我的javaFX应用程序在Web服务器（wamp）上运行，客户端通过浏览器访问此应用程序。我想在服务器端创建一个xml文件。我怎样才能做到这一点？因为目前如果我使用路径例如“/ Users / username / Desktop”，它将在客户端桌面上创建该文件。我想在服务器桌面上创建此文件。 我在netbeans 7.2.1上使用javaFX 2.2
抱歉我的英语不好！谢谢！
答案 0 :(得分：2)
您的JavaFX应用程序需要与您的Web服务进行通信。在这种情况下，我认为它是您网站上的一个简单表格。为此，您的客户端需要使用GET（灵活性较低）或POST（更灵活）方法来上载稍后由PHP脚本处理的文件。
根据jewelsea的建议，Apache HttpClient可以为您完成这项工作。但是，如果你像我一样，不喜欢为简单的东西添加依赖项，那么你可能决定卷起你的袖子并像我一样实现HttpPost类：
/**
* Project: jlib
* Version: $Id: HttpPost.java 463 2012-09-17 10:58:04Z dejan $
* License: Public Domain
*
* Authors (in chronological order):
* Dejan Lekic - http://dejan.lekic.org
* Contributors (in chronological order):
* -
*/
package co.prj.jlib;
import java.io.File;
import java.io.FileInputStream;
import java.io.InputStream;
import java.io.OutputStream;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLConnection;
import java.util.logging.Level;
import java.util.logging.Logger;
/**
* A class that uses HttpURLConnection to do a HTTP post.
*
* The main reason for this class is to have a simple solution for uploading files using the PHP file below:
*
* Example:
*
* <pre>
* <?php
* // In PHP versions earlier than 4.1.0, $HTTP_POST_FILES should be used instead
* // of $_FILES.
*
* $uploaddir = '/srv/www/lighttpd/example.com/files/';
* $uploadfile = $uploaddir . basename($_FILES['userfile']['name']);
*
* echo '<pre>';
* if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
* echo "File is valid, and was successfully uploaded.\n";
* } else {
* echo "Possible file upload attack!\n";
* }
*
* echo 'Here is some more debugging info:';
* print_r($_FILES);
*
* print "</pre>";
* }
* ?>
*
* </pre>
*
* TODO:
* - Add support for arbitrary form fields.
* - Add support for more than just one file.
* - Allow for changing of the boundary
*
* @author dejan
*/
public class HttpPost {
private final String crlf = "\r\n";
private URL url;
private URLConnection urlConnection;
private OutputStream outputStream;
private InputStream inputStream;
private String[] fileNames;
private String output;
private String boundary;
private final int bufferSize = 4096;
public HttpPost(URL argUrl) {
url = argUrl;
boundary = "---------------------------4664151417711";
}
public void setFileNames(String[] argFiles) {
fileNames = argFiles;
}
public void post() {
try {
System.out.println("url:" + url);
urlConnection = url.openConnection();
urlConnection.setDoOutput(true);
urlConnection.setDoInput(true);
urlConnection.setUseCaches(false);
urlConnection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
String postData = "";
String fileName = fileNames[0];
InputStream fileInputStream = new FileInputStream(fileName);
byte[] fileData = new byte[fileInputStream.available()];
fileInputStream.read(fileData);
// ::::: PART 1 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
String part1 = "";
part1 += "--" + boundary + crlf;
File f = new File(fileNames[0]);
fileName = f.getName(); // we do not want the whole path, just the name
part1 += "Content-Disposition: form-data; name=\"userfile\"; filename=\"" + fileName + "\""
+ crlf;
// CONTENT-TYPE
// TODO: add proper MIME support here
if (fileName.endsWith("png")) {
part1 += "Content-Type: image/png" + crlf;
} else {
part1 += "Content-Type: image/jpeg" + crlf;
}
part1 += crlf;
System.out.println(part1);
// File's binary data will be sent after this part
// ::::: PART 2 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
String part2 = crlf + "--" + boundary + "--" + crlf;
System.out.println("Content-Length"
+ String.valueOf(part1.length() + part2.length() + fileData.length));
urlConnection.setRequestProperty("Content-Length",
String.valueOf(part1.length() + part2.length() + fileData.length));
// ::::: File send ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
outputStream = urlConnection.getOutputStream();
outputStream.write(part1.getBytes());
int index = 0;
int size = bufferSize;
do {
System.out.println("wrote " + index + "b");
if ((index + size) > fileData.length) {
size = fileData.length - index;
}
outputStream.write(fileData, index, size);
index += size;
} while (index < fileData.length);
System.out.println("wrote " + index + "b");
System.out.println(part2);
outputStream.write(part2.getBytes());
outputStream.flush();
// ::::: Download result into the 'output' String :::::::::::::::::::::::::::::::::::::::::::::::
inputStream = urlConnection.getInputStream();
StringBuilder sb = new StringBuilder();
char buff = 512;
int len;
byte[] data = new byte[buff];
do {
len = inputStream.read(data);
if (len > 0) {
sb.append(new String(data, 0, len));
}
} while (len > 0);
output = sb.toString();
System.out.println("DONE");
} catch (Exception e) {
e.printStackTrace();
} finally {
System.out.println("Close connection");
try {
outputStream.close();
} catch (Exception e) {
System.out.println(e);
}
try {
inputStream.close();
} catch (Exception e) {
System.out.println(e);
}
}
} // post() method
public String getOutput() {
return output;
}
public static void main(String[] args) {
// Simple test, let's upload a picture
try {
HttpPost httpPost = new HttpPost(new URL("http://www.example.com/file.php"));
httpPost.setFileNames(new String[]{ "/home/dejan/work/ddn-100x46.png" });
httpPost.post();
System.out.println("=======");
System.out.println(httpPost.getOutput());
} catch (MalformedURLException ex) {
Logger.getLogger(HttpPost.class.getName()).log(Level.SEVERE, null, ex);
}
} // main() method
} // HttpPost class
如您所见，有许多需要改进的地方。此类使用HttpURLConnection并使用POST方法上载文件。我用它将图片上传到我的一个网站。
答案 1 :(得分：1)
看起来wamp是基于php的服务器。在这种情况下，服务器组件将需要一些PHP脚本来处理上载。 w3schools has a sample script for uploading via php（我不支持这个脚本，因为我从未使用过它，也没有使用php - 我只是提供它作为参考）。
文件上传的w3schools教程使用html将文件数据发布到服务器。使用JavaFX，您将使用Java编写文件post。 JavaFX客户端中的Java部分需要使用多部分表单帖子将文件从客户端发送到服务器。像apache httpclient这样的东西能够做到这一点。此帖子中有端到端解决方案的示例代码：How to upload a file using Java HttpClient library working with PHP。