使用POST將數據從Java（Android）發送到PHP

Question

我在Android中有一個應用程序，女巫需要拍照，並需要將其升級到使用php的服務器。

最初沒有問題，我可以上傳它，但是現在我需要傳遞將圖片上傳到服務器的路徑。 當我嘗試發送兩個值時，問題就來了。

我正在使用以下代碼：

private  void httpPostFileUpload(String fpath, String uPath){
        HttpURLConnection connection = null;
        DataOutputStream outputStream = null;
        DataInputStream inputStream = null;

        //Path where to upload
        String uploadPath = uPath;

        String pathToOurFile = fpath;
        Log.v("file path",fpath.toString());
        String urlServer = "http://leojg.alwaysdata.net/tests/androidConnect/handle_upload.php";

        String lineEnd = "\r\n";
        String twoHyphens = "--";
        String boundary =  "*****";
        String fname = "";

        int bytesRead, bytesAvailable, bufferSize;
        byte[] buffer;
        int maxBufferSize = 1*1024*1024;

        try
        {
            File f = new File(pathToOurFile);
            fname =  f.getName();

            FileInputStream fileInputStream = new FileInputStream(f);

            URL url = new URL(urlServer);
            connection = (HttpURLConnection) url.openConnection();

            // Allow Inputs & Outputs
            connection.setDoInput(true);
            connection.setDoOutput(true);
            connection.setUseCaches(false);

            // Enable POST method
            connection.setRequestMethod("POST");

            connection.setRequestProperty("Connection", "Keep-Alive");
            connection.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);

            outputStream = new DataOutputStream( connection.getOutputStream() );
            outputStream.writeBytes(twoHyphens + boundary + lineEnd);

            Log.d("path file", pathToOurFile);
            Log.d("path up", uploadPath);

        outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + pathToOurFile + "\"" + lineEnd);
            outputStream.writeBytes(lineEnd);

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

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

            while (bytesRead > 0)
            {
                outputStream.write(buffer, 0, bufferSize);
                bytesAvailable = fileInputStream.available();
                bufferSize = Math.min(bytesAvailable, maxBufferSize);
                bytesRead = fileInputStream.read(buffer, 0, bufferSize);
            }

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

            // Responses from the server (code and message)
            int serverResponseCode = connection.getResponseCode();
            String serverResponseMessage = connection.getResponseMessage();

            Log.d("amanda", "upload async finished, server response : " + serverResponseMessage);

            fileInputStream.close();
            fileInputStream = null;
            outputStream.flush();
            outputStream.close();
            outputStream = null;
            f= null;



            parent.invokeJs("pictureUploadEnded('" + fname + "')");
            this.delete(fpath);

        }
            catch (Exception ex)
        {
                Log.d("amanda", "exception uploading image : " + ex.getMessage());
        }     
}

我很確定問題出在這一行

outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + pathToOurFile + "\"" + lineEnd);

這是使職位成為主體的人。 我不知道在哪里上傳uploadPath

Answer 1

以下課程對我有用：

import java.io.DataOutputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import org.json.JSONException;
import org.json.JSONObject;
import org.json.JSONTokener;
import android.os.Environment;
import android.util.Log;

public class HttpPictureUpload implements Runnable {
private String TAG = "HTTPFILEUPLOAD";
private URL connectURL;
private FileInputStream fileInputStream = null;
private String pictureFileName;
private String myName;

public HttpPictureUpload(String urlString, String pictureFileNameString,
        String myNameString) {
    try {
        connectURL = new URL(urlString);

        pictureFileName = pictureFileNameString;
        myName = myNameString;

    } catch (Exception ex) {
        Log.i(TAG, "URL Malformatted");
    }
}

public void SendPictureNow() {

    Sending();
}

private void Sending() {
    int maxBufferSize = 1 * 1024*1024*1024;
    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "*****";

    try {

        fileInputStream = new FileInputStream(Environment
                .getExternalStorageDirectory().toString()
                + "/Picture/"
                + pictureFileName);

        Log.e(TAG, "Starting Http File Sending to URL");

        // Open a HTTP connection to the URL
        HttpURLConnection conn = (HttpURLConnection) connectURL
                .openConnection();
        conn.setChunkedStreamingMode(maxBufferSize);
        // conn.setchu

        // Allow Inputs
        conn.setDoInput(true);

        // Allow Outputs
        conn.setDoOutput(true);

        // Don't use a cached copy.
        conn.setUseCaches(false);

        // Use a post method.
        conn.setRequestMethod("POST");

        conn.setRequestProperty("Connection", "Keep-Alive");

        conn.setRequestProperty("Content-Type",
                "multipart/form-data;boundary=" + boundary);

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

        // sending appid
        dos.writeBytes(twoHyphens + boundary + lineEnd);
        dos.writeBytes("Content-Disposition: form-data; name=\"my_name\""
                + lineEnd);
        dos.writeBytes(lineEnd);
        dos.writeBytes(myName);
        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + lineEnd);


        dos.writeBytes("Content-Disposition: form-data; name=\"file\";filename=\""
                + System.currentTimeMillis() + "\"" + lineEnd);
        dos.writeBytes(lineEnd);

        int bytesAvailable = fileInputStream.available();

        byte[] buffer = new byte[bytesAvailable];

        int bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);

        while (bytesRead > 0) {
            dos.write(buffer, 0, bytesAvailable);
            bytesAvailable = fileInputStream.available();
            bytesAvailable = Math.min(bytesAvailable, maxBufferSize);
            bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);
        }
        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

        fileInputStream.close();

        dos.flush();

        Log.e(TAG,
                "File Sent, Response: "
                        + String.valueOf(conn.getResponseCode()));

        InputStream is = conn.getInputStream();

        // retrieve the response from server
        int ch;

        StringBuffer b = new StringBuffer();

        while ((ch = is.read()) != -1) {
            b.append((char) ch);

        }

        String s = b.toString();

        Log.i("Response", s);
        dos.close();
    } catch (MalformedURLException ex) {
        Log.e(TAG, "URL error: " + ex.getMessage(), ex);
    }

    catch (IOException ioe) {
        Log.e(TAG, "IO error: " + ioe.getMessage(), ioe);
    }
}

@Override
public void run() {
    // TODO Auto-generated method stub
}
 }

上面的類可以用來發送圖片文件或任何其他文件以及一個任意字符串。
在上面使用：

HttpPictureUpload uploadPicture=new HttpPictureUpload("http://serverurl.com/webservice.php", "tux.png", "my name is john");
uploadPicture.SendPictureNow();

Answer 2

請檢查此鏈接

本示例使用HttpClient和HttpPost類簡化了使用POST方法將數據發送到服務器並獲取響應的過程。