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我正在尝试通过 HttpURLConnection 编写图像。

我知道如何写文字，但我在尝试写图片时遇到了真正的问题

我已使用 ImageIO 成功写入本地 HD：

但我试图在 url 上通过 ImageIO 编写 Image 并且失败了

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);

uploadURL 是服务器上的 ASP 页面的 URL，该页面将使用“content-Disposition：part.xml”中给出的文件名上传图像。

现在当我发送这个然后asp页面找到请求并找到文件名。但没有找到要上传的文件。

问题是，当 ImageIO 在 URL 上写入时，ImageIO 正在写入的文件的名称是什么，

所以请帮助我 ImageIO 将如何在 URLConnection 上写入图像以及我如何知道我必须在 asp 页面中使用来上传文件的文件的名称

感谢您花时间阅读这篇文章 Dilip Agarwal

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2 回答 2

5

首先我相信你应该打电话io.flush()，然后io.close()在写图像之后。

第二种内容类型对我来说似乎很奇怪。您似乎正在尝试提交表单，而它实际上是图像。我不知道您的 asp 期望什么，但通常当我编写应该通过 HTTP 传输文件的代码时，我会发送适当的内容类型，例如image/jpeg.

以下是我从我编写的一个小实用程序中提取的示例代码片段，我在当前工作期间正在使用它：

    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);
    }

我在这里使用了从 Jakarta IO utils 获取的方法 copy()。这是供参考的代码：

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;
}

显然，服务器端必须准备好直接从 POST 正文中读取图像内容。我希望这有帮助。

于 2011-02-03T07:48:47.773 回答
0

OP 似乎被遗忘了，但为了风筝先生的利益：

// 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.
}

您可能想要添加 Gzip 压缩 -当我使用 GZIPOutputStream 将文件发布到 servlet 时，对于有或没有 Gzip 的工作类，您可能想要添加 Gzip 压缩。ImageIO 在这里没有位置- 只需将字节写入线路并将 ImageIO 用于您在服务器上的心脏内容。基于@BalusC 的回答

于 2013-09-23T08:48:30.207 回答