Java小程序上传文件(Java applet to upload a file)

我要寻找一个Java小程序来读取客户机中的文件，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正常文件上传提交。我使用下面的代码。该文件的内容传递给PHP脚本，但他们没有被正确转换为图像。

//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();

Answer 1:

下面是一些代码，可以帮助你，那是从我的旧项目，一堆去除不相关的东西之一，它采取了它的价值。基本上，我觉得你的问题的代码缺少某些部分的HTTP协议要求

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);
    }

}

Answer 2:

我建议你看一看画廊远程。这是将照片上传到后端PHP的开源项目。这是一个更全面一点比你可能需要什么功能的，但你应该能够非常轻松地修改代码以您的需求。

您还可以看看JUpload 。它并不像功能齐全，但是它是开源的，能任务。