走过海棠暮

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走过海棠暮 2024-12-21 19:56:20

我为你做了一些谷歌搜索

从 GitHub 获取 Simple-KML 解析库 并查看它附带的示例项目,看看它如何解析出纬度和经度(我相信是 SimpleKMLPoints?),然后您将能够将注释添加到 MapKit 中。

希望我为您指明了正确的方向!

I did a bit of Googling for you.

Grab the Simple-KML parsing library from GitHub and look at the sample project that comes with it and see how it parses out latitude and longitude (which I believe are SimpleKMLPoints?) and then you will be able to make your annotations into MapKit.

Hopefully I set you in the right direction!

KML 文件谷歌地球集成到 uiMapkit

走过海棠暮 2024-12-21 17:55:44

这取决于您的 SWF 包含的内容。如果是视频,则有不同的尺寸(例如高清与标清)。但是,如果它是应用程序或时间轴动画,您可以查看设置 舞台比例模式

通常,当您创建应用程序时,您会让它们变得流畅,以便它们可以根据它们所在的位置调整大小,可以是全屏,也可以是在设定的宽度和高度内。

It depends on what your SWF contains. If it's a video, then it makes sense to have different sizes (such as HD vs SD). However, if it's an application or timeline animation, you can look at setting Stage scale mode.

Usually when you create apps you would make them fluid so that they can resize depending on where they are, either full screen or within a set width and height.

我应该为每个屏幕尺寸创建一个 Flash 文件吗?

走过海棠暮 2024-12-21 16:58:17

使用 视口 标签

<meta name="viewport" content="initial-scale=1.0, user-scalable=no" />

Use the viewport meta tags.

<meta name="viewport" content="initial-scale=1.0, user-scalable=no" />

使用 HTML & 设置移动应用程序的大小音位间隙

走过海棠暮 2024-12-21 12:46:59

我已经弄清楚了。

配置服务的 .config 文件时,我使用 [UserName] 和 [Password] 等占位符来替换用户在安装程序中给出的实际值。

服务在这些值被换出之前启动,并且服务尝试使用用户名和密码作为 [UserName] 和 [Password] 进行连接。

我一开始没有想到这种可能性,因为我以为我会收到“访问被拒绝”错误,但由于某种原因,当用户名包含 [ 或 ] 时,连接返回“RPC 服务器不可用”。

I've figured it out.

When configuring the .config file of the service I use placeholders like [UserName] and [Password] to replace actual values given by the user in the installer.

The service got started before these values were swapped out, and the service tried connecting with the username and password as [UserName] and [Password].

I didnt think of this possibility at first because I thought I would get an "Access is denied" error, but for some reason when the username contains [ or ] the connection returns "RPC server not available".

使用 ServiceController 启动的 Windows 服务抛出“RPC 服务器不可用”执行 WMI 连接时

走过海棠暮 2024-12-21 12:43:34

这里有很多事情需要讨论。

有多少玩家可以连接到一场比赛?

您使用什么类型的数据库?它的写入速度快吗?

重新启动进程是最后一个解决方案,因为如果您的游戏中一切都应该快速发生,并且您重新启动进程,则玩家重新连接到它之前将需要几秒钟的时间。

我不认为每场比赛的一个进程是可扩展的,例如,当您同时有 50.000 场比赛时会发生什么?我想说,更好的解决方案是按两个标准对子进程上的匹配进行分组:
a) 通过匹配 id(某种分片算法)
b) 如果越来越多的玩家出现,则根据玩家的数量旋转另一个进程(甚至更多)。

在对游戏进行一些测试之前很难决定要做什么。您应该真正测试它,看看它在几次真实比赛中的表现如何(它“消耗”了多少 CPU、内存),并根据数据进行一些分析。没有人能确切地说出在这种情况下该怎么做,因为这取决于项目。

There are multiple things to discuss here.

How many players can connect to a match?

What kind of database are you using? Does it have fast writes?

Restarting a process is the last solution, because if you have a game where everything should happen fast and you restart the process it will take a couple of seconds before the players reconnect to it.

I don't think one process per match is scalable, what happens when you have 50.000 matches at the same time for example? I would say that a better solution would be to group matches on a child process by 2 criteria:
a) by the match id (some kind of sharding algorithm)
b) if more and more players come popping up spin another process (or even more), based on the number of players.

It's really hard to decide what to do before making some tests on your game. You should really test it to see how it behaves several real matches (how much CPU it "eats", memory) and make some analysis based on the data. Nobody can say exactly what to do in these kinds of situations, because it depends on the project.

每场比赛的 Node.js 进程

走过海棠暮 2024-12-21 12:36:58
class Number {
    enum ValType {DoubleType, IntType} CurType;
    union {
        double DoubleVal;
        int IntVal;
    };
public:
    Number(int n) : IntVal(n), CurType(int) { }
    Number(float n) : DoubleVal(n), CurType(DoubleType) { }
    Number(double n) : DoubleVal(n), CurType(DoubleType) { }

   // Assume copy constructors and assignment operators exist

    Number& add(const Number& other) {
        switch(CurType) {
        case DoubleType: DoubleVal += other.to_double(); break;
        case IntType: IntVal+= other.to_int(); break;
        }
        return *this;
    }

    int& to_int() { 
        switch(CurType) {
        case DoubleType: IntVal = DoubleVal; CurType = IntType; break;
        //case IntType: DoubleVal = IntVal; CurType = DoubleType; break;
        }
        return IntVal; 
    }
    const int to_int() const { 
        switch(CurType) {
        case DoubleType: return (int)DoubleVal;
        case IntType: return (int)IntVal;
        }
    }
    const float to_float() const { 
        switch(CurType) {
        case DoubleType: return (float)DoubleVal;
        case IntType: return (float)IntVal;
        }
    }

    double& to_double() { 
        switch(CurType) {
        //case DoubleType: IntVal = DoubleVal; CurType = IntType; break;
        case IntType: DoubleVal = IntVal; CurType = DoubleType; break;
        }
        return DoubleVal; 
    }
    const double to_double() const { 
        switch(CurType) {
        case DoubleType: return (double)DoubleVal;
        case IntType: return (double)IntVal;
        }
    }
};

void primitive_increment(int& n) {
    ++n;
}

int main() {
    Number pi(3.1415);
    primitive_increment(pi.to_int());
    //pi now is 4
    return 0;
}

我承认这很尴尬,而且不是理想的情况,但它解决了给定的问题。

class Number {
    enum ValType {DoubleType, IntType} CurType;
    union {
        double DoubleVal;
        int IntVal;
    };
public:
    Number(int n) : IntVal(n), CurType(int) { }
    Number(float n) : DoubleVal(n), CurType(DoubleType) { }
    Number(double n) : DoubleVal(n), CurType(DoubleType) { }

   // Assume copy constructors and assignment operators exist

    Number& add(const Number& other) {
        switch(CurType) {
        case DoubleType: DoubleVal += other.to_double(); break;
        case IntType: IntVal+= other.to_int(); break;
        }
        return *this;
    }

    int& to_int() { 
        switch(CurType) {
        case DoubleType: IntVal = DoubleVal; CurType = IntType; break;
        //case IntType: DoubleVal = IntVal; CurType = DoubleType; break;
        }
        return IntVal; 
    }
    const int to_int() const { 
        switch(CurType) {
        case DoubleType: return (int)DoubleVal;
        case IntType: return (int)IntVal;
        }
    }
    const float to_float() const { 
        switch(CurType) {
        case DoubleType: return (float)DoubleVal;
        case IntType: return (float)IntVal;
        }
    }

    double& to_double() { 
        switch(CurType) {
        //case DoubleType: IntVal = DoubleVal; CurType = IntType; break;
        case IntType: DoubleVal = IntVal; CurType = DoubleType; break;
        }
        return DoubleVal; 
    }
    const double to_double() const { 
        switch(CurType) {
        case DoubleType: return (double)DoubleVal;
        case IntType: return (double)IntVal;
        }
    }
};

void primitive_increment(int& n) {
    ++n;
}

int main() {
    Number pi(3.1415);
    primitive_increment(pi.to_int());
    //pi now is 4
    return 0;
}

I will admit this is quite awkward, and not the ideal situation, but it solves the given problem.

原始数据类型的包装类

走过海棠暮 2024-12-21 12:15:50

MSDN 页面还说:

组装: WindowsBase

您引用了该程序集吗?您的项目是 WPF 项目吗? (WindowsBase 是一个 WPF 库。)

The MSDN page also says:

Assembly: WindowsBase

Did you reference this assembly? Is your project even a WPF project? (WindowsBase is a WPF library.)

DispatcherTimer 类的命名空间是什么?

走过海棠暮 2024-12-21 10:11:04

您可以使用remote show,尽管它会询问您的密钥:

git remote show origin | grep 'Fetch URL'

You can use remote show, although it's gonna ask your key:

git remote show origin | grep 'Fetch URL'

如何获取我之前克隆的git仓库?

走过海棠暮 2024-12-21 09:22:41

为什么不能在代码隐藏中保留对“ paymentExpected.Load_All()”数据源的引用并从按钮单击事件中调用 paymentExpected.Save() ?

Why can't you just keep a reference to the "paymentsExpected.Load_All()" data source in the code-behind and call paymentExpected.Save() from the button click event?

如何在没有 SQLAdapter 的情况下使用 DatagridView 进行更新

走过海棠暮 2024-12-21 07:46:53

Exchange 自己检查重复邮件的方法是这样的:对于给定的邮件,收集元组(日期标头、消息 ID 标头、收件人)。如果多次记录此元组,重复的邮件将被静默删除。也许您可以使用类似的方法。

Exchange's own way to check for duplicate mails is this: For a given message, gather the tuple (Date Header, Message-Id Header, Recipient). If this tuple is recorded multiple times, the duplicate mails are silently dropped. Maybe you can use a similar approach.

交换重复邮件

走过海棠暮 2024-12-21 06:00:56

Grid2 实际上绑定到 IEnumerable(Of Contact) 而不是 Observable Collection。这就是为什么更改没有反映在 Grid2 中。您需要使用事件或 INotifyPropertyChanged 来重新执行 Linq 查询。

Grid2 is actually binding to an IEnumerable(Of Contact) instead of an Observable Collection. That's why the change isn't reflected in Grid2. You need to cause your Linq query to reexecute using an event or INotifyPropertyChanged.

LINQ 和静态 ObservableCollection

走过海棠暮 2024-12-21 03:41:05

在版本 1.8.3(至少)中,您可以单击“持久”按钮以在重新加载后保留控制台信息。

In version 1.8.3 (at least) you can click the "Persist" button to keep console information around after reload.

页面重新加载之前无法读取 Firebug 控制台中的 JavaScript 错误

走过海棠暮 2024-12-21 00:31:31

我尝试了这个解决方案,到目前为止它有效。

UTC 2024 年 5 月 8 日,Visual Studio 2022 17.9.2

但警告:

此解决方案仍会撤消文件重命名。

我发现了这个:TFS 撤消未编辑文件的签出< /a>

本文说您只需要在签出文件上“撤消挂起的更改”。当撤消第一个“实际编辑”的文件时,会弹出一个询问窗口询问您“是否要放弃更改”。选择“不全部”,所有未编辑的文件将被撤消检出,并保留您实际编辑的所有文件。

如果您对此解决方案有疑虑或者担心由于误操作而导致更改丢失,您可以先复制备份您将撤消更改的代码。

I tried this solution and it works so far.

UTC May 8, 2024, on Visual Studio 2022 17.9.2

But WARNING:

This solution will still undo the file renaming.

I've found this : TFS undo checkout of unedited files

This article says that you just need to "Undo pending changes" on checkout files. When undo to a first "actually edited" files, will pop up a asking window to ask you "if you want to discard the changes". Select "Not to all" and all unedited files will be undo the checkout, and leaving all the files you actually edited.

If you have concerns about this solution or are worried that changes will be lost due to misoperation, you can first copy to back up the code you will undo changes.

Visual Studio TFS 在待更改列表中显示未更改的文件

走过海棠暮 2024-12-20 16:23:06

尝试 CTRL + Y & CTRL + E 滚动而不是移动该行
使用 CTRL + B 滚动到页面底部,使用 CTRL + F 滚动到页面顶部
使用H(高度)、M(中)、L(低)将光标移动到屏幕的顶部、中间和

底部还可以提高 CTRL + Y 和其他键的速度,正如这个答案所解释的https://stackoverflow.com/a/7990810/10539792

try CTRL + Y & CTRL + E to scroll instead of moving the line
use the CTRL + B to scroll to bottom of page and CTRL + F to scroll to top of page
use the H (Height), M (middle), L (low) to move the cursor to top, middle and bottom of screen

you can also increase the speed of CTRL + Y and other keys as of this answer explained https://stackoverflow.com/a/7990810/10539792

Vim:提高文档遍历速度

走过海棠暮 2024-12-20 15:36:47
using (var rdr = new StreamReader(fs, Encoding.GetEncoding(1252))) {
     result = rdr.ReadToEnd();
}
using (var rdr = new StreamReader(fs, Encoding.GetEncoding(1252))) {
     result = rdr.ReadToEnd();
}

c# 编码问题?

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