php - Android、PHP 和 JSON 之间的多个查询问题

Question

我必须执行一些查询才能从数据库中获取我需要的所有信息。大约有 7 个查询。如果我只用一个，它会完美地工作，但是当我尝试添加更多时，我会出错。

这是我的 PHP 代码。

<?php
  //connection etc

$sql="SELECT * FROM paramedic";
$result=mysql_query($sql)  or die(mysql_error()); 

$sql2="SELECT * FROM doctor";
$result2=mysql_query($sql2) or die(mysql_error());

while($row=mysql_fetch_array($result)) 
 $output[]=$row;

 while($rows=mysql_fetch_array($result2)) 
 $output2[]=$rows;


   print(json_encode(array($output, $output2)));

   mysql_close();
  ?>

这是我的安卓代码:

 btnLogin.setOnClickListener(new OnClickListener() {
        @Override
        public void onClick(View v) {
            // Check Login
            String username = etUsername.getText().toString().trim();
            String passwrd = etPassword.getText().toString().trim();

            try{
                httpclient=new DefaultHttpClient();
                httppost= new HttpPost("websitegoesher.php");

                nameValuePairs = new ArrayList<NameValuePair>();
                nameValuePairs.add(new BasicNameValuePair("user1",username));
                nameValuePairs.add(new BasicNameValuePair("pass1",passwrd));                   

                httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                response=httpclient.execute(httppost);
                HttpEntity entity = response.getEntity();
                is = entity.getContent();
            }catch(Exception e){
                Log.e("log_tag", "Eror at httpost "+e.toString());
            }

           //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 "+e.toString());
            }

            //Parse JSON data              
            try{
                jArray = new JSONArray(result);
                for(int i =0;i<jArray.length();i++){
                    JSONObject json_data = jArray.getJSONObject(i);

                    if(!json_data.getString("paramedic_serial").equals(null))
                    Log.i("log_tag","paramedic_license:           "+json_data.getString("paramedic_serial"));
                //  if(!json_data.getString("dr_serial").equals(null))           Log.i("log_tag","dr_serial:           "+json_data.getString("dr_serial"));



                }
            }catch(JSONException e){
                Log.e("log_tag", "Error parsing data "+e.toString());
            }

所以当我收到并尝试运行它时，我得到了这个错误

11-18 02:26:28.905: E/log_tag(27401): Error parsing data org.json.JSONException: Value [{"3":"1","2":"passwrd","inst_serial":"1","passwrd":"passwrd","username":"carlis","1":"carlis","paramedic_license":"123443","paramedic_serial":"100","0":"100","email":"gusti@upr.edu","5":"123443","4":"gusti@upr.edu"},{"3":"23","2":"passwrd","inst_serial":"23","passwrd":"passwrd","username":"paramedic","1":"paramedic","paramedic_license":"123111","paramedic_serial":"111","0":"111","email":"paramedic@aki.com","5":"123111","4":"paramedic@aki.com"},{"3":"23","2":"bb","inst_serial":"23","passwrd":"bb","username":"aa","1":"aa","paramedic_license":"1234","paramedic_serial":"138","0":"138","email":"email@email.com","5":"1234","4":"email@email.com"}] at 0 of type org.json.JSONArray cannot be converted to JSONObject

提前致谢。

Answer 1

你的 php 代码:

 print(json_encode(array($output, $output2)));

你在这里犯了错误。我认为您的输出类似于:{{{"":"","":""}}}

你需要这样写你的代码:

 $output3[]=array_merge($output,$output2);
 print(json_encode($output3));

用这段代码代替上面的代码试试。现在退房。希望这会有所帮助。