php - 需要发送图片和文字到服务器

Question

在我的应用程序中，我需要将图像和注释（180 个文本）字符发送到服务器。

现在可以单独发送图像和文本，但我现在需要一起发送。

方法是什么？

目前正在使用线程发送图像。

// open a URL connection to the Servlet
                FileInputStream fileInputStream = new FileInputStream(
                        sourceFile);
                URL url = new URL(
                        "http://xxx.com/image.php");

                // Open a HTTP connection to the URL
                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("uploaded_file", fileName);
                // conn.setRequestProperty("id", imei);

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

                dos.writeBytes(twoHyphens + boundary + lineEnd);
                dos.writeBytes("Content-Disposition: form-data; name=uploaded_file; filename="
                        + fileName + imei + lineEnd);


                dos.writeBytes(lineEnd);

                // create a buffer of maximum size
                bytesAvailable = fileInputStream.available();

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

                // read file and write it into form...
                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);

                }

                // send multipart form data necesssary after file data...
                dos.writeBytes(lineEnd);
                dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

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

                Log.i("uploadFile", "HTTP Response is : "
                        + serverResponseMessage + ": " + serverResponseCode);

                if (serverResponseCode == 200) {

                    runOnUiThread(new Runnable() {
                        public void run() {

                            Toast.makeText(getBaseContext(),
                                    "File Upload Complete.", Toast.LENGTH_SHORT)
                                    .show();
                            File delfile = new File(currentfile);
                            //delfile.delete();
                        }
                    });
                }

                // close the streams //
                fileInputStream.close();
                dos.flush();
                dos.close();

            } catch (MalformedURLException ex) {

                // dialog.dismiss();
                ex.printStackTrace();

                runOnUiThread(new Runnable() {
                    public void run() {
                        // messageText.setText("MalformedURLException Exception : check script url.");
                        Toast.makeText(getBaseContext(),
                                "MalformedURLException", Toast.LENGTH_SHORT)
                                .show();
                    }
                });

                Log.e("Upload file to server", "error: " + ex.getMessage(), ex);
            } catch (Exception e) {

                // dialog.dismiss();
                e.printStackTrace();

                runOnUiThread(new Runnable() {
                    public void run() {
                        // messageText.setText("Got Exception : see logcat ");
                        Toast.makeText(getBaseContext(),
                                "upload failed Please, try after some time ",
                                Toast.LENGTH_SHORT).show();
                    }
                });
                Log.e("Upload file to server Exception",
                        "Exception : " + e.getMessage(), e);
            }
            // dialog.dismiss();
            return serverResponseCode;

        } // End else block

使用简单的 http post 方法发送文本。

我的应用程序需要这个，我们将在其中拍摄快照并在编辑文本框中输入一些内容并发送。

score 5 · Accepted Answer

首先，您需要这个org.apache.httpcomponents.httpclient_4.2.1 jar 文件并导入到您的构建路径。

你可以这样做。考虑以下

import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONException;
import org.json.JSONObject;

和声明：

 private List<NameValuePair> nameValuePairs;

private HttpURLConnection connection = null;
private DataOutputStream outputStream = null;
private String lineEnd = "\r\n";
private String twoHyphens = "--";
private String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;

源代码如下：

try {

            FileInputStream fileInputStream = new FileInputStream(StickerToSend);
            URL url = new URL("Your URL");
            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());

            // ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
                   // Here you can add parameters as follows :

            outputStream.writeBytes(twoHyphens + boundary + lineEnd);

                    // The keyword "type" is the key value and 

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

                    // You can assign values as like follows : 

            outputStream.writeBytes("Your value");
            outputStream.writeBytes(lineEnd);
            outputStream.writeBytes(twoHyphens + boundary + lineEnd);

            ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

            outputStream.writeBytes(twoHyphens + boundary + lineEnd);
            outputStream
                    .writeBytes("Content-Disposition: form-data; name=\"sticker\";filename=\""
                            + StickerToSend + "\"" + 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);

            // int serverResponseCode = connection.getResponseCode();
            String serverResponseMessage = connection.getResponseMessage();

            Log.d("The server response message ", " Server Response"
                    + serverResponseMessage);

            fileInputStream.close();
            outputStream.flush();
            // ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
            InputStream is = connection.getInputStream();
            // retrieve the response from server
            int ch;
            StringBuffer b = new StringBuffer();
            while ((ch = is.read()) != -1) {
                b.append((char) ch);
            }
            String result = b.toString();
            Log.i("Response", result);

            JSONObject jsonobject;
            try {
                jsonobject = new JSONObject(result);
                boolean isPosted = jsonobject.getJSONObject("response")
                        .getString("httpCode").equals("200") ? true : false;

                if (isPosted)
                    mHandler.sendEmptyMessage(POST_SUCCESS);
                else
                    mHandler.sendEmptyMessage(POST_FAILURE);
            } catch (JSONException e) {
                Log.d("the xceptions ",
                        "Xcep in posting status messages are : "
                                + e.getMessage());
            }
            outputStream.close();

        } catch (Exception e) {
            Log.d("Xceptions",
                    "Xceptions are  upload video file  " + e.getMessage());
        }

这段代码真的很好用。我也检查过这个。如果有任何问题，请随意。

score 0 · Accepted Answer

试试这个可能对你有帮助

public void WSCall(final String filePath, final String textTosend) {

        try {

            DefaultHttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://xxx.com/image.php");
            StringBody data = new StringBody(textTosend, Charset.forName(HTTP.UTF_8));
            MultiPartEntity entity = new MultiPartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
            entity.addPart("uploaded_file", new FileBody(new File(filePath), "image/png"));
            entity.addPart("text", data);
            httppost.setEntity(entity);
            HttpResponse WSresponse = httpclient.execute(httppost);
        } catch (Throwable e) {

        }

    }

score 0 · Accepted Answer

您可以使用 base64 编码器将图像编码为字符串（Base 64 编码和解码示例代码），并创建一个 Web 服务将其发送到服务器。因此，在该 Web 服务中，您可以添加多个输入

php - 需要发送图片和文字到服务器

3 回答 3

Related

Reference