Question

我是Android编程的新手，但对PHP有很好的了解。

我正在尝试通过邮寄请求将数据发送到我的服务器。我有以下代码：

ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("year","1980"));
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs,"UTF-8"));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
InputStream inps = entity.getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(inps));
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
    sb.append(line);
}
inps.close();
result=sb.toString();
Log.v("data-received",result);

我在互联网上的许多地方都找到了代码。

在PHP方面，我写这个：

<?php
    echo "something".$_REQUEST['year'];
?>

但我只是在我的应用程序结束时输出了“某些东西”（在日志值“数据接收”中）？

我做错了什么？我是否需要设置任何环境变量等？

Answer 1

$_REQUEST并非始终受支持，请使用$_GET或$_POST，具体取决于参数所在的位置。

Answer 2

我使用这段代码将日期发送到php文件。这是使用POST。我希望你能用它做点什么......

URL url;
            try {
                url = new URL("url.nl/phppage.php");


              InputStream myInputStream =null;
                StringBuilder sb = new StringBuilder();
                        //adding some data to send along with the request to the server

    // Data below:  
    sb.append("email="+email.getText().toString()+"&name="+name.getText().toString()+"&message="+om.getText().toString());


                    //url = new URL(url);
                    HttpURLConnection conn = (HttpURLConnection) url.openConnection();
                    conn.setDoOutput(true);
                    conn.setRequestMethod("POST");
                    OutputStreamWriter wr = new OutputStreamWriter(conn
                            .getOutputStream());
                                // this is were we're adding post data to the request
                                wr.write(sb.toString());
                    wr.flush();
                    myInputStream = conn.getInputStream();
                    wr.close();
                    Context context = getApplicationContext();
                     Log.i("TAG", "Message send successful");
                CharSequence text = "Message send successful";
                int duration = Toast.LENGTH_SHORT;

                Toast toast = Toast.makeText(context, text, duration);
                toast.show(); 
                finish();
                } catch (Exception e) {
                      Context context = getApplicationContext();
                      Log.e("TAG", "Message not sended - SERVER POST ERROR");
                            CharSequence text = "Message couldn't be send.";
                            int duration = Toast.LENGTH_SHORT;

                            Toast toast = Toast.makeText(context, text, duration);
                            toast.show(); 


                }

Answer 3

在代码中已经是一个检查，他在成功时给出了一个Toast消息和一个日志。在php服务器上你可以使用print_r($_POST);这应该没问题。你现在使用的代码是什么？

Answer 4

我终于开始工作了。唯一的变化是在第一行：

List<BasicNameValuePair> nameValuePairs = new ArrayList<BasicNameValuePairs>();

而不是

ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();

有人知道为什么会有效吗？

在服务器端未使用httppost发布

4 个答案: