小编Jon*_*han的帖子

我发帖是因为我已经完成了其他问题的解决方案,但他们没有帮助.

我想要做的是通过Google Apps for Business Gmail使用Swiftmailer发送电子邮件,但我一直收到此错误:

无法通过主机smtp.gmail.com建立连接[连接超时#110]

我知道代码很好,因为它可以在我的本地机器上运行,但不能在生产服务器上运行.

到目前为止我尝试了什么:

• 启用OpenSSL.
• 来自Google验证码的未屏蔽帐户.
• 使用应用程序专用密码.
• 列入白名单的Gmail SMTP IP地址

组态:

# Swiftmailer Configuration
swiftmailer:
    transport:  smtp
    encryption: ssl
    auth_mode:  login
    host:       smtp.gmail.com
    username:   contact@mydomain.com
    password:   applicationspecificpassword
    port:       465

我还能尝试什么？这可能是一个DNS问题,因为我使用的是Gmails SMTP MX记录而不是服务器.

我是JSF的新手,在学习构建在线书店应用程序的过程中.

我有1个班级和1个豆子:Book.java和BookCatelogBean.java.Book类有3个属性:id,title和author它相应的getter和setter.该BookCatelogBean包含ArrayList<Book>在那里我用填充它Books(在将来,我将它连接到数据库).

我有两页:index.xhtml和book.xhtml.我想显示index.xhtml每个格式为REST链接的书名列表及其ID book.xhtml,如下所示:<h:link outcome="book?id=#{bookCatelogBean.id}" value="#{bookCatelogBean.title}" />

我知道如何使用BookCatelogBean显示1,book但我想显示所有这些？我有一个想法,调用一个方法来BookCatelogBean调用getAllBooks(),返回每个书籍标题,但我如何将它们中的每一个返回到index.xhtml作为JavaserverFace链接而不是字符串？

谢谢

这是我的代码:

Book.java

package bookshop;

import java.io.Serializable;

public class Book implements Serializable {

    private int id;
    private String title;
    private String author;

    public Book(int id, String title, String author){
        this.title = title;
        this.id = id;
        this.author = author;
    }

    public …

我正在尝试运行此命令:

git filter-branch --force --index-filter 'git rm --cached --ignore-unmatch filename.js' --prune-empty --tag-name-filter cat -- --all

但我一直收到这个错误:

fatal: ambiguous argument 'rm': unknown revision or path not in the working tree
.
Use '--' to separate paths from revisions, like this:
'git <command> [<revision>...] -- [<file>...]'

我正在尝试从匹配表中计算联赛积分榜.

+----------------------------------+
|              Matches             |
+----------------------------------+
| id                               |
| league_id (FK League)            |   
| season_id (FK Season)            |
| home_team_id (FK Team)           |
| away_team_id (FK Team)           | 
| home_score                       |
| away_score                       |
| confirmed                        |
+----------------------------------+

我可以使用此查询正确计算Home League Standings:

SELECT team.name, home_team_id AS team_id,
    COUNT(*) AS played,
    SUM((CASE WHEN home_score > away_score THEN 1 ELSE 0 END)) AS won,
    SUM((CASE WHEN away_score > home_score THEN 1 ELSE 0 END)) AS lost,
    SUM((CASE WHEN home_score = away_score THEN …

var query = Session
   .find({ player: player, logout: null })
   .sort({ 'login.date': -1 })
   .limit(1);

query.exec(function(err, data) {
  var session = new Session(data);
  session.logout = logoutEvent;
  session.save();
});

如何将其转换为findOneAndUpdate？

以下是使用findOneAndUpdate的示例,但我不确定如何添加sort和limit:

var query = {'username':req.user.username};
req.newData.username = req.user.username;
MyModel.findOneAndUpdate(query, req.newData, {upsert:true}, function(err, doc){
    if (err) return res.send(500, { error: err });
    return res.send("succesfully saved");
});

为什么JavaScript将parseInt(0000000101126)转换为33366而不是101126？

var example = parseInt(0000000101126);
console.log(example); //33366 

目前我正在使用一个名为的表来实现下面的结果league_standing,并在每次比赛后更新它.我希望能够对表进行一次查询matches.

Teams在主场和客场两次互相比赛.请注意team_id两列home_team_id和away_team_id

+----------------------------------+
|              Matches             |
+----------------------------------+
| id                               |
| league_id (FK League)            |   
| season_id (FK Season)            |
| home_team_id (FK Team)           |
| away_team_id (FK Team)           | 
| home_score                       |
| away_score                       |
| confirmed                        |
+----------------------------------+

这是我尝试但失败的原因:

select team.name, HomePoints + AwayPoints points
from team join (
    select team.id, 
        sum(case when home.home_score > home.away_score then 3
            when home.home_score = home.away_score then 1 else 0 end) HomePoints, …

<?php

// Page class

class Page {

// Declare a class member variable
var $page;
var $title;
var $year;
var $copyright;

// The Constructor function
function Page($title, $year, $copyright){
// Assign values to member variables
$this->page = '';
$this->title = $title;
$this->year = $year;
$this->copyright = $copyright;

// Call the addHeader() method
$this->addHeader();
}

// Generates the top of the page
function addHeader(){
$this->page .= <<<EOD
                <html>
                <head>
                <title>$this->title</title>
                </head>
                <body>
                <h1 align="center">$this->title</h1>
                </body>
                EOD;
    }
}




?>

我有一个名为table_date当前我now()用来插入当前日期(2011-02-23)的列.我知道我可以用sql/php操纵这个来显示年份和月份名称.但是,我想知道是否可以table_date像这样插入当前日期作为年月2011-02？谢谢

我对设计模式比较陌生,在下面的例子中,我使用的是我认为的策略模式.但是,我在一些,而不是所有的具体策略中重复自己,并想知道有没有办法避免这种情况？注意ACommand和CCommand在做一些独特的事情之前是如何拥有相同的代码的.

public interface Command 
{
    public boolean execute(CommandSender sender, String[] args);
    public String getName();
    //...
}

public abstract class PlayerCommand implements Command 
{
    protected BukkitPlugin plugin = BukkitPlugin.getInstance();

    private String name;
    //...

    public PlayerCommand(String name) 
    {
        this.name = name;
    }

    public String getName() 
    {
        return this.name;
    }

    //...
}

ACommand

    public class ACommand extends PlayerCommand
    {
        public ACommand()
        {
            super("A");
        }

        public boolean execute(CommandSender sender, String[] args)
        {
            Player player = (Player) sender;
            PlayerInventory inventory = …