands: If you are going to embed strings (such as arguments) to external commands, and call them with exec, then you must use escapeshellcmd and escapeshellarg.

Also, a very compelling example is JSON. The rules are so numerous and complicated that you would never be able to follow them all manually. That's why you should never ever create a JSON string manually, but always use a dedicated function, json_encode() that will correctly format every bit of data.

And so on and so forth ...

The only case where you need to actively filter data, is if you're accepting preformatted input. For example, if you let your users post HTML markup, that you plan to display on the site. However, you should be wise to avoid this at all cost, since no matter how well you filter it, it will always be a potential security hole.

Php – How to get a file extension in PHP

People from other scripting languages always think theirs is better because they have a built-in function to do that and not PHP (I am looking at Pythonistas right now :-)).

In fact, it does exist, but few people know it. Meet pathinfo():

$ext = pathinfo($filename, PATHINFO_EXTENSION);

This is fast and built-in. pathinfo() can give you other information, such as canonical path, depending on the constant you pass to it.

Remember that if you want to be able to deal with non ASCII characters, you need to set the locale first. E.G:

setlocale(LC_ALL,'en_US.UTF-8');

Also, note this doesn't take into consideration the file content or mime-type, you only get the extension. But it's what you asked for.

Lastly, note that this works only for a file path, not a URL resources path, which is covered using PARSE_URL.

Enjoy

Best Answer

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