软糯酥胸

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软糯酥胸 2025-02-20 15:11:48

来自包装可以使用'azure.messaging.servicebus v7.8.1'。

servicebusadministrationclient.getqueueasync 返回queueproperties含量。状态:

https://learn.microsoft.com/en-us/dotnet/api/api/azure.messaging.servicebus.servicebus.administration.queeproperties.status.status?

公共azure.messaging.servicebus.administration.entitystatus状态{get;放; }

队列的当前状态(启用 /禁用)。< / p>

ServiceBusAdministrationClient serviceBusAdministrationClient = new ServiceBusAdministrationClient();

var status = await serviceBusAdministrationClient.GetQueueAsync("myQueue").Status;

ServiceBusAdministrationClient from Package 'Azure.Messaging.ServiceBus v7.8.1' can be used.

ServiceBusAdministrationClient.GetQueueAsync returns the QueueProperties incl. Status:

https://learn.microsoft.com/en-us/dotnet/api/azure.messaging.servicebus.administration.queueproperties.status?view=azure-dotnet#azure-messaging-servicebus-administration-queueproperties-status

public Azure.Messaging.ServiceBus.Administration.EntityStatus Status { get; set; }

The current status of the queue (Enabled / Disabled).

ServiceBusAdministrationClient serviceBusAdministrationClient = new ServiceBusAdministrationClient();

var status = await serviceBusAdministrationClient.GetQueueAsync("myQueue").Status;

如何检查其他区域中的Azure Service Bus是否正在使用Azure功能进行工作?

软糯酥胸 2025-02-20 05:20:32

验证您没有任何重复的列名。如果有另一个实例为2839238作为标题名称,PANDAS将添加.0或.1。

请参阅

说:
重复的列将被指定为“ x”,“ x.1”,…'x.n',而不是'x'…'x'。如果列中有重复的名称,则传递错误将导致数据被覆盖。

Verify that you don't have any duplicate column names. Pandas will add .0 or .1 if there is another instance of 2839238 as a header name.

See description of mangle_dupe_colsbool

which says:
Duplicate columns will be specified as ‘X’, ‘X.1’, …’X.N’, rather than ‘X’…’X’. Passing in False will cause data to be overwritten if there are duplicate names in the columns.

请勿将数字列名称转换为浮动pandas read_excel

软糯酥胸 2025-02-20 03:29:41

您的问题是,Symfony不知道ReadingData服务,正如Cerad New Symfony版本所提到的,使用Autowire来处理此服务,但在ShopWaree 6中不使用。您需要在Services.xml中注册它。就像这里描述的一样。

https://developer.shopware.com/docs/guides/plugins/plugins/plugin-fundamentals/add-custom-service

或使用建议的tag Controller.service.service_arguments。

Your problem is that the ReadingData service is not known by symfony, as mentioned by Cerad new symfony versions handle this with autowire, but not in shopwaree 6. You need to register it in your services.xml. Just like described here.

https://developer.shopware.com/docs/guides/plugins/plugins/plugin-fundamentals/add-custom-service

Or use the suggested tag controller.service_arguments.

也许您忘记将控制器注册为服务或错过使用“ controller.Service_arguments”将其标记为

软糯酥胸 2025-02-20 02:02:15

#我也有同样的问题。我添加了第一个答案中的这些行

spring:
  cloud:
    gateway:
      default-filters:
        - DedupeResponseHeader=Access-Control-Allow-Origin Access-Control-Allow-Credentials, RETAIN_UNIQUE
      globalcors:
        corsConfigurations:
          '[/**]':
            allowedOrigins: "http://localhost:8100"(or frontend link)
            allowedHeaders: "*"
            allowedMethods: "*"

#I had the same issue. I added these lines which are in the first answer

spring:
  cloud:
    gateway:
      default-filters:
        - DedupeResponseHeader=Access-Control-Allow-Origin Access-Control-Allow-Credentials, RETAIN_UNIQUE
      globalcors:
        corsConfigurations:
          '[/**]':
            allowedOrigins: "http://localhost:8100"(or frontend link)
            allowedHeaders: "*"
            allowedMethods: "*"

否访问控制;标题存在(CORS) - Spring Boot(Spring Security)微服务&#x2B; vue.js

软糯酥胸 2025-02-19 21:09:32

通过使用Linq

string[][] result = source.Split('\n')
                          .Select(line => line.Split(new char[] { ' ', '   ' }, StringSplitOptions.RemoveEmptyEntries));

Linq 具有更好的性能和更快的结果。

By using Linq

string[][] result = source.Split('\n')
                          .Select(line => line.Split(new char[] { ' ', '   ' }, StringSplitOptions.RemoveEmptyEntries));

linq has better performance and a faster result.

如何在C#中尽可能快地将字符串分为一系列字符串?

软糯酥胸 2025-02-19 17:58:28

您可以硬编码参数名称。但是请确保使用参数名称,而不是参数提示。例如:

You can hard-code the parameter name. But make sure you use the Parameter Name, not the Parameter Prompt. eg:

enter image description here

与参数的分页报告的电源自动化

软糯酥胸 2025-02-19 15:16:45

文件名是正确的。它没有单引号;这仅仅是 ls 显示它的方式。您可以使用 ls -n 告诉 ls 不要引用特殊字符的文件名:

❯ touch 'Los Angeles.txt'
❯ ls
'Los Angeles.txt'
❯ ls -N
Los Angeles.txt

The file name is correct as is. It doesn't have single quotes; that's simply how ls displays it. You can use ls -N to tell ls not to quote file names with special characters:

❯ touch 'Los Angeles.txt'
❯ ls
'Los Angeles.txt'
❯ ls -N
Los Angeles.txt

用空格创建来自可变值的文件

软糯酥胸 2025-02-19 02:23:30

看来不,您不能使用PDFMake做到这一点。

PDFMake用于创建/制作PDF文件(以声明性的方式),您似乎想做的就是阅读一个。如果您想在文档定义对象中添加一个斑点,我相信您不幸的是..

我建议您使用其他内容从二进制数据中读取PDF文件。 this 可能会做您想实现的目标。

It seems that no, you can't do that with pdfmake.

Pdfmake is used to create/make a pdf file (in a declarative way), what you seem to want to do is to read one. If you want to add a blob to your document-definition-object I believe you can't unfortunately..

I would suggest you to use something else to read a pdf file from binary data. This may do what you want to achieve.

使用pdfmake.createpdf()可以从现有斑点创建createPDF吗?

软糯酥胸 2025-02-18 23:01:09

您可以使用 flex 属性来实现。

为包装div提供弹性,然后按照您的要求将其在其中的孩子对齐

是一个例子

.maindiv{
  display:flex;
  flex-direction:row;
  align-items:center;
  justify-content:center;
}

.profileimg {
    top: 10px;
    right: 160px;
    width: 40px;
    height: auto;
    border-radius: 500px;
}
.text {
   margin-left:5px;
}
<div class="maindiv">
   <img class="profileimg" src="http://codeskulptor-demos.commondatastorage.googleapis.com/pang/IHXoEES.png" alt="profile">
   <h2 class="text">ASPECTER</h2>
</div>

You can use flex properties to do so.

Give flex to the wrapping div and then align the child inside it as per your requirements

Below is the example

.maindiv{
  display:flex;
  flex-direction:row;
  align-items:center;
  justify-content:center;
}

.profileimg {
    top: 10px;
    right: 160px;
    width: 40px;
    height: auto;
    border-radius: 500px;
}
.text {
   margin-left:5px;
}
<div class="maindiv">
   <img class="profileimg" src="http://codeskulptor-demos.commondatastorage.googleapis.com/pang/IHXoEES.png" alt="profile">
   <h2 class="text">ASPECTER</h2>
</div>

如何将文本和图像并排

软糯酥胸 2025-02-18 19:31:44

错误指出,Heroku无法确定您项目的语言。

您应该在项目中使用的Python软件包名称添加 sumplion.txt 文件,以允许Heroku将其检测为Python项目。
其他选项包括使用 setup.py 文件或 pipfile

手动设置buildpack,

heroku buildpacks:set heroku/python

您还可以在Heroku CLI中

以获取更多信息,请查看 document

The error states that heroku is unable to determine the language of your project.

You should add a requirements.txt file with the python package names used in your project to allow heroku to detect it as a python project.
Other options include using a setup.py file or Pipfile.

You can also manually set the buildpack using

heroku buildpacks:set heroku/python

in the heroku CLI

For more information check the documentation

将我的Python Discord Bot推向Heroku

软糯酥胸 2025-02-18 16:57:51

就我而言,就我而言,升级Lombok软件包的版本无效。

<dependency>
    <groupId>org.projectlombok</groupId>
    <artifactId>lombok</artifactId>
    <version>1.18.26</version>
</dependency>

我怀疑缓存问题或其他内容,我使用文件下的无效缓存并重新启动几次,但它不起作用。

最后,Build Runner打开了代表IDE构建/运行动作,仅此而已。但是,在打开它之后,当我本地运行项目调试时,它将非常慢,然后再次将其关闭。 Lombok没有报告错误,因此已解决。

In my case, In my case, upgrading the version of the Lombok package did not work.

<dependency>
    <groupId>org.projectlombok</groupId>
    <artifactId>lombok</artifactId>
    <version>1.18.26</version>
</dependency>

I suspect the cache problem or something, I use the invalid cache under the file and restart it several times, but it doesn’t work.

Finally, the build runner turns on the delegate ide build/run actions to maven, and that's it. However, after it is turned on, it will be very slow when I run the project debug locally, and then turn it off again. Lombok did not report an error, so it was solved.

enter image description here

Intellij Idea 2022.1.3最终显示错误Java:JPS增量注释处理被禁用

软糯酥胸 2025-02-18 15:15:10

您已将 *TMP分配给了头。但是我可以看到那头一直是无效的。因此,循环永远不会输入,并且什么都没有插入列表。

因此,我们需要做什么,首先将头放到节点实例,

typdef struct node 
{
    char* data;
    struct node *prev;
    struct node *next;
}node;

head = malloc(sizeof(node));

然后分配数据。

head->attribute = value;

最后,我们也将此指针位置值也设置为TMP指针。

tmp = head;

不,我们可以在C中进行循环

字符串表示为一系列字符,其存储的位置指向Char数组的第一个元素。数组还必须以零聊天'\ 0'结束。请注意,strlen(str)将返回无效char的字符串长度,因此您必须在杂交时添加1个才能将其归为sconsistration。除非绝对必要,否则我建议不要弄乱字符串。我的意思是试图手动操纵它们。这将引入另一组与我们通常从事的问题。我们应该只使用strncar(),strncpy()方法,直到C样式的字符串变得完全直观。

You have assigned *tmp to head. But I can see that head had always been NULL. So the loop is never entered, and nothing is being inserted to the list.

So what we'd need to do it first initialize head to a node instance

typdef struct node 
{
    char* data;
    struct node *prev;
    struct node *next;
}node;

head = malloc(sizeof(node));

then we assign it data.

head->attribute = value;

finally we set this pointer location value to our tmp pointer as well.

tmp = head;

no we can proceed with our loop

strings in C are represented as an array of chars, whose stored location points to the first element of the char array. the array must also end with a NULL chat '\0'. note that strlen(str) will return length of string without the NULL char so you must add 1 while mallocing to take this into sconsideration. i would advise not messing with strings unless absolutely necessary. by that I mean trying to manually manipulate them. this will introduce another set of problems not related to what we're working on in general. we should just use strncar(), strncpy() methods until c style strings become completely intuitive.

如何将C代码打印到输出文件?

软糯酥胸 2025-02-18 06:33:10

使用 [... new Set(FinalArray)] 删除重复。

详细信息在示例中评论

// Utility function
const log = data => console.log(JSON.stringify(data));

/**
 * @desc Given a number, list a multiples of 3 and 5 that are less
 * than the given number. If the given number is less than 4 return 0
 * @param {number} number - The number that serves as the limit
 * @return {array<number>} The array of multiples of 3 and 5 up to but
 * not including the given number.
 */
function threeFive(number) {
  // Return 0 if number is less than 4
  if (number < 4) return 0;
  /*
  Create an array of <number>-1 empty slots...
  .map() returns the <i>nterval times 3 if it is less than the <number>...
  ...otherwise returns null...
  ...all nulls and zeros will be .filter()'ed out.
  The next expression is identical to the previous with the exception
  of multiplying by 5 instead of 3
  */  
  const three = [...new Array(number-1)].map(
    (_, i) => (i * 3) < number ? i * 3 : null)
    .filter(n => n);
  
  const five = [...new Array(number-1)].map(
    (_, i) => (i * 5) < number ? i * 5 : null)
    .filter(n => n);
  /*
  merge the two arrays with .concat()...
  ...convert it into a Set...
  ...convert the Set back into an array thereby eliminating all
  duplicates...
  ...and finally .reduce() the sum of all numbers
  */
  return [...new Set(three.concat(five))].reduce((sum, add) => sum + add)
}
log(threeFive(10));
log(threeFive(44));
log(threeFive(98));
log(threeFive(3));
log(threeFive(-9));

Use [...new Set(finalArray)] to remove duplicates.

Details are commented in example

// Utility function
const log = data => console.log(JSON.stringify(data));

/**
 * @desc Given a number, list a multiples of 3 and 5 that are less
 * than the given number. If the given number is less than 4 return 0
 * @param {number} number - The number that serves as the limit
 * @return {array<number>} The array of multiples of 3 and 5 up to but
 * not including the given number.
 */
function threeFive(number) {
  // Return 0 if number is less than 4
  if (number < 4) return 0;
  /*
  Create an array of <number>-1 empty slots...
  .map() returns the <i>nterval times 3 if it is less than the <number>...
  ...otherwise returns null...
  ...all nulls and zeros will be .filter()'ed out.
  The next expression is identical to the previous with the exception
  of multiplying by 5 instead of 3
  */  
  const three = [...new Array(number-1)].map(
    (_, i) => (i * 3) < number ? i * 3 : null)
    .filter(n => n);
  
  const five = [...new Array(number-1)].map(
    (_, i) => (i * 5) < number ? i * 5 : null)
    .filter(n => n);
  /*
  merge the two arrays with .concat()...
  ...convert it into a Set...
  ...convert the Set back into an array thereby eliminating all
  duplicates...
  ...and finally .reduce() the sum of all numbers
  */
  return [...new Set(three.concat(five))].reduce((sum, add) => sum + add)
}
log(threeFive(10));
log(threeFive(44));
log(threeFive(98));
log(threeFive(3));
log(threeFive(-9));

在输入数字以下的倍数之和遇到困难吗?

软糯酥胸 2025-02-17 18:26:57

替代方法是通过

这为您提供了最终的控制 - 您可以在更精细的层面上过滤出不需要的东西(例如,您可以滤除详细级别的轨迹)。

Alternative way is to do filtering through telemetry processor.

This gives you ultimate control - you can filter out what you don't need at more granular level (for instance, you might filter out verbose level traces).

应用程序见解 - 我可以禁用其他所有内容,但是错误记录和异常吗?

软糯酥胸 2025-02-17 13:07:36

在进行PCIE挑选时,您确实需要一个已知的外部主机来进行JTAG,并在启动时捕获Linux序列输出。远程访问的电源板非常方便,您当然可以引起有趣的锁定。如果您无法访问PCIE分析仪,则可能很难查明这些核心问题。

通常,我会禁用DMA核心,直到您列举了基本的栏并正确响应。

没有物理访问交换机似乎是理由是“注释” sys_rst_n - 这需要正确绑定才能使内核工作。故障的核心肯定会降低点对点链接,虽然您希望PCIE Bridges可以隔离单个行为不当的设备,但您的​​结果表明可能并非如此。

我不遵循您的调试时钟与PCIE核心的交互方式,但是我的建议是按照董事会供应商的默认值离开PCIE核心时钟和重置安排,直到您真的知道您在做什么,并且可以制作和测试孤立的小型小更改。

When doing pcie bringup, you really need an known-good external host for JTAG, and to capture linux serial output at boot time. A remotely accessible power strip is handy as well as you can certainly cause interesting lock-ups. If you don't have access to a pcie analyser it can be very difficult to pinpoint problems with these cores.

Generally, I would disable the DMA core until you have the basic BAR enumerating and responding properly.

Not having physical access to switches doesn't seem like justification to "comment out" sys_rst_n - this needs to be tied properly for the cores to work. A faulty core can certainly bring down point-to-point links and while you would hope that the PCIe bridges would isolate single misbehaving devices your results demonstrate this might not be the case.

I don't follow how your debug clock is interacting with the pcie core, but my advice would be to leave the pcie core clocking and reset arrangements per the board vendors defaults until you really know what you are doing and can make and test isolated small changes.

FPGA PCIE端点可防止主机重新启动

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