【问题标题】:json parsing showing url and title in textviewjson解析在textview中显示url和标题
【发布时间】:2012-08-20 16:09:20
【问题描述】:

我有一点关于 json / android 解析的事情。

我正在使用 json 将数据库中的数据解析到 android textview。

但它也显示了 url 和标题。

有没有办法隐藏它或修改代码使其不显示?

http://www.wvvzondag2.nl/android/leesverslag.php{"introtext":"我的数据库值"}

我想看到它显示了我在数据库中的值,所以没有 url 和括号。

这是我正在使用的代码。

package wvv.zondag.app;

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.text.Html;
import android.util.Log;
import android.widget.LinearLayout;
import android.widget.TextView;

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

TextView txt;

public static final String KEY_121 = "http://www.wvvzondag2.nl/android/leesverslag.php"; 

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.leesverslag);
    LinearLayout mLinearlayout= (LinearLayout) findViewById(R.id.Linear);
        txt = new TextView(this);
        txt.setText("Connecting..."); 
        mLinearlayout.addView(txt);

        Runnable postRunnable = new Runnable() {

            @Override
            public void run() {
                txt.setText(Html.fromHtml(getServerData(KEY_121))); 

            }
        };
        txt.post(postRunnable);
}

private String getServerData(String returnString) {
   InputStream is = null;

   String result = "";
    //Value from last activity for database
   String titelhoofd = getIntent().getStringExtra("titelverslag");
    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("titel",titelhoofd));

    //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),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);
                    json_data.getString("introtext");
                    //Get an output to the screen
                    returnString += jArray.getJSONObject(i); 
            }
    }catch(JSONException e){
            Log.e("log_tag", "Error parsing data "+e.toString());
    }
    return returnString; 
}    

}

有人知道怎么解决这个问题吗?

亲切的问候, 帕特里克

【问题讨论】:

    标签: php android json url


    【解决方案1】:

    创建一个变量

    String strData = "";
    

    然后使用

    //parse json data
    try{
            JSONArray jArray = new JSONArray(result);
            for(int i=0;i<jArray.length();i++){
                    JSONObject json_data = jArray.getJSONObject(i);
    
                    //Get an output to the screen
                    strData += json_data.getString("introtext"); 
            }
    }catch(JSONException e){
            Log.e("log_tag", "Error parsing data "+e.toString());
    }
    return strData ; 
    

    【讨论】:

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