在将文件上传到PHP源代码时发送参数

时间:2018-09-08 08:11:12

标签: java php android http

我正在使用此类将图像从我的android应用上传到php服务器。

但是我想发送一些参数,例如自定义文件名,上传文件的用户等。

上传文件时有什么方法可以发送一些参数吗?

示例:name = newimage&uploadedby =用户名

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";
}

}

是否可以通过标头请求执行此操作,如下所示:

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

1 个答案:

答案 0 :(得分:1)

是的,可能

我不是Java开发人员,但是在PHP中,您可以通过以下两种方式接收此类数据:作为URL中的参数,该参数将在$ _GET数组中的PHP中提供 例如,将您的URL更改为:

Vibration mode

然后在PHP中:String upLoadServerUri = "http://website.com/abc.php?filename=blabla&anotherparam=1234";。如果要发布请求,也可以使用$ _GET。

或/,您可以在POST中执行此操作。您的POST请求可以包含所需数量的额外参数。 Here is an example如何做到这一点。