使用POST将数据从Java（Android）发送到PHP

Question

我在Android中有一个应用程序，女巫需要拍照，并需要将其升级到使用php的服务器。

最初没有问题，我可以上传它，但是现在我需要传递将图片上传到服务器的路径。 当我尝试发送两个值时，问题就来了。

我正在使用以下代码：

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());
        }     
}

我很确定问题出在这一行

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

这是使职位成为主体的人。 我不知道在哪里上传uploadPath

Answer 1

以下课程对我有用：

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
}
 }

上面的类可以用来发送图片文件或任何其他文件以及一个任意字符串。
在上面使用：

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

Answer 2

请检查此链接

本示例使用HttpClient和HttpPost类简化了使用POST方法将数据发送到服务器并获取响应的过程。