做个少女永远怀春

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做个少女永远怀春 2025-02-20 15:10:09

好的,我找到了解决方案。在这种情况下,AddEventListener需要在“ true”上设置第三个参数(opt_capture)。感谢此组范围侦听事件侦听器。

        routeLine.addEventListener('pointerenter', function(evt) {
            console.log('test 123', evt) 
        }, true) //<- this true is a key

OK, I found a solution. In this case addEventListener require third parameter (opt_capture) set on "true".Thanks to this group scope listen event listener.

        routeLine.addEventListener('pointerenter', function(evt) {
            console.log('test 123', evt) 
        }, true) //<- this true is a key

这里地图指针无法与polyline grupe一起使用,但TAP正在工作

做个少女永远怀春 2025-02-20 00:51:29

您要实现的目标有点复杂,可以通过CSS实现。您可能需要考虑设置每个子元素的不透明度,而不是设置父母的不透明度。

查看此线程

What you're trying to achieve is a bit complex to achieve with CSS. You may want to take a look at setting the opacity of each child element than setting the opacity of the parent.

Take a look at the answers in this thread

将单独的分离不透明度设置为1,而父母不透明度设置为.7在React中

做个少女永远怀春 2025-02-19 16:39:09

我可以在本地运行纱线以生成yarn.lost并使用npm安装在CI上复制包装树?

不,如果要使用NPM V7复制包装树,您仍然需要一个package-lock.json文件。 V7仅将YARN.LOCK文件用作包装元数据的来源解决指南

这是因为NPM V7使用新的lockfileversion生成了锁紧files:2(向后兼容),以提高性能并允许具有确定性树形状的可重复性构建;它还试图从package.json元数据文件中减少读数。

当我使用yarn.lock安装NPM时,该怎么办?它会更新吗?它可以与(一种) - Frozen-Lockfile一起运行吗? NPM CI命令根本不提及Yarn.lock。

是的,yarn.lock文件将被更新,而NPM仍将创建package> package-lock.json文件,如果package> package> package> package-lock.json 文件,它将用作创建树​​形的权威定义。

按照他们不支持yarn.lock文件的原因是

一个常见的问题,我们现在已经遇到了几次,一旦我们宣布NPM V7将包括对Yarn.lock文件的支持,那就是“为什么要保留package-lock.json,那么?为什么不只使用yarn.lock?”

简单的答案是:因为Yarn.lock并未完全满足NPM的需求,并且仅依靠它将限制我们将来生产最佳包装安装或添加功能的能力。

最后,

它是否支持V1和Berry?

这尚不清楚,并且暗示为berry更改,因为YARN使用Yarn.lock文件和纱线版本的组合来保证确定性分辨率。

我建议阅读此blogpost 上述帖子来自 npm 用于更多的in-depth for-depth。

Can I run yarn locally to generate yarn.lock and reproduce the package tree on CI with npm install?

No, you still need a package-lock.json file if you want to reproduce the package tree using npm v7. v7 only uses the yarn.lock file as a source of package metadata and resolution guidance.

This is because npm v7 generates the lockfiles with the a new lockFileVersion: 2 (backward compatible) to improve performance and allow reproducible builds with deterministic tree shape; it also tries to reduce reading from package.json file for metadata.

What exactly happens when I run npm install with yarn.lock? Does it get updated? Can it be run with (sort of) --frozen-lockfile? The npm ci command doesn't mention yarn.lock at all.

Yes, the yarn.lock file will be updated and npm will still create a package-lock.json file, and if a package-lock.json file is present, it’ll be used as the authoritative definition of the tree shape to create.

As per this NPM blog post, the reason they don't support yarn.lock file yet is

One common question we’ve gotten a few times now, once we announce that npm v7 will include support for yarn.lock files, is “Why keep package-lock.json at all, then? Why not just use yarn.lock only?”

The simple answer is: because yarn.lock doesn’t fully address npm’s needs, and relying on it exclusively would limit our ability to produce optimal package installs or add features in the future.

Finally,

Does it support both v1 and berry?

This is not clear, and is implied to change for berry as yarn uses a combination of yarn.lock file and yarn version to guarantee deterministic resolution.

I suggest reading this blogpost along with the above post from npm for more in-depth explanation.

NPM和Yarn.lock

做个少女永远怀春 2025-02-19 07:51:37

使用JSON_ENCODE()创建JSON和urlencode()将其正确编码以在URL中使用。

$param = urlencode(json_encode(['need' => 'help']));
curl_setopt($ch, CUROPT_URL, "example.com/api/?params=$param");

Use json_encode() to create the JSON, and urlencode() to encode it properly for use in a URL.

$param = urlencode(json_encode(['need' => 'help']));
curl_setopt($ch, CUROPT_URL, "example.com/api/?params=$param");

我可以通过php curl中的get方法发送一个包含卷曲括号的参数

做个少女永远怀春 2025-02-19 04:07:37

我已经在不同的项目中使用了这两种方法。

无论哪种情况,您都可以使用其中之一,因为另一个已经用于其他东西。如果您的RTOS想要Systick,请使用调试计数器。如果您的调试器想要调试计数器,请使用Systick。

Systick的主要缺点是它只有24位,而调试计数器有32位。

主要缺点是调试计数器是每个部分都不可用(Systick也是可选的,但几乎没有任何硅供应商认为它出去)。

启用整个调试块仅用于柜台也浪费了一点功率,如果您要从电池跑步,您可能会关心。

I have used both these methods in different projects.

In either case, you might use one of these because the other was already used for something else. If your RTOS wants the systick, use the debug counter. If your debugger wants the debug counter, use the systick.

The main disadvantage of the systick is that it only has 24 bits, whereas the debug counter has 32.

The main disadvantage is the debug counter is it is not available on every part (the systick is optional too, but hardly any silicon vendors take it out).

Enabling the whole debug block just for a counter also wastes a little bit of power, which you might care about if you are running from batteries.

如何测量皮质-M4中的经过时间

做个少女永远怀春 2025-02-18 21:28:20

Firebase Cloud Messaging的后端系统中主题的粉丝针对总系统吞吐量进行了优化,而不是任何单个主题的延迟。

它是否足够快地是主观的,只有您才能确定项目足够快,但是对映射的性能进行控制的唯一方法就是自己做。

The fanout of topics in Firebase Cloud Messaging's backend systems is optimized for total system throughput and not for the latency of any individual topic.

Whether it is "fast enough" is subjective and only you can determine what is fast enough for your project, but the only way to have control over the performance of the mapping is to do it yourself.

firebase fcm,主题为客户平均交付时间

做个少女永远怀春 2025-02-18 08:00:19

这个问题已经在

虽然可以将二进制文件存储在CQL blob列中,但这样做不是一个好主意。有关详细信息,请参阅重复问题中的建议。干杯!

This question has already been asked and answered in Is it possible to store PDF files in a CQL blob type in Cassandra?.

While it is possible to store binary files in a CQL blob column, it's not a good idea to do so. For details, see my recommendation in the duplicate question. Cheers!

如何将PDF存储在Cassandra中?

做个少女永远怀春 2025-02-18 03:55:33

不同发票之间的关系尚不清楚。

可能很简单:

>>> df.drop_duplicates('InvoiceNumber', keep='last')
   Reference Number InvoiceNumber  InvoiceAmount DocType      Date
2                 3          A123            100      IN  1/3/2022
3                 4          A342            123      IN  1/3/2022
4                 5          A444            345      IN  1/3/2022

The relation between different invoices are not clear.

It could be simply:

>>> df.drop_duplicates('InvoiceNumber', keep='last')
   Reference Number InvoiceNumber  InvoiceAmount DocType      Date
2                 3          A123            100      IN  1/3/2022
3                 4          A342            123      IN  1/3/2022
4                 5          A444            345      IN  1/3/2022

当3行类似

做个少女永远怀春 2025-02-17 03:36:06
const data = {name: 'Ronn', age: 27};              //sample json
const a = document.createElement('a');
const blob = new Blob([JSON.stringify(data)]);
a.href = URL.createObjectURL(blob);
a.download = 'sample-profile';                     //filename to download
a.click();

在此处查看BLOB文档 - blob mdn 类型。默认情况下它将制作.txt文件

const data = {name: 'Ronn', age: 27};              //sample json
const a = document.createElement('a');
const blob = new Blob([JSON.stringify(data)]);
a.href = URL.createObjectURL(blob);
a.download = 'sample-profile';                     //filename to download
a.click();

Check Blob documentation here - Blob MDN to provide extra parameters for file type. By default it will make .txt file

是否可以仅使用JavaScript编写数据来文件?

做个少女永远怀春 2025-02-16 11:53:45

带有浮点数的陷阱是它们看起来像十进制,但在二进制中起作用。

唯一的主要因素是2,而10个的主要因素为2和5。结果是,可以完全写成二进制分数的每个数字也可以完全写成小数分数,但只有一个子集可以写入小数分数的数字可以写成二进制分数。

浮点数本质上是二进制分数,其数量有限。如果您超越了这些重要数字,那么结果将被舍入舍入。

当您在代码中键入文字或调用该函数以将浮点数分解为字符串时,它会期望十进制数字,并且在变量中将该小数号的二进制近似值存储。

当您打印浮点号或调用函数以将一个函数转换为字符串时,它会打印浮点数号的小数近似值。它是 可以准确地将二进制号转换为十进制的,但是我默认不知道在转换为字符串*时这样做。某些语言使用固定数量的大数字,而另一些语言则使用最短的字符串,该字符串将“往返”回到相同的浮点值。

* python dis 将浮点数转换为“ Decimal.decimal”时,请准确转换。这是我知道获得浮点数的确切小数等效的最简单方法。

The trap with floating point numbers is that they look like decimal but they work in binary.

The only prime factor of 2 is 2, while 10 has prime factors of 2 and 5. The result of this is that every number that can be written exactly as a binary fraction can also be written exactly as a decimal fraction but only a subset of numbers that can be written as decimal fractions can be written as binary fractions.

A floating point number is essentially a binary fraction with a limited number of significant digits. If you go past those significant digits then the results will be rounded.

When you type a literal in your code or call the function to parse a floating point number to a string, it expects a decimal number and it stores a binary approximation of that decimal number in the variable.

When you print a floating point number or call the function to convert one to a string it prints a decimal approximation of the floating point number. It is possible to convert a binary number to decimal exactly, but no language I'm aware of does that by default when converting to a string*. Some languages use a fixed number of significant digits, others use the shortest string that will "round trip" back to the same floating point value.

* Python does convert exactly when converting a floating point number to a "decimal.Decimal". This is the easiest way I know of to obtain the exact decimal equivalent of a floating point number.

浮点数学破裂了吗?

做个少女永远怀春 2025-02-15 16:29:39

您的测试不使用ReadInt()的结果,并且不接收输入,因为您正在设置system.setin(in);在调用readint() readint()代码>。

class InputTest {
   @Test
   public void testMethod() throws IOException {
      // arrange
      String input = "4";
      InputStream in = new ByteArrayInputStream(input.getBytes());
      System.setIn(in);

      // act
      MyClass myclass = new MyClass();
      int num = myclass.readInt();

      // assert
      assertEquals(4, num);
   }

   @Test
   public void testMethodInject() throws IOException {
      // arrange
      String input = "4";
      InputStream in = new ByteArrayInputStream(input.getBytes());
      Scanner scanner = new Scanner(in);

      // act
      MyClass myclass = new MyClass();
      int num = myclass.readIntInjected(scanner);

      // assert
      assertEquals(4, num);
   }
}

如果您注入scanner对象而不是创建它,则可能会更严重的测试。

class MyClass {
   public int readInt() {
      Scanner scan = new Scanner(System.in);
      System.out.print("Enter number: ");
      int num = scan.nextInt();
      scan.close();
      return num;
   }

     // more easly testable
   public int readIntInjected(Scanner scanner) {
      System.out.print("Enter number: ");
      int num = scanner.nextInt();
      scanner.close();
      return num;
   }
}

Your test does not use the result of readInt() and receives no input because you are setting System.setIn(in); after calling readInt().

class InputTest {
   @Test
   public void testMethod() throws IOException {
      // arrange
      String input = "4";
      InputStream in = new ByteArrayInputStream(input.getBytes());
      System.setIn(in);

      // act
      MyClass myclass = new MyClass();
      int num = myclass.readInt();

      // assert
      assertEquals(4, num);
   }

   @Test
   public void testMethodInject() throws IOException {
      // arrange
      String input = "4";
      InputStream in = new ByteArrayInputStream(input.getBytes());
      Scanner scanner = new Scanner(in);

      // act
      MyClass myclass = new MyClass();
      int num = myclass.readIntInjected(scanner);

      // assert
      assertEquals(4, num);
   }
}

It may be more easly testable if you inject Scanner object instead of creating it.

class MyClass {
   public int readInt() {
      Scanner scan = new Scanner(System.in);
      System.out.print("Enter number: ");
      int num = scan.nextInt();
      scan.close();
      return num;
   }

     // more easly testable
   public int readIntInjected(Scanner scanner) {
      System.out.print("Enter number: ");
      int num = scanner.nextInt();
      scanner.close();
      return num;
   }
}

测试一种方法,中间需要用户输入-Junit

做个少女永远怀春 2025-02-15 15:45:55

您可以尝试

$ data = ceartryAtereQuest :: where('id',$ type) - &gt; get();

edit

CertificateRequest::select($field)->where('id',$type);

也许这可以有所帮助!

You can try

$data = CertificateRequest::where('id',$type)->get();

EDIT

CertificateRequest::select($field)->where('id',$type);

Maybe this can help !

Laravel通过主键查找并返回单列值

做个少女永远怀春 2025-02-15 15:45:22

您可以使用ticklabel_format()设置tick标签格式。您可以添加以下行 - &gt; plt.ticklabel_format(style ='sci',axis ='x',scilimits =( - 6,-6)) 在e-06中具有x轴。请注意,-6至-6告诉Matplotlib将E -06的格式设置为E -06。更多信息

”相同...

代码

import numpy as np 
import matplotlib.pyplot as plt
import pylab
import style

xdata = np.array([1.21, 1.32, 2.54]) * (1e-5)
ydata = [1, 4, 15]

#Plot
plt.figure(1,dpi=800)
plt.title('Force Spectroscopy')
plt.ylabel('Vertical Deflection')
plt.xlabel('Measured Height')

plt.plot(xdata,ydata, color='green',label=('Experimental data'))

#Theoritical Plot
new= -(0.107e-5)*np.array(xdata)
plt.plot(xdata,new, color= 'purple',label='Theoritical')
plt.ticklabel_format(style='sci', axis='x', scilimits=(-6,-6))

#Legend Modification
plt.legend('upper right',fontsize=20)
plt.legend()

输出

“在此处输入图像说明”

you can use ticklabel_format() to set the tick label format. You can add the following line -> plt.ticklabel_format(style='sci', axis='x', scilimits=(-6,-6)) to your code to have the x-axis in e-06. Note that the -6 to -6 is telling matplotlib to set the format from e-06 to e-06. More info here

Your modified code sample here to demonstrate the same...

Code

import numpy as np 
import matplotlib.pyplot as plt
import pylab
import style

xdata = np.array([1.21, 1.32, 2.54]) * (1e-5)
ydata = [1, 4, 15]

#Plot
plt.figure(1,dpi=800)
plt.title('Force Spectroscopy')
plt.ylabel('Vertical Deflection')
plt.xlabel('Measured Height')

plt.plot(xdata,ydata, color='green',label=('Experimental data'))

#Theoritical Plot
new= -(0.107e-5)*np.array(xdata)
plt.plot(xdata,new, color= 'purple',label='Theoritical')
plt.ticklabel_format(style='sci', axis='x', scilimits=(-6,-6))

#Legend Modification
plt.legend('upper right',fontsize=20)
plt.legend()

Output

enter image description here

我想将X轴科学符号从E-05更改为E-06

做个少女永远怀春 2025-02-15 08:20:44

快速清单

  • 不正确的凹痕最常见的是indentationError,但也可以导致taberrorsub-type indentation> indentation error)或> SyntaxError(凹痕本身是合法的,但导致其他代码具有语法错误)。对Python代码有效的凹痕,但对于程序员的意图是错误的,会导致逻辑错误(该代码不会引起异常,但做错了什么)。
  • 强烈建议不要使用选项卡进行凹痕。在2.x中,使用-TT命令行参数运行Python会导致其提高相同的TabEror,这对于查找问题很有用。
  • 社区标准,在
  • 之类的语句,,, while , def 和class等使用COLON 需要一个缩进块之后:
     如果x:
        do_y()#&lt;  - 必须缩进
     
  • blocks 不能为空,并且注释不算 算作使块“不是空”。使用 Pass语句如果什么都不发生:
     如果x:
        通过#此陈述什么都不做,除了使块非空块
     
  • 块内的代码必须具有相同的缩进:
     如果x:
        do_y()
        do_z()#这需要与do_y()对齐
     
  • 标记之前,占用整个行的注释可能具有任何数量的空格。
  • DocStrings 不是注释,必须以与以下代码相同的方式进行缩进。
  • 与开头语句或较低缩进水平对齐的第一行代码在块之外:
     如果x:
        do_y()
    do_z()#无论x值如何
    #不能写``else':在这里;那是语法错误
    
    对于我在范围(10)中:
        如果x:
            do_y()
    do_z()#`
     
  • python 将选项卡解释为扩展到下一个第8列 ;但是在3.x中,混合空间和选项卡的压痕必须具有完全相同的模式以相同的凹痕计数(并且要进一步需要与前缀具有相同的模式)。未能正确获得此结果Taberror
  • 和 for python中的循环可能具有 AN else else rape ,如果循环正常完成,而不是通过break 执行。这是逻辑错误的常见来源:
     在范围(10)中为i:
        如果我%2:
            print(f'{i}是奇数')
    否则:#错误的缩进
        print(f'{i}是什至')#仅发生一次,索赔9甚至是!
     

Quick checklist

  • Incorrect indentation most commonly results in IndentationError, but it can also result in TabError (a sub-type of IndentationError) or SyntaxError (the indentation itself was legal, but it caused other code to have a syntax error). Indentation that is valid for Python code, but wrong for the programmer's intent, causes logical errors (the code doesn't raise an exception, but does something wrong).
  • It is strongly recommended not to use tabs for indentation. In 2.x, running Python with the -tt command line argument causes it to raise the same TabError, which is useful for finding problems.
  • The community standard, laid out in PEP 8, is to use four spaces per level of indentation.
  • Statements like if, for, while, def, and class etc that end with a colon : need an indented block after them:
    if x:
        do_y() # <- this must be indented
    
  • Blocks cannot be empty, and a comment does not count as making the block "not empty". Use the pass statement if nothing should happen:
    if x:
        pass # this statement does nothing, except make the block non-empty
    
  • Code within a block must have the same indentation:
    if x:
        do_y()
        do_z() # this needs to line up with do_y()
    
  • Comments that take up the entire line may have any amount of whitespace before the # mark.
  • Docstrings are not comments, and must be indented the same way as the following code.
  • The first line of code that lines up with the opening statement, or a lower level of indentation, is outside the block:
    if x:
        do_y()
    do_z() # this happens regardless of the x value
    # cannot write `else`: here; that is a syntax error
    
    for i in range(10):
        if x:
            do_y()
    do_z() # both the `if` and `for` blocks are ended
    
  • Python interprets tabs as expanding to the next 8th column; but in 3.x, indentation that mixes spaces and tabs must have the exact same pattern to count as the same indentation (and to indent further requires having the same pattern as a prefix). Failure to get this right results in TabError.
  • while and for loops in Python may have an else clause, which executes if the loop completes normally rather than via break. This is a common source of logical errors:
    for i in range(10):
        if i % 2:
            print(f'{i} is odd')
    else: # wrongly indented
        print(f'{i} is even') # only happens once, and claims 9 is even!
    

我得到了一个凹痕(或taberror)。我该如何修复?

做个少女永远怀春 2025-02-15 04:44:56

它可能与AWS_USER_POOLS_WEB_CLIENT_ID有关。请确保您使用正确的用户池。

It may be related with aws_user_pools_web_client_id. Please be sure you use the right user pool.

如果我在AWS Incognito中配置了5分钟,为什么我的Idtoken在1天内到期?

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