半城柳色半声笛

文章 评论 浏览 26

半城柳色半声笛 2024-12-08 10:13:42

我认为您使用的方案没有问题。
请注意,它不一定是临时表,您也可以使用某种内存表。

通过照常创建一个表来执行此操作,然后执行

ALTER TABLE <table_name> CACHE;  

此操作将优先将该表存储在内存中。

只要您在短时间内填充清空表格,您就不需要执行第 1 步和第 2 步。 2.
请记住 cache 修饰符只是一个提示。该表仍然在缓存中老化,最终将被推出内存。

只需执行以下操作:

  1. 通过存储过程使用数据填充缓存表

  2. 在另一个存储过程中使用数据,但不要等待太久。
    2a.清除缓存表中的数据。

I see no problem in the scheme your are using.
Note that it doesn't have to be a temp-table, you can use a sort of kind of memory table as well.

Do this by creating a table as usual, then do

ALTER TABLE <table_name> CACHE;  

This will prioritize the table for storage in memory.

As long as you fill and empty the table in short order you don't need to do step 1 & 2.
Remember the cache modifier is just a hint. The table still ages in the cache and will be pushed out of memory eventually.

Just do:

  1. Populate cache-table with data through a stored procedure

  2. Use the data in another stored procedure, but don't wait to long.
    2a. Clear the data in the cache table.

Oracle 中 MySQL 临时表的替代方案

半城柳色半声笛 2024-12-08 08:53:21

Servlet 容器示例包括 TomcatJetty

Solr 与任何 Java Web 应用程序一样,需要托管和运行 servlet 容器。 Solr 安装文档 包括每个容器的一般信息和特定信息。

当您运行 java -jar start.jar (start.jar 随 Solr 发行版一起提供)时,您实际上是在 嵌入式 Jetty实例

Servlet container examples include Tomcat and Jetty.

Solr, just like any Java web app, needs a servlet container to be hosted and run. The Solr installation docs include general and specific information for each container.

When you run java -jar start.jar (start.jar comes with the Solr distribution) you're actually running Solr in an embedded Jetty instance.

JVM、JAVA 二进制文件和 Tomcat..哪一个提供 servlet 容器?

半城柳色半声笛 2024-12-08 08:45:02

您不能严格/硬限制它找到的位置,尽管系统中有一个功能请求这样做,但您可以对结果设置“偏差”。它作为谷歌地图边界对象的参数传递给自动完成方法。然后,自动完成将优先考虑这些边界内的位置。但请注意,由于这不是硬边界,因此如果边界之外的搜索存在匹配项,它将返回这些匹配项。

从我的使用情况来看,它似乎有点问题,可以进行一些改进 - 特别是考虑到边界之外的任何东西都不会被接近度标记,因此边界之外一个街区的东西与 1000 英里之外的东西一样有可能显示,所以确保你努力让边界发挥作用。

You can't strictly/hard limit the locations that it finds, although there is a feature request in the system to do so, but you can set a 'bias' on the results. It's passed in as an argument to the autocomplete method as a google maps bounds object. Autocomplete will then favor locations within those boundaries. Note, however, that since this isn't a hard boundary, if there are matches for the search outside the boundaries it will return those.

From my usage it seems a bit buggy and can use some improvement - especially considering that anything outside your boundary is not tagged by proximity at all, so something one block outside the boundary is just as likely to show as something 1000 miles outside, so make sure you play around with getting the boundaries working right.

将谷歌地图自动填充限制为仅英国地址

半城柳色半声笛 2024-12-07 21:19:40

你还没有说你是否想使用winforms,wpf或其他东西。不管怎样,下面的代码适用于 winforms - 只需在表单中添加一个文本框:

    private void button1_Click(object sender, EventArgs e)
    {
        // Create reader & open file
        using (TextReader tr = new StreamReader(@"C:\myfile.txt"))
        {            
            textBox1.Text += tr.ReadToEnd();                
        }

    }

You haven't said if you wanted to use winforms, wpf or something else. Anyway, the code below will work for winforms - just add a textbox to your form:

    private void button1_Click(object sender, EventArgs e)
    {
        // Create reader & open file
        using (TextReader tr = new StreamReader(@"C:\myfile.txt"))
        {            
            textBox1.Text += tr.ReadToEnd();                
        }

    }

如何在C#中单击按钮时显示txt文件

半城柳色半声笛 2024-12-07 20:13:31

您可以将circle.js放入app/assets/javascripts中。它不需要被称为circle.js.erb,但它可以是。

You can put circle.js into app/assets/javascripts. It doesn't need to be called circle.js.erb, but it could be.

初学者问题:如何将processing.js 与Rails 3.1 应用程序集成?

半城柳色半声笛 2024-12-07 18:45:18

其实很简单。使用分配器的容器的构造函数获取分配器并存储它的副本。为了做到这一点,它需要分配器是CopyConstructible。就这样。请注意,除非分配器类型的 propagate_on_container_copy_assignment 特征为 true(这种情况很少见),否则分配器类型不需要 CopyAssignable

C++11 规范还规定“这些类型上的构造函数、比较运算符、复制操作、移动操作或交换操作不得通过异常退出”。异常规则允许您创建分配器的(堆栈)副本(尤其是在构造或销毁期间),而不必担心复制分配器会抛出异常。在存在可能引发复制、移动、交换或比较的分配器的情况下设计异常安全的容器几乎是不可能的。在实践中,分配器只能容纳指向某些资源的指针,因此允许分配器进行复制等操作会增加很多痛苦,而几乎没有任何好处。

It's actually pretty simple. The constructor of a container using an allocator takes the allocator and stores a copy of it. In order to do that, it needs the allocator to be CopyConstructible. That's all. Note that an allocator type not required to CopyAssignable unless its propagate_on_container_copy_assignment trait is true (which is rare).

The C++11 specification also states that "No constructor, comparison operator, copy operation, move operation, or swap operation on these types shall exit via an exception." The exception rules allow you to make a (stack) copy of an allocator (especially during construction or destruction) without worrying that copying the allocator will throw. Designing containers that are exception-safe in the presence of allocators that might throw on copy, move, swap, or compare is nearly impossible. In practice, an allocator cannot hold much more than a pointer to some resource, so allowing allocators to throw on copy, etc., would add a lot of pain for virtually no gain.

为什么 c++ 中的分配器需要复制构造函数吗?

半城柳色半声笛 2024-12-07 10:44:32

试试这个。

BOOL contentIsInLargerArray = NO;
for (id *object in symbols) {
    if ([fetchedSymbolsArray containsObject:object])
        contentIsInLargerArray = YES;
}
return contentIsInLargerArray;

这使用 -(BOOL)containsObject:(id)anObject 而不是 -[indexObObject:]。

try this.

BOOL contentIsInLargerArray = NO;
for (id *object in symbols) {
    if ([fetchedSymbolsArray containsObject:object])
        contentIsInLargerArray = YES;
}
return contentIsInLargerArray;

This uses -(BOOL)containsObject:(id)anObject instead of -[indexObObject:].

如何检查较小数组中的值是否在较大数组中?

半城柳色半声笛 2024-12-07 06:41:55

就我而言,分配给实例和负载均衡器的安全组规则不允许流量在两者之间传递。这导致健康检查失败。

In my case, the rules on security groups assigned to the instance and the load balancer were not allowing traffic to pass between the two. This caused the health check to fail.

为什么 Elastic Load Balancing 报告“停止服务”?

半城柳色半声笛 2024-12-07 03:17:13

这对我有用。您确定 imagesTurnedOn/imagesTurnedOff 返回正确的值吗?

这个解决方案在时间安排上还有很多不足之处——它会非常不平衡。也许这样的事情会更好(使用 AsyncTask),

public void deplayedPlay2() {
    if (mTaskHandler == null) {
        mTaskHandler = new AsyncTask<Void, Void, Void>() {
            @Override
            public Void doInBackground(Void... params) {
                long now = System.currentTimeMillis();
                try {
                    for (final int btnid : mScenario) {
                        Log.d(TAG,
                                "ON: " + btnid + " (" + (System.currentTimeMillis() - now) + ")");
                        mButtons[btnid].post(new Runnable() {
                            public void run() {
                                mButtons[btnid]
                                        .setBackgroundDrawable(GoodbyeAndroidActivity.this
                                                .getResources()
                                                .getDrawable(
                                                        R.drawable.on_icon));

                            }
                        });
                        Thread.sleep(1000);
                        Log.d(TAG,
                                "OFF: " + btnid + " (" + (System.currentTimeMillis() - now) + ")");
                        mButtons[btnid].post(new Runnable() {
                            public void run() {
                                mButtons[btnid]
                                        .setBackgroundDrawable(GoodbyeAndroidActivity.this
                                                .getResources()
                                                .getDrawable(
                                                        R.drawable.off_icon));
                            }
                        });
                    }
                } catch (InterruptedException ex) {
                    Log.d(TAG, "Interrupted.");
                }
                return null;
            }

            @Override
            public void onPostExecute(Void param) {
                Log.d(TAG, "Done!");
                mTaskHandler = null;
            }
        };
        mTaskHandler.execute();
    }
}

不要忘记在 onPause() 中处理这个问题:

public void onPause() {
    super.onPause();
    if (mTaskHandler != null) {
        mTaskHandler.cancel(true);
        // May want to reset buttons too?
    }
}

It worked for me. Are you sure that imagesTurnedOn/imagesTurnedOff are returning the correct values?

This solution leaves a lot to be desired in terms of timing -- it will be quite uneven. Perhaps something like this would work better (using an AsyncTask)

public void deplayedPlay2() {
    if (mTaskHandler == null) {
        mTaskHandler = new AsyncTask<Void, Void, Void>() {
            @Override
            public Void doInBackground(Void... params) {
                long now = System.currentTimeMillis();
                try {
                    for (final int btnid : mScenario) {
                        Log.d(TAG,
                                "ON: " + btnid + " (" + (System.currentTimeMillis() - now) + ")");
                        mButtons[btnid].post(new Runnable() {
                            public void run() {
                                mButtons[btnid]
                                        .setBackgroundDrawable(GoodbyeAndroidActivity.this
                                                .getResources()
                                                .getDrawable(
                                                        R.drawable.on_icon));

                            }
                        });
                        Thread.sleep(1000);
                        Log.d(TAG,
                                "OFF: " + btnid + " (" + (System.currentTimeMillis() - now) + ")");
                        mButtons[btnid].post(new Runnable() {
                            public void run() {
                                mButtons[btnid]
                                        .setBackgroundDrawable(GoodbyeAndroidActivity.this
                                                .getResources()
                                                .getDrawable(
                                                        R.drawable.off_icon));
                            }
                        });
                    }
                } catch (InterruptedException ex) {
                    Log.d(TAG, "Interrupted.");
                }
                return null;
            }

            @Override
            public void onPostExecute(Void param) {
                Log.d(TAG, "Done!");
                mTaskHandler = null;
            }
        };
        mTaskHandler.execute();
    }
}

Don't forget to handle this in onPause():

public void onPause() {
    super.onPause();
    if (mTaskHandler != null) {
        mTaskHandler.cancel(true);
        // May want to reset buttons too?
    }
}

如何在 ImageButtons 上一张一张地更改图像?

半城柳色半声笛 2024-12-06 21:25:14

您可以在“总计”栏中重复计算。

SELECT 
    SUM(b.bet_win * cy.fx_rate) as dr, 
    SUM(b.bet_loss * cy.fx_rate) as cr, 
    SUM(b.bet_win * cy.fx_rate) + SUM(b.bet_loss * cy.fx_rate) as total
FROM ....
WHERE ....

You can repeat the calculations in the "total" column.

SELECT 
    SUM(b.bet_win * cy.fx_rate) as dr, 
    SUM(b.bet_loss * cy.fx_rate) as cr, 
    SUM(b.bet_win * cy.fx_rate) + SUM(b.bet_loss * cy.fx_rate) as total
FROM ....
WHERE ....

在 SELECT 中重用别名

半城柳色半声笛 2024-12-06 16:50:30

感谢大家,最后我使用 CAST(expr AS type) 方法找到了解决方案。

CAST() 函数采用任何类型的表达式,并生成指定类型

SELECT 名称、CAST(date AS CHAR(10)) FROM table WHERE 条件的结果值。

感谢扎克·科尔马诺和亨内尔·科尔马诺。

thanks to all of you, finaly i have made the solution using CAST(expr AS type) method.

The CAST() function takes an expression of any type and produces a result value of a specified type

SELECT name, CAST(date AS CHAR(10)) FROM table WHERE condition.

thanx to Zack and Hennele Kormano.

[错误]:返回的对象不是 JSON 可序列化的

半城柳色半声笛 2024-12-06 16:15:14

在编辑器的上下文菜单中选择“显示在 - 包资源管理器”。

Select 'Show In - Package Explorer' in the context menu of your editor.

如何在 Eclipse 中复制和编辑文件?

半城柳色半声笛 2024-12-06 14:19:36

我不确定开源 SDK 是否可以解决您的任务。根据您的描述,我发现您需要一个具有文档逻辑结构重建功能的复杂 ocr 应用程序。如果您正在规划商业软件,您可以查看ABBYY FineReader Engine。它有一套文档分析和重建功能,提供c#的api,而且是免费的去尝试。对于免费使用的程序来说,价格昂贵,但对于商业软件而言,ABBYY OCR 技术可以为您的产品增加巨大的价值,因此请考虑尝试一下。我在 ABBYY 工作,如有必要,可以为您提供更多信息。

最好的问候,尼古拉。

I’m not sure whether opensource SDKs can solve your tasks. Based on what you describe I see that you need a complex ocr application with document logical structure reconstruction functions. If you are planning business software you may look at ABBYY FineReader Engine. It has a set of document analyzing and reconstruction features, provides api for c# and it’s free to try. It’s not affordable for free-to-use programs, but when it comes to business software – ABBYY OCR technologies can add a serious value to your product, so consider trying it out. I work @ ABBYY and can provide you additional info if necessary.

Best regards, Nikolay.

免费的 OCR SDK for .net,可以将文本、带格式的表格和图像提取到 Office Word 文档中

半城柳色半声笛 2024-12-06 12:14:16

Eclipse 显示的源代码和实际用于编译可执行文件的源代码似乎很可能是同一文件的不同版本。

您可以使用 info source GDB 命令询问 GDB(Eclipse 有一个 GDB 控制台窗口)实际使用什么源来构建可执行文件,以及 GDB 在哪里找到它正在显示的源(向 Eclipse)。我敢打赌编译目录+当前源文件!=位于

It seems pretty likely that the source that Eclipse is showing, and the source that was actually used to compile the executable are different versions of the same file.

You can ask GDB (Eclipse has a console window for GDB) what source was actually used to build the executable, and where GDB found the source it is showing (to Eclipse) with info source GDB command. I bet the Compilation directory + Current source file != Located in.

使用 Eclipse CDT 进行调试时出现问题 - 执行和 GUI 未对齐

半城柳色半声笛 2024-12-06 10:56:33

如果删除所有非数字字符,则只剩下电话号码。然后,如果您愿意,您可以将该字符串解析为 ###-###-####

$phone = preg_replace('/\D/', '', 'Some test with 123-456-7890 that phone number');
//$phone is now 1234567890
echo substr($phone, 0, 3);//123
echo subsr($phone, 3, 3);//456
echo substr($phone, 6);//7890

不确定这是否是您正在寻找的。

If you delete all of the non numeric characters, you will only be left with the phone number. You can then take that string and parse it into ###-###-#### if you wish.

$phone = preg_replace('/\D/', '', 'Some test with 123-456-7890 that phone number');
//$phone is now 1234567890
echo substr($phone, 0, 3);//123
echo subsr($phone, 3, 3);//456
echo substr($phone, 6);//7890

Not sure if that is what you are looking for or not.

查找字符串中的电话号码

更多

推荐作者

丶视觉

文章 0 评论 0

蓝礼

文章 0 评论 0

birdxs

文章 0 评论 0

foonlee

文章 0 评论 0

微信用户

文章 0 评论 0

    我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
    原文