Java小程序上传文件 [英] Java applet to upload a file
本文介绍了Java小程序上传文件的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我要寻找一个Java小程序读取客户机的文件，并创造了PHP服务器上载POST请求。
在服务器上的PHP脚本应接受的形式文件作为普通文件上传提交。
我使用下面的code。文件内容传递给PHP脚本
但它们不正确地转换成图像。
//的uploadURL将PHP脚本像一个网址
// http://www.example.com/uploadfile.php网址URL =新的URL（的uploadURL）;
HttpURLConnection的CON =（HttpURLConnection类）url.openConnection（）;
con.setRequestMethod（POST）;
con.setDoInput（真）;
con.setDoOutput（真）;InputStream为=新的FileInputStream（C：//img.jpg）;
OutputStream的OS = con.getOutputStream（）;
字节[] = B1新的字节[千万]
INT N;
而（（N = is.read（B1））！= - 1）{
os.write（你好，0，5）;
测试+ = B1;}
con.connect（）;
解决方案
下面是一些code，可以帮助你，那是从我的旧项目，去掉一堆不相关的东西之一，把它用于它的价值。基本上，我认为你的问题code缺少一些零件HTTP协议要求
公共类UploaderExample
{
私有静态最后弦乐边界=--7d021a37605f0; 公共无效上传（URL网址，列表与LT;文件＆gt;文件）抛出异常
{
HttpURLConnection的theUrlConnection =（HttpURLConnection类）url.openConnection（）;
theUrlConnection.setDoOutput（真）;
theUrlConnection.setDoInput（真）;
theUrlConnection.setUseCaches（假）;
theUrlConnection.setChunkedStreamingMode（1024）; theUrlConnection.setRequestProperty（内容类型，的multipart / form-data的;边界=
+边界）; DataOutputStream类httpOut =新的DataOutputStream类（theUrlConnection.getOutputStream（））; 的for（int i = 0; I＆LT; files.size（）;我++）
{
文件f = files.get（ⅰ）;
字符串str = - +边界+\\ r \\ n
+内容处置：表格数据;名称= \\文件+ I +\\;文件名= \\+ f.getName（）+\\\\ r \\ n
+内容类型：图像/ PNG \\ r \\ n
+\\ r \\ n; httpOut.write（str.getBytes（））; 的FileInputStream uploadFileReader =新的FileInputStream（F）;
INT numBytesToRead = 1024;
INT availableBytesToRead;
而（（availableBytesToRead = uploadFileReader.available（））大于0）
{
字节[] bufferBytesRead;
bufferBytesRead = availableBytesToRead＆GT; = numBytesToRead？新的字节[numBytesToRead]
：新的字节[availableBytesToRead]
uploadFileReader.read（bufferBytesRead）;
httpOut.write（bufferBytesRead）;
httpOut.flush（）;
}
httpOut.write（（ - +边界+ - \\ r \\ n）的getBytes（））; } httpOut.write（（ - +边界+ - \\ r \\ n）的getBytes（））; httpOut.flush（）;
httpOut.close（）; //读取和放大器;解析响应
InputStream为= theUrlConnection.getInputStream（）;
StringBuilder的响应=新的StringBuilder（）;
字节[] = respBuffer新的字节[4096];
而（is.read（respBuffer）GT; = 0）
{
response.append（新的String（respBuffer）.trim（））;
}
is.close（）;
的System.out.println（response.toString（））;
} 公共静态无效的主要（字串[] args）抛出异常
{
清单＆LT;文件＆gt;名单=新的ArrayList＆LT;文件＆gt;（）;
list.add（新文件（C：\\\\ square.png））;
list.add（新文件（C：\\\\ narrow.png））;
UploaderExample上传=新UploaderExample（）;
uploader.upload（新URL（http://systemout.com/upload.php），清单）;
}}
I am looking for a Java applet to read a file from client machine and creat a POST request for PHP server uploading.
PHP script on server should receive the file as normal file upload in FORM submit. I am using the following code. The file contents are passed to PHP script but they are not correctly converted to an image.
//uploadURL will be a url of PHP script like
// http://www.example.com/uploadfile.php
URL url = new URL(uploadURL);
HttpURLConnection con = (HttpURLConnection)url.openConnection();
con.setRequestMethod("POST");
con.setDoInput(true);
con.setDoOutput(true);
InputStream is = new FileInputStream("C://img.jpg");
OutputStream os = con.getOutputStream();
byte[] b1 = new byte[10000000];
int n;
while((n = is.read(b1)) != -1) {
os.write("hello" , 0, 5);
test += b1;
}
con.connect();
解决方案
Here is some code that might help you it's from one of my old projects with a bunch of unrelated stuff removed, take it for what it's worth. Basically, I think the code in your question is missing some parts that the HTTP protocol requires
public class UploaderExample
{
private static final String Boundary = "--7d021a37605f0";
public void upload(URL url, List<File> files) throws Exception
{
HttpURLConnection theUrlConnection = (HttpURLConnection) url.openConnection();
theUrlConnection.setDoOutput(true);
theUrlConnection.setDoInput(true);
theUrlConnection.setUseCaches(false);
theUrlConnection.setChunkedStreamingMode(1024);
theUrlConnection.setRequestProperty("Content-Type", "multipart/form-data; boundary="
+ Boundary);
DataOutputStream httpOut = new DataOutputStream(theUrlConnection.getOutputStream());
for (int i = 0; i < files.size(); i++)
{
File f = files.get(i);
String str = "--" + Boundary + "\r\n"
+ "Content-Disposition: form-data;name=\"file" + i + "\"; filename=\"" + f.getName() + "\"\r\n"
+ "Content-Type: image/png\r\n"
+ "\r\n";
httpOut.write(str.getBytes());
FileInputStream uploadFileReader = new FileInputStream(f);
int numBytesToRead = 1024;
int availableBytesToRead;
while ((availableBytesToRead = uploadFileReader.available()) > 0)
{
byte[] bufferBytesRead;
bufferBytesRead = availableBytesToRead >= numBytesToRead ? new byte[numBytesToRead]
: new byte[availableBytesToRead];
uploadFileReader.read(bufferBytesRead);
httpOut.write(bufferBytesRead);
httpOut.flush();
}
httpOut.write(("--" + Boundary + "--\r\n").getBytes());
}
httpOut.write(("--" + Boundary + "--\r\n").getBytes());
httpOut.flush();
httpOut.close();
// read & parse the response
InputStream is = theUrlConnection.getInputStream();
StringBuilder response = new StringBuilder();
byte[] respBuffer = new byte[4096];
while (is.read(respBuffer) >= 0)
{
response.append(new String(respBuffer).trim());
}
is.close();
System.out.println(response.toString());
}
public static void main(String[] args) throws Exception
{
List<File> list = new ArrayList<File>();
list.add(new File("C:\\square.png"));
list.add(new File("C:\\narrow.png"));
UploaderExample uploader = new UploaderExample();
uploader.upload(new URL("http://systemout.com/upload.php"), list);
}
}
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