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所以我建立一個URL來調用來獲得JSON響應,但面臨一個奇怪的問題。如下所示構建URL返回「未找到」,但出於測試目的,我只是將URL構建爲「http://api.themoviedb.org/3/search/person?api_key=XXX & query = brad」,並且沒有「 t附加任何東西,並返回正確的響應。也嘗試不編碼「文本」和相同的東西...沒有找到。有任何想法嗎?Android Java奇怪的問題建設URL
StringBuilder url = new StringBuilder();
url.append("http://api.themoviedb.org/3/search/person?api_key=XXX&query=").append(URLEncoder.encode(text, ENCODING));
Log.v("URL", url.toString());
try {
HttpGet httpRequest = null;
httpRequest = new HttpGet(url.toString());
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response = (HttpResponse) httpclient.execute(httpRequest);
HttpEntity entity = response.getEntity();
BufferedHttpEntity bufHttpEntity = new BufferedHttpEntity(entity);
InputStream input = bufHttpEntity.getContent();
String result = toString(input);
//JSONObject json = new JSONObject(result);
return result;
行Log.v(「URL」,url.toString());'打印出來? – 2012-02-10 17:55:40
它打印出正確的地址,我甚至複製/粘貼到瀏覽器,它的工作... – Paul 2012-02-10 18:00:27
遠射,但它可能需要編碼您的API密鑰? – Peter 2012-02-10 18:10:53