Android Mysql JSON輸出格式不正確

Question

嗨，我正在使用此代碼通過PHP將Android連接到Mysql

PHP文件

    <?php

      mysql_connect("127.0.0.1","root","xxpasswordxx");

      mysql_select_db("peopledata");

      $q=mysql_query("SELECT * FROM people WHERE birthyear>'".$_REQUEST['year']."'");

      while($e=mysql_fetch_assoc($q))

              $output[]=$e;

           print(json_encode($output));

    mysql_close();
?>

這是java

    package com.connector;

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import android.app.Activity;
import android.os.Bundle;
import android.util.Log;
import android.widget.LinearLayout;
import android.widget.TextView;


public class whitehat extends Activity {
/** Called when the activity is first created. */

   TextView txt;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    // Create a crude view - this should really be set via the layout resources 
    // but since its an example saves declaring them in the XML. 
    LinearLayout rootLayout = new LinearLayout(getApplicationContext()); 
    txt = new TextView(getApplicationContext()); 
    rootLayout.addView(txt); 
    setContentView(rootLayout); 

    // Set the text and call the connect function. 
    txt.setText("Connecting...");
  //call the method to run the data retreival
    txt.setText(getServerData(KEY_121));



}
public static final String KEY_121 = "http://xx.xx.xxx.xxx/hellomysql/mysqlcon.php"; //i use my real ip here



private String getServerData(String returnString) {

   InputStream is = null;

   String result = "";
    //the year data to send
    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("year","1970"));

    //http post
    try{
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost(KEY_121);
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();

    }catch(Exception e){
            Log.e("log_tag", "Error in http connection "+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 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","id: "+json_data.getInt("id")+
                            ", name: "+json_data.getString("name")+
                            ", sex: "+json_data.getInt("sex")+
                            ", birthyear: "+json_data.getInt("birthyear")
                    );
                    //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;
}   

}

輸出是這樣的

{“ birthyear”：“ 1973”，“ id”：“ 1”，“ sex”：“ 1”，“ name”：“ vlizzard”}

但這不是正確的JSON格式！ 如果我這樣評論Java的一部分

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

              /*  Log.i("log_tag","id: "+json_data.getInt("id")+
                        ", name: "+json_data.getString("name")+
                        ", sex: "+json_data.getInt("sex")+
                        ", birthyear: "+json_data.getInt("birthyear") */

                );
                //Get an output to the screen
                returnString += "\n\t" + jArray.getJSONObject(i); 

無論如何都可以工作，這意味着PHP文件的結果仍然可以顯示在屏幕上...這是怎么回事？ 有人可以幫我嗎？

Answer 1

更改為以下代碼

mysql_connect("127.0.0.1","root","xxpasswordxx"); 

代替

mysql_connect("localhost","root","");

我希望它有用

Answer 2

是的，那是有效的JSON。 你為什么認為不是？ 您讀取JSON值的方式可能有問題。 它們似乎都以字符串形式返回。 嘗試使用getString（）而不是getInt（），然后根據需要轉換為int或其他數據類型。