我有一個爲我的android應用程序提供數據的url,我從教程中學習並編寫了一些代碼。它完全適用於其他的URL,但這個無法檢索JSON響應
http://acolyteserv.appspot.com/Products/getProductMatchedList/?format=json&p0=galaxy&p1=4&p2=all
代碼:
private void testere()
{
InputStream is = null;
String result;
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://acolyteserv.appspot.com/Products/getProductMatchedList/?format=json&p0=galaxy&p1=4&p2=all");
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
result=sb.toString();
Toast t=Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG);
t.show();
}catch(Exception e){
Log.e("log_tag", "Error converting result "+e.toString());
}
}
catch(Exception e){
Log.e("log_tag", "Error in http connection "+e.toString());
}
}
如果我用這個URL敬酒,結果給我一個空字符串,但如果我使用例如URL像
http://api.geonames.org/earthquakesJSON?north=44.1&south=-9.9&east=-22.4&west=55.2&username=demo
請告訴我我在做什麼錯了,我是JSON和整個Web服務世界的新手。
感謝您指出了這一點!愚蠢的錯誤,它的工作很好。改變了它。我實際上使用了一個例子,它具有我刪除的那些參數,因爲我知道它不會對我有用。 – 2011-12-24 18:04:52