Question

我正在寻找一个Java applet来从客户端机器读取文件，并为PHP服务器上传创建一个POST请求。

服务器上的PHP脚本应该在FORM提交中接收文件作为普通文件上传。我使用以下代码。文件内容传递给PHP脚本但它们没有正确转换为图像。

//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

这里有一些代码可以帮助你从我的一个旧项目中删除一堆不相关的东西，把它当作它的价值。基本上，我认为你的问题中的代码缺少HTTP协议所需的一些部分

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

我建议你看看Gallery Remote。这是一个用于将照片上传到PHP后端的开源项目。它比您可能需要的功能更全面，但您应该能够相当容易地修改代码以满足您的需求。

您还可以查看JUpload。它不是功能齐全，但它是开源的，能够完成任务。

用于上传文件的Java小程序

2 个答案: