PHP session 在 Android 中不起作用？

Question

我的 Android 应用程序和 php 文件有问题。

我有两个 php 文件:* 第一个是在长 if/else block 之后连接:

[...]
else{

                $response["message"]="Connexion OK";

                session_start();
                $_SESSION['id'] = $id;
                $_SESSION['pseudo'] = $pseudo;
                $_SESSION['options'] = $options;
                $response['session'] = $_SESSION['id'];
            }
[...]
echo json_encode($response);

我的第二个文件只在连接与否时返回:

<?php

session_start();
$response =0;

    if(isset($_SESSION['id'])){
        $response= 1;
    }



echo $response;

?>

当我在 Firefox 中执行文件的 URL 时，顺序是:1/第二个，我有 02/第一个3/又是第二个，我得到 1

Php 文件构建良好!

现在，我想在 AsyncTasks Android 中执行此操作:无论我在哪个 Activity 中，我首先要测试用户与第二个文件的连接:但返回始终为 0...未连接，甚至在我使用第一个在 Connexion Activity 中调用的记录后。

Android 中还有什么可以打开 Php session 的吗？

感谢您的回答，

---

这是我的代码:我有一个用于连接的 AsyncTask:

private class AddDataAsyncTask extends AsyncTask<Void, Void, Void> {
        String message = "";
        [...] //preExecute
        @Override
        protected Void doInBackground(Void... params) {


            Log.i("add", " start doInBackground");

            // Making a request to url and getting response

            String urlt = url;
            if (nameValuePair != null) {
                String paramString = URLEncodedUtils.format(nameValuePair, "utf-8");

            }
            String response = "";
            try {

                DefaultHttpClient httpClient = new DefaultHttpClient();

                CookieStore cookieStore = new BasicCookieStore();
                                    HttpContext localContext = new BasicHttpContext();
                                    localContext.setAttribute(ClientContext.COOKIE_STORE, cookieStore);

                HttpGet post = new HttpGet(urlt);
                HttpResponse rp = httpClient.execute(post);
                if (rp.getStatusLine().getStatusCode() != HttpStatus.SC_OK) {
                    // Server is unavailable
                }
                response = EntityUtils.toString(rp.getEntity()); //here is your response

            } catch (ClientProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
           /* } catch (UnsupportedEncodingException e) {
                e.printStackTrace();
            */
            if (response != null) {
                try {

                    JSONObject JsonResponse = new JSONObject(response);
                    message = JsonResponse.getString("message");

                    if (message.equals("Connexion OK"))
                        session = Integer.parseInt(JsonResponse.getString("session"));
                        Log.i("connnnnnexionC", response);
                } catch (JSONException e) {

                    e.printStackTrace();
                }
            }

            return null;

[...]//后执行

还有我要检查连接的 AsyncTask:

        private class VerifierConnexionAsyncTask extends AsyncTask<Void, Void, Void> {
                [...] //preExecute

                @Override
                protected Void doInBackground(Void... params) {

                    String response = "";
                    try {

                        DefaultHttpClient httpClient = new DefaultHttpClient();
                        HttpEntity httpEntity = null;
                        HttpResponse httpResponse = null;


                        CookieStore cookieStore = new BasicCookieStore();
                        HttpContext localContext = new BasicHttpContext();
                        localContext.setAttribute(ClientContext.COOKIE_STORE, cookieStore);

                        HttpPost httpPost = new HttpPost(url);
                        httpResponse = httpClient.execute(httpPost);
                        httpEntity = httpResponse.getEntity();
                        response = EntityUtils.toString(httpEntity);
                        Log.i("connnnnexion",response);
                    } catch (ClientProtocolException e) {
                        e.printStackTrace();
                    } catch (IOException e) {
                        e.printStackTrace();
                    }

                    return null;
                }

    [...]//postExecute
            }

首先我输入检查结果是:

我/connnnnexion:0

我转到 Connexion Activity 并使用第一个 AsyncTask 登录:我得到:

I/connnnnnexionC:{"message":"Connexion OK","session":"27"}

所以连接的很好

我重新输入支票，我又一次:

我/connnnnexion:0

---

解决方案我写了一个新的 Class Connecte:

    import org.apache.http.client.CookieStore;
import org.apache.http.client.protocol.ClientContext;
import org.apache.http.impl.client.BasicCookieStore;
import org.apache.http.protocol.BasicHttpContext;
import org.apache.http.protocol.HttpContext;
public class Connecte {
    static CookieStore cookieStore;
    static HttpContext localContext;

    public static void makeCookie(){
        cookieStore = new BasicCookieStore();
        localContext = new BasicHttpContext();
        localContext.setAttribute(ClientContext.COOKIE_STORE, cookieStore);
    }


}

在连接 AsyncTask 中我添加:Connexion.makeCookie():

HttpGet post = new HttpGet(urlt);
                Connecte.makeCookie();
                HttpResponse rp = httpClient.execute(post, Connecte.localContext);

在检查异步任务中我刚刚添加:

HttpPost httpPost = new HttpPost(url);
                httpResponse = httpClient.execute(httpPost, Connecte.localContext);

Answer 1

您必须确保您正在使用的连接正在重复使用相同的 cookie。 PHP 通过给每个客户端一个 PHP_SESSION cookie 来管理 session ，并根据您的 PHP_SESSION 跟踪各种值。如果您不在每个请求之间跟踪它，那么 php 会认为您每次都是不同的客户端。

CookieHandler.setDefault( new CookieManager( null, CookiePolicy.ACCEPT_ALL ) );

---

DefaultHttpClient httpClient = new DefaultHttpClient();
HttpEntity httpEntity = null;
HttpResponse httpResponse = null;

CookieStore cookieStore = new BasicCookieStore();
HttpContext localContext = new BasicHttpContext();
localContext.setAttribute(ClientContext.COOKIE_STORE, cookieStore);

HttpPost httpPost = new HttpPost(url);

httpResponse = httpClient.execute(httpPost, localContext);
httpEntity = httpResponse.getEntity();

确保在所有 future 请求中重用相同的 localContext 对象，以便它们都共享相同的 cookie 并可以保持 PHP session 。