Question

嗨，我有这段代码：

protected void onPostExecute(String response) {
            String res=response.toString();
           // res = res.trim();
            res= res.replaceAll("\\s+","");   
            if(!res.equals("0")){
                un.setVisibility(View.GONE);
                ok.setVisibility(View.GONE);
                error.setVisibility(View.GONE);
                key.setVisibility(View.GONE);
                merchant.setVisibility(View.VISIBLE);
                String data = getServerData(res); 
                merchant.setText(data);
            }
            else
                error.setText("Incorrect Password");

        }


    }
    private String getServerData(String returnString) {

           InputStream is = null;

           String result = "";




            //convert response to string
            try{
                    BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
                    StringBuilder sb = new StringBuilder();
                    String line = null;
                    while ((line = reader.readLine()) != null) {
                            sb.append(line + "\n");
                    }
                    is.close();
                    result=sb.toString();
            }catch(Exception e){
                    Log.e("log_tag", "Error converting result "+e.toString());
            }
            //parse json data
            try{
                    JSONArray jArray = new JSONArray(result);
                    for(int i=0;i<jArray.length();i++){
                            JSONObject json_data = jArray.getJSONObject(i);
                            Log.i("log_tag","merchant_id: "+json_data.getInt("merchant_id")+
                                    ", merchant_name: "+json_data.getString("merchant_name")
                            );
                            //Get an output to the screen
                            returnString += "\n\t" + jArray.getJSONObject(i); 
                    }
            }catch(JSONException e){
                    Log.e("log_tag", "Error parsing data "+e.toString());
            }
            return returnString; 
        }

和这个PHP代码：

include 'config.php';
include 'opendb.php';
$key = $_POST['username'];
$key = mysql_real_escape_string($key);

$sql = "SELECT merchant_id, merchant_name FROM merchants WHERE merchant_key = '$key'";
$number = mysql_query($sql);
$rows = mysql_num_rows($number);

if ($rows > 0){
    while($result=mysql_fetch_assoc($number)){
        $output[]=$result;
        print(json_encode($result));
    }
}
else{
    echo 0;
}

基本上在android模拟器屏幕上我得到了json数组：

{"merchant_id":"1",
"merchant_name":
"ChowRestaurant"}

我想以这种方式在屏幕上显示：

“1 Chow Resstaurant”

只需要对此进行一些指导

Answer 1

因为你必须用库Gson，Jackson等解析Json，或者你也可以手动解析它

手动方式：http://www.androidcompetencycenter.com/2009/10/json-parsing-in-android/

Gson：https://sites.google.com/site/gson/gson-user-guide

杰克逊：http://wiki.fasterxml.com/JacksonInFiveMinutes

我建议使用 Gson

android将json转换为字符串

1 个答案: