一个天气，json哪里出错，谢谢

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
init();
}

//对spinner赋值
private void init()
{
Spinner city_spr = (Spinner) findViewById(R.id.Spinner01);//取得xml中的按钮
ArrayAdapter adapter = new ArrayAdapter(this, android.R.layout.simple_spinner_item, ConstData.city);
adapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item);
city_spr.setAdapter(adapter);//放上去

    Button submit = (Button) findViewById(R.id.Button01);//确认按钮

    //设置它的监听
    submit.setOnClickListener(new OnClickListener() {
        @Override
        public void onClick(View v)
        {
            // TODO Auto-generated method stub

            Spinner spr = (Spinner) findViewById(R.id.Spinner01);
            Long l = spr.getSelectedItemId();//被选择项的id 
            int index = l.intValue();
            String cityParamString = ConstData.cityCode[index];//得到城市代码                 

            String str = "http://www.weather.com.cn/data/cityinfo/"+cityParamString+".html";
            str=getWebContent(str); 
            try {

             JSONObject json;
             String str1 = null,str2 = null,str3 = null;

                json=new JSONObject(str).getJSONObject("weatherinfo");
                str1=json.getString("city");
                str2=json.getString("temp1");
                str3=json.getString("weather1");
              TextView tex1=(TextView)findViewById(R.id.city_name);
               tex1.setText(str1);
               ///tex1.setText("1");
               TextView tex2=(TextView)findViewById(R.id.wind);
               tex2.setText(str2);
               //tex2.setText("2");
               TextView tex3=(TextView)findViewById(R.id.temp);
               tex3.setText(str3);
               //tex3.setText("3");
            } catch (JSONException e1) {
                // TODO Auto-generated catch block
                e1.printStackTrace();
            }

        }
    });

}

public  String getWebContent(String url) {
    //创建一个http请求对象
    HttpGet request = new HttpGet(url);
    //创建HttpParams以用来设置HTTP参数
    HttpParams params=new BasicHttpParams();
    //设置连接超时或响应超时
    HttpConnectionParams.setConnectionTimeout(params, 3000);
    HttpConnectionParams.setSoTimeout(params, 5000);
    //创建一个网络访问处理对象
    HttpClient httpClient = new DefaultHttpClient(params);
    try{
        //执行请求参数项
        HttpResponse response = httpClient.execute(request);
        //判断是否请求成功
        if(response.getStatusLine().getStatusCode() == HttpStatus.SC_OK) {
            //获得响应信息
            String content = EntityUtils.toString(response.getEntity());
            return content;
        } else {
            //网连接失败，使用Toast显示提示信息

        }

    }catch(Exception e) {
        e.printStackTrace();
    } finally {
        //释放网络连接资源
        httpClient.getConnectionManager().shutdown();
    }
    return null;
}

有什么错误？

试了一下，返回的json string是
{"weatherinfo":{"city":"北京","cityid":"101010100","temp1":"3℃","temp2":"-6℃","weather":"晴转多云","img1":"d0.gif","img2":"n1.gif","ptime":"08:00"}}

没有weather1，应该是weather

查看json格式是否正确
解析对了吗

参考：http://blog.csdn.net/hello_haozi/article/details/7564223
http://www.bubuko.com/infodetail-469483.html
http://my.oschina.net/xiaopihailaotou/blog/160960

报的什么错误？解析json的话 推荐使用GSON