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Question

我尝试将图像（base64编码）与其他值一起上传到PHP网络服务器，该服务器获取此信息并进行进一步处理。

以下函数是AsyncTask的一部分，它执行请求。

我通过手工连接不同的值来构建postData。

JsonUrlLook类帮助我找到发送数据的正确URL。我验证了它。这是对的。

当我尝试执行创建的HttpPost时，我的响应为null。有谁知道这是为什么？

private String getPostData(String arrayString, String base64Image, String boundary) {
    String remoteMethod = "uploadPic";
    String postData = "";

    // Add remote method name
    postData += "--" + boundary + "\r\n";
    postData += "Content-Disposition: form-data; name=\"" + boundary + "[method]\"\r\n\r\n";
    postData += remoteMethod;

    postData += "--" + boundary + "\r\n";
    postData += "Content-Disposition: form-data; name=\"" + boundary + "[params]\"\r\n\r\n";
    postData += arrayString;

    postData += "--" + boundary + "\r\n";
    postData += "Content-Disposition: form-data; name=\"" + boundary + "\"\r\n\r\n";
    postData += "filename=\"image.jpg\"\r\n\r\n";
    postData += base64Image;

    postData += "--" + boundary + "--";

    return postData;
}

private String uploadPicture(String arrayString, String base64Image) {

    String boundary = "tx_fpoemobile_pi2";

    // Get post data
    String postData = getPostData(arrayString, base64Image, boundary);

    // Get webservice uri
    JsonUrlLookup lookup = new JsonUrlLookup(context);
    URL webserviceUrl = lookup.buildUrl(context.getResources().getString(R.string.page_uid_webservice));
    URI webserviceUri = null;
    try {
        webserviceUri = new URI(webserviceUrl.toString());
    } catch(Exception e) {
        // TODO
    }

    // Build client and post
    HttpClient httpClient = new DefaultHttpClient();
    HttpPost httpPost = new HttpPost(webserviceUri);

    // Set header of post
    httpPost.setHeader("Content-Type", "multipart/form-data; boundary=" + boundary);
    httpPost.setHeader("Content-Length", String.valueOf(postData.length()));

    // Set body
    try {
        httpPost.setEntity(new StringEntity(postData));
    } catch (Exception e) {
        // TODO
    }

    HttpResponse response = null;
    try {
        response = httpClient.execute(httpPost);
    } catch (ClientProtocolException e) {
        // TODO
    } catch (IOException e) {
        // TODO
    }

    return "uploadPictureSuccess";
}

Answer 1

你必须像这样工作

public void connectForMultipart() throws Exception {
    con = (HttpURLConnection) ( new URL(url)).openConnection();
    con.setRequestMethod("POST");
    con.setDoInput(true);
    con.setDoOutput(true);
    con.setRequestProperty("Connection", "Keep-Alive");
    con.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
    con.connect();
    os = con.getOutputStream();
}

public void addFormPart(String paramName, String value) throws Exception {
    writeParamData(paramName, value);
}

public void addFilePart(String paramName, String fileName, byte[] data) throws Exception {
    os.write( (delimiter + boundary + "\r\n").getBytes());
    os.write( ("Content-Disposition: form-data; name=\"" + paramName +  "\"; filename=\"" + fileName + "\"\r\n"  ).getBytes());
    os.write( ("Content-Type: application/octet-stream\r\n"  ).getBytes());
    os.write( ("Content-Transfer-Encoding: binary\r\n"  ).getBytes());
    os.write("\r\n".getBytes());

    os.write(data);

    os.write("\r\n".getBytes());
}   
public void finishMultipart() throws Exception {
    os.write( (delimiter + boundary + delimiter + "\r\n").getBytes());
}

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httpPost.setEntity(entity);
HttpResponse response = httpClient.execute(httpPost);

BufferedReader reader = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"));



String sResponse;
while ((sResponse = reader.readLine()) != null) 
 {
     s = s.append(sResponse);
 }

 if(response.getStatusLine().getStatusCode() == HttpStatus.SC_OK)
 {
     return s.toString();
 }else
 {
     return "{\"status\":\"false\",\"message\":\"Some error occurred\"}";
 }   
} catch (Exception e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

代替图像，您可以发送任何内容。

在这种情况下，您需要httpmime-4.2.1-1.jar

使用HttPost和HttpClient上传base64编码图像

1 个答案:

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