小编Mat*_*bul的帖子

我希望菜单中的标题会因点击的片段名称而改变.我得到的代码是,每个片段中的实际标题是"Home",它不会改变.但我发现,当我点击菜单中的一个项目时,标题会改变一秒并返回"主页"标题.我实施了拉杆机,但我仍然不知道是什么原因造成的.

我的代码:

 package com.example.matant.gpsportclient;

import android.app.Fragment;
import android.app.FragmentManager;
import android.app.ProgressDialog;
import android.content.res.Configuration;
import android.os.Bundle;
import android.support.v4.app.ActionBarDrawerToggle;
import android.support.v4.widget.DrawerLayout;
import android.support.v7.app.AppCompatActivity;
import android.util.Log;
import android.view.MenuItem;
import android.view.View;
import android.widget.AdapterView;
import android.widget.ListView;
import android.widget.Toast;

import com.example.matant.gpsportclient.Controllers.Fragments.CreateEventFragmentController;
import com.example.matant.gpsportclient.Controllers.DBcontroller;
import com.example.matant.gpsportclient.Controllers.Fragments.GoogleMapFragmentController;
import com.example.matant.gpsportclient.Controllers.Fragments.ManageEventFragmentController;
import com.example.matant.gpsportclient.Controllers.Fragments.ProfileFragmentController;
import com.example.matant.gpsportclient.InterfacesAndConstants.AsyncResponse;
import com.example.matant.gpsportclient.InterfacesAndConstants.Constants;
import com.example.matant.gpsportclient.Utilities.DrawerItem;
import com.example.matant.gpsportclient.Utilities.DrawerItemCustomAdapter;
import com.example.matant.gpsportclient.Utilities.SessionManager;

import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONException;
import org.json.JSONObject;

import java.util.ArrayList;
import java.util.List;


public class MainScreen extends AppCompatActivity implements AsyncResponse {

    private String[] mNavigationDrawerItemTitles;
    private DrawerLayout mDrawerLayout;
    private ListView …

我是Google Api位置的新手.我实现了一个处理所有位置回调的类.现在它在位置服务开启时工作.但如果关闭我创建了一个函数LocationAlertDialog(),它应该处理此错误.我不明白的问题我应该在我的代码中给她打电话,因为我不知道我在哪里检查GPS或网络提供商是否可用.

这是我的代码:

public class LocationTool implements GoogleApiClient.ConnectionCallbacks, GoogleApiClient.OnConnectionFailedListener{
    private Fragment frag;
    private GoogleApiClient mGoogleApiClient;
    private LocationRequest mLocationRequest;
    private Location mLastLocation;
    private boolean mRequestingLocationUpdates;
    public OnLocationChangedListener delegate= null;

    public LocationTool (Fragment frag, OnLocationChangedListener onLocationChangedListener)
    {
        mRequestingLocationUpdates = true;
        delegate = onLocationChangedListener;
        this.frag = frag;
        buildGoogleApiClientAndCreateLocationRequest();
    }

    public boolean ismRequestingLocationUpdates() {
        return mRequestingLocationUpdates;
    }

    public LocationRequest getmLocationRequest() {
        return mLocationRequest;
    }

    public GoogleApiClient getmGoogleApiClient() {
        return mGoogleApiClient;
    }

    public Location getmLastLocation() {
        return mLastLocation;
    }

    public void setmLastLocation(Location mLastLocation) {
        this.mLastLocation …

我正在尝试使用基于imap协议的java mail api连接到我的邮箱.我检查并确定我插入了正确的参数.这是我得到的以下异常:

[警告]请通过您的网络浏览器登录:https://support.google.com/mail/accounts/answer/78754(失败)

我不知道为什么会发生这种情况我在我的Gmail帐户设置中启用了imap选项.

我是我的代码:

 Properties protocol = new Properties();
 protocol.setProperty("mail.store.protocol", "imaps");

 try{
      Session session = Session.getInstance(protocol, null);
      Store store = session.getStore();
      String host = prop.getProperty("host");
      String email = prop.getProperty("username");
      String password = prop.getProperty("password");
      store.connect(host, email, password);
      Folder inbox = store.getFolder("INBOX");
      inbox.open(Folder.READ_ONLY);
      int messageCount = inbox.getMessageCount();
      model.addAttribute("msg","number of mails"+" "+messageCount);
      Message[] messages = inbox.getMessages();
      PrintWriter writer = new PrintWriter(username+".txt", "UTF-8");

      for(int i=0;i<messageCount || prop.getProperty("status").equals(status.RUNNING.toString()) ;i++ ){
          model.addAttribute("msg","Reading Mails");
          Multipart mp = (Multipart) messages[i].getContent();
          BodyPart bp …

我是 PowerShell 的初学者，我的主要困难是我有以下脚本，我想从 URL 的输出创建一个日志文件。现在脚本正在运行，但我不知道如何将浏览器输出保存到文件中。只有在未创建文件的情况下，我才需要将输出写入新文件。我尝试了 |Out-File 方法，但没有任何反应。

我的脚本：

start "http://myurl"
Start-Sleep -s 5
get-process iexplore | stop-process
get-process powershell | stop-process

这是我的 php 代码：

    <?php
/**
 * Created by PhpStorm.
 * User: matant
 * Date: 11/17/2015
 * Time: 1:46 PM
 */
define ('mysql_host','host');
define ('mysql_user','user');
define ('mysql_password','password');
define ('myDB','DBname');


$dblink= mysqli_connect(mysql_host, mysql_user, mysql_password,myDB);


if (!$dblink)
{
    $message = sprintf(
        "Could not connect to local database: %s",
        mysql_error()
    );
    trigger_error($message);
    echo $message;
    return;
}
else{
    echo "connection success"."<br/>";
}
$c_date = …