我有一个为我的android应用程序提供数据的url,我从教程中学习并编写了一些代码。它完全适用于其他的URL,但这个无法检索JSON响应
http://acolyteserv.appspot.com/Products/getProductMatchedList/?format=json&p0=galaxy&p1=4&p2=all
代码:
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());
}
}
如果我用这个URL敬酒,结果给我一个空字符串,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
请告诉我我在做什么错了,我是JSON和整个Web服务世界的新手。
感谢您指出了这一点!愚蠢的错误,它的工作很好。改变了它。我实际上使用了一个例子,它具有我删除的那些参数,因为我知道它不会对我有用。 – 2011-12-24 18:04:52