我正在调用我的登录网络服务,如下所示
String url = "http://mydomaim.com/login.php";
UserFunctions userFunction = new UserFunctions();
JSONObject json = userFunction.loginUser(userEmail, password, url);
它工作正常并发送给我如下所示的响应
{
"userName":"a",
"login_success":1,
"user_id":"3",
"session_id":"1067749aae85b0e6c5c5e697b61cd89d",
"email":"a"
}
我解析了这个响应,并成功获得了会话ID,并将其存储在静态变量中(在MainActivity.java中),现在我必须将此会话ID作为cookie值发送到我的其他Web服务。
我这样做了
public JSONObject getJSONQuestion(String url, List<NameValuePair> params) {
// Making HTTP request
try {
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
CookieStore cookieStore = new BasicCookieStore();
Cookie cookie = new BasicClientCookie("PHPSESSID",
MainActivity.SESSION_ID);
// MainActivity.SESSION_ID has my session id
cookieStore.addCookie(cookie);
HttpContext localContext = new BasicHttpContext();
localContext.setAttribute(ClientContext.COOKIE_STORE, cookieStore);
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet,
localContext);
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, "iso-8859-1"), 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;
}
但我没有得到理想的回应。请查看我的cookie代码。我犯了哪个错误?
修改的
我明白了,
{"logout_success":1,"logout_reason":"no_activity_since_15_min"}
但我必须得到这个,
{
"random_questions":[
{
"questionid":"31",
"question":"I hate V-",
"answer1":"some answer 1",
"answer2":"another answer 2",
"answer3":"yet another answer 3",
"answer4":"a great answer 4",
"correct":"3"
},
… (more follow) …
]
}
编辑2
这是我的会话和cookie的PHP代码
<?php
$min = 15 * 60;
//$min = 60;
if (time() - $_SESSION['LAST_ACTIVITY'] > $min) {
session_destroy();
setcookie("PHPSESSID", "", time() - 3600, "/"); // delete session cookie
echo json_encode(array('logout_success' => 1, 'logout_reason' => 'no_activity_since_15_min'));
die;
} else {
$_SESSION['LAST_ACTIVITY'] = time();
mysqli_query($link, "UPDATE logged_in_users SET last_activity = " . $_SESSION['LAST_ACTIVITY'] . " WHERE session_id = '" . session_id() . "' ");
}
答案 0 :(得分:1)
使用:来自How to save session_id as cookie value and send it to server in android
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
List<Cookie> cookies = httpClient.getCookieStore().getCookies();
for (Cookie cookie : cookies) {
System.out.println("Cookie: " + cookie.toString());
if (cookie.getName().contains("PHPSESSID"))
guid = cookie.getValue();
}
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();