我有一个android的登录应用程序,它使用JSON将数据从数据库解析到应用程序。接受HTTP请求通过标签识别的PHP API,要么是“登录”或“注册”,像这样:解析JSONobject时出错
if (isset($_POST['tag']) && $_POST['tag'] != '') {
"Do stuf
} else {
echo "access denied";
的应用程序已经工作正常,但现在我只是得到响应“访问拒绝”。
public JSONObject loginUser(String email, String password){
// Building Parameters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("tag", "login"));
params.add(new BasicNameValuePair("email", email));
params.add(new BasicNameValuePair("password", password));
JSONObject json = jsonParser.getJSONFromUrl(loginURL, params);
// return json
// Log.e("JSON", json.toString());
return json;
}
这是发送请求的JSON对象,并且我怀疑它没有正确发送标记。有没有人知道发生了什么?
更新:新增JSONparser.class
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) {
// Making HTTP request
try {
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "UTF-8"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "n");
}
is.close();
json = sb.toString();
Log.e("JSON", json);
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
为什么你不只是调试自己陷入更多的一些信息仅通过在你的PHP脚本中看到$ _POST数组的实际内容 – 2013-02-28 14:00:23
我回应了$ _POST [“tag”],它返回Null,所以由于某种原因JSONparser没有发送数据。 – 2013-02-28 14:09:11
你可以用strlen($ _ POST ['tag'])> 0替换所有的东西(isset($ _ POST ['tag'])&& $ _POST ['tag']!='') jsonParser.getJSONFromURL使用$ _GET而不是$ _POST – Dave 2013-02-28 14:32:53