Question

我正在创建一个Android应用程序，允许用户通过在所述服务器上运行php脚本将他们希望销售的项目上传到服务器我通过namevaluepair将参数传递给PHP脚本

但我收到的错误是

未处理的异常类型IOException

未处理的例外类型



ClientProtocolException

在我尝试向数组添加更多参数之后出现这些错误

Btn_Submit.setOnClickListener(new OnClickListener() {



            @Override
            public void onClick(View v) {
                // TODO Auto-generated method stub

                    }
                byte[] data;
                HttpPost httppost;
                StringBuffer buffer;
                HttpResponse response;
                HttpClient httpclient;
                InputStream inputStream;
                List<NameValuePair> nameValuePairs;

                String ItemDescription = TxtItemDescription.getText().toString();
                String ItemTitle = TxtTitle.getText().toString();
                String StartTimeAndDate = TxtStartTimeAndDate.getText().toString();
                String StartPrice = TxtStartPrice.getText().toString();

class MyAsyncTask extends AsyncTask<String, Integer, Double>{

                    protected Double doInBackground(String...  params){
                        return (Double) null;
                        }       
                    protected void onPostExecute(Double result){
                        Toast.makeText(getApplicationContext(), "Data Sent", Toast.LENGTH_LONG).show();

                httpclient = new DefaultHttpClient();
                httppost = new HttpPost(
                        "http://www.albarrett.co.uk/PS_Auctions/test.php");
                // Add your data
                List<NameValuePair> params = new ArrayList<NameValuePair>();
                params.add(new BasicNameValuePair("name", ItemDescription));
                params.add(new BasicNameValuePair("price", ItemTitle));
                params.add(new BasicNameValuePair("description", StartTimeAndDate));


                // Execute HTTP Post Request
                response = httpclient.execute(httppost);
                inputStream = response.getEntity().getContent();

                data = new byte[256];

                buffer = new StringBuffer();
                int len = 0;
                while (-1 != (len = inputStream.read(data))) {
                    buffer.append(new String(data, 0, len));
                }

                inputStream.close();
            }
}
        });

Answer 1

未处理并不一定意味着抛出错误，而是在抛出错误时需要执行某些操作。

您可以使用try / catch执行此操作。如果你使用eclipse，你可以点击错误并自动应用它，当然你也可以手动执行：

try {  
    .. your code that might throw something ..
} catch(IOException ex) {
    //writes to your log
    ex.printStackTrace();
}

您可以将IOException替换为任何其他例外，或使用Exception来捕获任何例外。

注意：顺便说一句，如果您这样做，请阻止主线程上的网络连接。

android中不支持的异常

1 个答案: