我正在编写一个打算在 Android设备上运行的应用程序.应用程序通过PHP读取MysqL数据库中的信息,但是当我运行应用程序时,Log cat会提示错误’错误解析数据org.json.JSONException：Value

我得到的代码是从教程中下载的,请耐心等待我已经掌握了一些基本的PHP知识和很少的java知识.我已经测试了PHP脚本,它运行完美,所以我不会费心去附上它.

main.java代码：

 package test.an2MysqL;

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 main 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://10.1.1.19/cms/test/android2MysqL/read.PHP"; //i use my real ip here



private String getServerData(String returnString) {

   InputStream is = null;

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

    //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")+
                            ",country: "+json_data.getString("country")+
                            ",documentn: "+json_data.getInt("documentn")
                    );
                    //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;
}   

}

我很感激你能给我的任何帮助.

正如您对问题的评论中所指出的那样,您的服务器似乎正在返回XML而不是JSON.您只需输出结果即可轻松确认：

}catch(JSONException e){
    Log.e("log_tag","Error parsing data "+e.toString());
    Log.e("log_tag","Failed data was:\n" + result);
}

如果它是XML,它几乎肯定是必须的,那么你需要让服务器输出JSON,或者你需要解析它发送给你的XML.

是的,所以我只是进入指令,他们似乎非常棒.我遇到了一个问题：

我需要将一个图像数组传递给一个指令,以便我可以按照某些标准过滤它们.这是我的html调用指令：

<img cover="{{challenge.images}}">

这是我的指示：

myproject.directive('cover',function() {
    return {
        link: function ($scope,element,attrs) {
            console.debug("attrs.cover",Array(attrs.cover));
        }
    };
});

输出是一个字符串.有没有办法防止attr变成String？

我在这里假设您不想创建隔离范围,因此：

myproject.directive('cover',function($parse) {
    return {
        link: function ($scope,attrs) {

            var covers = $parse(attrs.cover)($scope);
            console.debug("attrs.cover",covers);
        }
    };
});

然后像这样使用指令：

<img cover="challenge.images">

我正在尝试将图像(png)存储到 Windows 8应用程序中的sqlite数据库,我发现它可以通过将其转换为base64字符串并将字符串存储到数据库来完成.稍后在应用程序中我想将该base64字符串转换为png图像并将其存储到指定位置.问题是我不知道如何将图像转换为base64和base64图像并将其存储到c#windows 8 app中的指定位置.任何帮助,将不胜感激.

public string ImagetoBase64(Image image,System.Drawing.Imaging.ImageFormat format)
{
  using (MemoryStream ms = new MemoryStream())
  {
    // Convert Image to byte[]
    image.Save(ms,format);
    byte[] imageBytes = ms.ToArray();

    // Convert byte[] to Base64 String
    string base64String = Convert.ToBase64String(imageBytes);
    return base64String;
  }
}



    public Image Base64ToImage(string base64String)
    {
        // Convert Base64 String to byte[]
        byte[] imageBytes = Convert.FromBase64String(base64String);
        using (var ms = new MemoryStream(imageBytes,imageBytes.Length))
        {

            // Convert byte[] to Image
            ms.Write(imageBytes,imageBytes.Length);
            Image image = Image.FromStream(ms,true);
            return image;
        }
    }