上传文件到PHP源码时发送参数

Question

我使用这个类将图片从我的android应用上传到php服务器。

但我想发送一些参数，如自定义文件名，上传文件的用户等。

有没有办法在上传文件的同时发送一些参数？

示例: name=newimage&uploadedby=username

private class UploadFileAsync extends AsyncTask<String, Void, String> {

@Override
protected String doInBackground(String... params) {

    try {
        String sourceFileUri = "/mnt/sdcard/abc.png";

        HttpURLConnection conn = null;
        DataOutputStream dos = null;
        String lineEnd = "\r\n";
        String twoHyphens = "--";
        String boundary = "*****";
        int bytesRead, bytesAvailable, bufferSize;
        byte[] buffer;
        int maxBufferSize = 1 * 1024 * 1024;
        File sourceFile = new File(sourceFileUri);

        if (sourceFile.isFile()) {

            try {
                String upLoadServerUri = "http://website.com/abc.php?";

                // open a URL connection to the Servlet
                FileInputStream fileInputStream = new FileInputStream(
                        sourceFile);
                URL url = new URL(upLoadServerUri);

                conn = (HttpURLConnection) url.openConnection();
                conn.setDoInput(true); // Allow Inputs
                conn.setDoOutput(true); // Allow Outputs
                conn.setUseCaches(false); // Don't use a Cached Copy
                conn.setRequestMethod("POST");
                conn.setRequestProperty("Connection", "Keep-Alive");
                conn.setRequestProperty("ENCTYPE",
                        "multipart/form-data");
                conn.setRequestProperty("Content-Type",
                        "multipart/form-data;boundary=" + boundary);
                conn.setRequestProperty("bill", sourceFileUri);

                dos = new DataOutputStream(conn.getOutputStream());

                dos.writeBytes(twoHyphens + boundary + lineEnd);
                dos.writeBytes("Content-Disposition: form-data; name=\"bill\";filename=\""
                        + sourceFileUri + "\"" + lineEnd);

                dos.writeBytes(lineEnd);

                bytesAvailable = fileInputStream.available();

                bufferSize = Math.min(bytesAvailable, maxBufferSize);
                buffer = new byte[bufferSize];

                bytesRead = fileInputStream.read(buffer, 0, bufferSize);

                while (bytesRead > 0) {

                    dos.write(buffer, 0, bufferSize);
                    bytesAvailable = fileInputStream.available();
                    bufferSize = Math
                            .min(bytesAvailable, maxBufferSize);
                    bytesRead = fileInputStream.read(buffer, 0,
                            bufferSize);

                }

                dos.writeBytes(lineEnd);
                dos.writeBytes(twoHyphens + boundary + twoHyphens
                        + lineEnd);

                serverResponseCode = conn.getResponseCode();
                String serverResponseMessage = conn
                        .getResponseMessage();

                if (serverResponseCode == 200) {

                    //Toast.makeText(ctx, "File Upload Complete.",
                    //      Toast.LENGTH_SHORT).show();



                }

                fileInputStream.close();
                dos.flush();
                dos.close();

            } catch (Exception e) {

                e.printStackTrace();

            }

        }


    } catch (Exception ex) {

        ex.printStackTrace();
    }
    return "Executed";
}

}

有没有办法对header请求这样做：

conn.setRequestProperty("parameters","name=filename&uploadedby=username");

Answer 1

是的，这是可能的

我不是java开发人员，但在PHP中，您可以通过两种方式接收此类数据:将URL中的参数更改为：

String upLoadServerUri = "http://website.com/abc.php?filename=blabla&anotherparam=1234";

然后在PHP中：echo $_GET['filename']。如果您正在POSTing您的请求，也可以使用$_GET。

或者/你可以在你的帖子中这样做。POST请求可以包含所需的任意多个额外参数。Here is an example你怎么做到这一点。