Android:登录应用程序后无法维护会话

我试图在

WebView

中打开一个URL，但我无法打开，我想这是因为会话没有维护。 我在活动中将用户名、密码和用户id发送到服务器。这是密码

public class ServiceActivity extends Activity {
private Button button_back;

private Button button_submit_user_pass;
private EditText edit_id_code;
private String contents;
private String format;
private String username;
private String password;
private String id;
public static HttpClient client;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.qr_code_view);

    edit_id_code = (EditText) findViewById(R.id.editText_id);

    button_back = (Button) findViewById(R.id.button_back);
    button_submit_user_pass = (Button) findViewById(R.id.button_submit_user_pass);

    Intent user_pass = getIntent();

    username = user_pass.getStringExtra("user");
    password = user_pass.getStringExtra("pass");


    button_submit_user_pass
            .setOnClickListener(user_pass_qr_submit_listener);
    button_back.setOnClickListener(back_listener);

}

private View.OnClickListener user_pass_qr_submit_listener = new View.OnClickListener() {

    @Override
    public void onClick(View v) {
        // TODO Auto-generated method stub
        try {
            id = edit_id_code.getText().toString();

            client = new DefaultHttpClient();
            HttpPost post1 = new HttpPost(
                    "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx");
            List<NameValuePair> nvp = new ArrayList<NameValuePair>();
            nvp.add(new BasicNameValuePair("uname", username));
            nvp.add(new BasicNameValuePair("password", password));
            nvp.add(new BasicNameValuePair("id", id));
            post1.setEntity(new UrlEncodedFormEntity(nvp));
            HttpResponse resp = client.execute(post1);
            String responseText = inputStreamTOString(
                    resp.getEntity().getContent()).toString();
            Log.i("response", responseText);
            int num = Integer.parseInt(responseText);

            if (num == 0) {
                Toast.makeText(getApplicationContext(),
                        "Response" + responseText, 0).show();
            } else if (num == 1) {
                Intent survey = new Intent(ServiceActivity.this,
                        WebViewActivity.class);
                startActivity(survey);
            }
        } catch (Exception e) {
            Log.e("error", "ERROR" + e);
        }

    }

};
private View.OnClickListener back_listener = new View.OnClickListener() {

    @Override
    public void onClick(View v) {
        // TODO Auto-generated method stub
    }
};

private StringBuilder inputStreamTOString(InputStream is) {
    String line = "";
    StringBuilder total = new StringBuilder();

    // read response until the end
    try {
        BufferedReader rd = new BufferedReader(new InputStreamReader(is,
                "iso-8859-1"), 8);
        while ((line = rd.readLine()) != null) {
            total.append(line);

        }
    } catch (Exception e) {
        // TODO: handle exception
    }
    return total;

}
}

---

但是我无法加载用户对应的URL，我在webviewactivity上收到php错误，这是因为我无法维护登录用户的会话。请给我一些解决办法

尝试使用SharedReferences为应用程序保留会话

为了保存价值

SharedPreferences prefs;
prefs = PreferenceManager.getDefaultSharedPreferences(this);
Editor editor = prefs.edit();
editor.putString("key", value);
editor.commit();

在其他活动中重审

SharedPreferences prefs;
prefs = PreferenceManager.getDefaultSharedPreferences(this);
String variable = prefs.getString("key","default value");

---

您需要做的是保存PHP使用的cookie，以便在您发出post请求返回会话时跟踪会话，并将其用于以下请求。根据您如何执行post请求，有几种方法

---

这个答案描述了如何处理cookie：

如何处理？请详细说明。这不是一个答案。这应该是一个评论。编辑的答案。希望这会有帮助。希望我能给你一个10分。

SharedPreferences prefs;
prefs = PreferenceManager.getDefaultSharedPreferences(this);
String variable = prefs.getString("key","default value");