Question

这里是代码，但它无法获取json数据，虽然浏览器显示json数据时我打开它为什么？php代码工作正常但是当android试图获取它时说我我取得空白

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    StrictMode.enableDefaults();
    resultview=(TextView)findViewById(R.id.resulttext);
    getdata();

}

public void getdata()
{
    String result=null;
    InputStream inputstream=null;
    try{
        HttpClient httpclient=new DefaultHttpClient();
        HttpPost httppost=new HttpPost("http://127.0.0.1/getallcustomers.php");
        HttpResponse response=httpclient.execute(httppost);
        HttpEntity httpentity=response.getEntity();
        inputstream= httpentity.getContent();
    }
    catch(Exception ex)
    {
        resultview.setText(ex.getMessage()+"at 1st exception");
    }
    try{
        BufferedReader reader=new BufferedReader(new InputStreamReader(inputstream,"iso-8859-1"),8);
        StringBuilder sb=new StringBuilder();
        String line=null;
        while((line=reader.readLine())!= null)                  
        {
            sb.append(line +"\n");
        }
        inputstream.close();
        result=sb.toString();
    }
    catch(Exception ex)
    {
        resultview.setText(ex.getMessage()+"at 2nd exception");
    }
    try{
        String s="test :";
        JSONArray jarray=new JSONArray(result);
        for(int i=0; i<jarray.length();i++)
        {
            JSONObject json=jarray.getJSONObject(i);
            s= s +
                    "Name : "+json.getString("firstname")+" "+json.getString("lastname")+"\n"+
                    "age :"+json.getString("age")+"\n"+
                    "phone : "+json.getString("phone");
        }
        resultview.setText(s);
    }
    catch(Exception ex)
    {
        this.resultview.setText(ex.getMessage()+"at 3rd exception");
    }

}

}

Answer 1

您确定可以从模拟器访问本地主机吗？你应该看看这个： Accessing localhost:port from Android emulator

Answer 2

您正在UI线程上启动http请求。可能它崩溃了。

尝试记录异常，而不是放在resultview对象中并确认

Answer 3

是你的PHP版本5.5？

你应该使用mysqli而不是使用mysql

php代码是正确的，也许访问数据库没有正确的数据

Answer 4

从服务器获取json数据的代码。它对我来说很好。

private void get_valueFromServer(String php) {
String responseString = null;
try{    
    HttpClient httpclient = new DefaultHttpClient();
    String url ="your_url"; 
    HttpPost httppost = new HttpPost(url);
    HttpResponse response = httpclient.execute(httppost);
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    response.getEntity().writeTo(out);
    out.close();
    responseString = out.toString();
    Toast.makeText(getApplicationContext(),responseString,1000).show();

    //The below code is for Separating values.It may varies according with your result from server.

    JSONArray ja = new JSONArray(responseString);
    int x=Integer.parseInt(ja.getJSONObject(0).getString("your_string_name"));

} catch(Exception e) {
}
}

我的android应用程序没有从服务器获取json数据

4 个答案: