通過URLConnection寫圖像

Question

我想通過HttpURLConnection寫圖像。

我知道怎麼寫的文字，但我有試圖 寫的圖像

我已經成功地寫入本地HD使用ImageIO的實際問題：

但我想通過寫圖片ImageIO的關於URL和失敗

URL url = new URL(uploadURL); 
connection = (HttpURLConnection) url.openConnection(); 
connection.setDoOutput(true); 
connection.setRequestMethod("POST"); 
connection.setDoInput(true); 
connection.setUseCaches(false); 
connection.setRequestProperty("Content-Type", "multipart/form-data; 
              boundary=" + boundary); 
output = new DataOutputStream(connection.getOutputStream()); 
output.writeBytes("--" + boundary + "\r\n"); 
output.writeBytes("Content-Disposition: form-data; name=\"" + FIELD_NAME + "\"; 
              filename=\"" + fileName + "\"\r\n"); 
output.writeBytes("Content-Type: " + dataMimeType + "\r\n"); 
output.writeBytes("Content-Transfer-Encoding: binary\r\n\r\n"); 
ImageIO.write(image, imageType, output); 

的的uploadURL是鏈接直接將上傳的圖片在「內容處置指定的文件名在服務器上的ASP頁：部分

現在，當我發送這個然後asp頁面找到請求並找到文件的名稱。但沒有找到要上傳的文件。

的問題是，ImageIO的URL上書寫時會出現什麼文件的名稱在其上的ImageIO是寫作，

所以，請幫助我的ImageIO如何寫上URLConnection的圖像，我怎麼能知道我有在ASP頁面中使用上傳文件

感謝您抽出寶貴的時間來閱讀這篇文章 迪利普·阿加瓦爾

Answer 1

首先，我認爲，你應該叫io.flush()的文件名，然後寫入後io.close()圖片。

第二種內容類型對我來說似乎很奇怪。看起來你試圖提交表單，而實際上是圖像。我不知道你的期望是什麼，但通常當我編寫應通過HTTP傳輸文件的代碼時，我會發送適當的內容類型，例如image/jpeg。

下面是示例代碼片段，我從一個小工具，我寫的提取和我我目前的工作中使用：

URL url = new URL("http://localhost:8080/handler"); 
    HttpURLConnection con = (HttpURLConnection)url.openConnection(); 
    con.setDoInput(true); 
    con.setDoOutput(true); 
    con.setUseCaches(false); 
    con.setRequestProperty("Content-Type", "image/jpeg"); 
    con.setRequestMethod("POST"); 
    InputStream in = new FileInputStream("c:/temp/poc/img/mytest2.jpg"); 
    OutputStream out = con.getOutputStream(); 
    copy(in, con.getOutputStream()); 
    out.flush(); 
    out.close(); 
    BufferedReader r = new BufferedReader(new InputStreamReader(con.getInputStream())); 


      // obviously it is not required to print the response. But you have 
      // to call con.getInputStream(). The connection is really established only 
      // when getInputStream() is called. 
    System.out.println("Output:"); 
    for (String line = r.readLine(); line != null; line = r.readLine()) { 
     System.out.println(line); 
    } 

我這裏使用的方法複製（），我從雅加達IO utils的了。下面是引用代碼：

protected static long copy(InputStream input, OutputStream output) 
     throws IOException { 
    byte[] buffer = new byte[12288]; // 12K 
    long count = 0L; 
    int n = 0; 
    while (-1 != (n = input.read(buffer))) { 
     output.write(buffer, 0, n); 
     count += n; 
    } 
    return count; 
} 

顯然，服務器端必須準備好直接讀POST體內的圖像內容。 我希望這可以幫助。

Answer 2

的OP似乎失去被遺忘但對於風箏先生的利益：

// main method 
URL url = new URL(uploadURL); 
connection = (HttpURLConnection) url.openConnection(); 
connection.setDoOutput(true); // triggers "POST" 
// connection.setDoInput(true); // only if needed 
connection.setUseCaches(false); // dunno 
final String boundary = Long.toHexString(System.currentTimeMillis()); 
connection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" 
                   + boundary); 
output = new DataOutputStream(connection.getOutputStream()); 
try { 
    // image must be a File instance 
    flushMultiPartData(image, output, boundary); 
} catch (IOException e) { 
    System.out.println("IOException in flushMultiPartData : " + e); 
    return; 
} 
// ... 
private void flushMultiPartData(File file, OutputStream serverOutputStream, 
      String boundary) throws FileNotFoundException, IOException { 
    // SEE https://stackoverflow.com/a/2793153/281545 
    PrintWriter writer = null; 
    try { 
     // true = autoFlush, important! 
     writer = new PrintWriter(new OutputStreamWriter(serverOutputStream, 
       charsetForMultipartHeaders), true); 
     appendBinary(file, boundary, writer, serverOutputStream); 
     // End of multipart/form-data. 
     writer.append("--" + boundary + "--").append(CRLF); 
    } finally { 
     if (writer != null) writer.close(); 
    } 
} 

private void appendBinary(File file, String boundary, PrintWriter writer, 
     OutputStream output) throws FileNotFoundException, IOException { 
    // Send binary file. 
    writer.append("--" + boundary).append(CRLF); 
    writer.append(
     "Content-Disposition: form-data; name=\"binaryFile\"; filename=\"" 
      + file.getName() + "\"").append(CRLF); 
    writer.append("Content-Type: " 
      + URLConnection.guessContentTypeFromName(file.getName())) 
      .append(CRLF); 
    writer.append("Content-Transfer-Encoding: binary").append(CRLF); 
    writer.append(CRLF).flush(); 
    InputStream input = null; 
    try { 
     input = new FileInputStream(file); 
     byte[] buffer = new byte[1024]; 
     for (int length = 0; (length = input.read(buffer)) > 0;) { 
      output.write(buffer, 0, length); 
     } 
     output.flush(); // Important! Output cannot be closed. Close of 
     // writer will close output as well. 
    } finally { 
     if (input != null) try { 
      input.close(); 
     } catch (IOException logOrIgnore) {} 
    } 
    writer.append(CRLF).flush(); // CRLF is important! It indicates end of 
    // binary boundary. 
} 

您可能需要添加Gzip壓縮 - 見file corrupted when I post it to the servlet using GZIPOutputStream有或沒​​有Gzip已工人階級。 ImageIO有這裏沒有地方 - 只需寫出通過線路的字節並將ImageIO用於服務器上的內容。基於@BalusC answer