Question

如何通过gallary或相机在服务器上传图像？

我尝试了this link的代码，但它显示的错误就像这样

错误：（329,24）错误：无法访问AbstractBody 找不到org.apache.james.mime4j.message.AbstractBody的类文件

Answer 1

如果您使用async-http作为您的http客户端click-here，那么您可以使用以下方法[如果您选择async-http，您可以使用此方法] 此方法不仅可以上传图像，还可以用于对服务器进行正常的帖子调用。

YourFragment.class [此类是您调用http请求]

 public class YourFragment extends CommanAbstract 
    {


    @Override
        public View onCreateView(LayoutInflater inflater, ViewGroup container,
                Bundle savedInstanceState) {
            // TODO Auto-generated method stub

            View v = inflater.inflate(R.layout.your_layout, container, false);
            ButterKnife.inject(this, v);

            return v;
        }


    private void linkCall() {
            // TODO Auto-generated method stub

            RequestParams params = new RequestParams();
            params.put("params1", "value1");
            params.put("params2", "value2");
//note that "image" tag will be the tag which accepts the file to the server
            params.put("image", new File("your image path from gallery or camera"));
            parse(params, 1, your link here, true);

        }
    @Override
        public void parseresult(String response, boolean success, int value) {
            // TODO Auto-generated method stub

            switch (value) {
            case 1:
                //here you can parse the link response
                break;
            }
        }

    @Override
        public void error(String response) {
            // TODO Auto-generated method stub
            //error will come here
        }

    }

CommanAbstract类[扩展这个类，其中有一个类，你有一个http请求]

public abstract class CommanAbstract extends Fragment{

private static final String Url = "your link head here";

    public abstract void parseresult(String response, boolean success, int value);

    public abstract void error(String response);

    public void parse(RequestParams params, final int value, String link,
            boolean progrss) {
        // TODO Auto-generated method 

        AsyncHttpClient client = new AsyncHttpClient();
        GlobalFunctions.postApiCall(getActivity(), link, params, client,
                new HttpResponseHandler() {

                    @Override
                    public void handle(String response, boolean success) {
                        // TODO Auto-generated method stub

                        if (success) {
                            parseresult(response, true, value);
                        } else {
                            error(response);
                            toast("Connection error");
                        }
                    }
                });
    }
}

GlobalFunctions类[此类调用http请求]

public class GlobalFunctions {
    // static ProgressDialog progress;

    public interface HttpResponseHandler {
        void handle(String response,boolean failre);
    }


    public static void postApiCall(final Context context, final String url,
            RequestParams params, AsyncHttpClient httpClient,
            final HttpResponseHandler handler) {

        httpClient.post(url, params, new AsyncHttpResponseHandler() {
            @Override
            public void onFailure(Throwable arg0, String failureResponse) {
                // TODO Auto-generated method stub

                super.onFailure(arg0, failureResponse);
                System.out.println("fail" + failureResponse + "url is" + url);
                handler.handle(failureResponse,false);

                // errorToast(context);
            }

            @Override
            public void onSuccess(String response) {



                handler.handle(response,true);

            }
        });
    }

}

Answer 2

您可以使用http post multi-part将图像上传到服务器。这是一个可以帮助你的链接。

https://vikaskanani.wordpress.com/2011/01/11/android-upload-image-or-file-using-http-post-multi-part/

http://www.codejava.net/java-se/networking/upload-files-by-sending-multipart-request-programmatically

上传图片的另一种方法是将其转换为Base64而不是上传。请查看以下链接。

Android upload image to server using base64

通过Httppost将图像上传到android中的服务器

2 个答案: