脸赞

文章 评论 浏览 27

脸赞 2025-02-08 09:44:25

这是另一种做到这一点的方式,使用 TAC + awk + TAC 组合。

tac Input_file |
awk -v RS="" '{sub(/<\/Polygon>/,"</MultiGeometry>\n&")} 1' | 
tac

说明: 简单说明将是,首先使用 tac 首先将文件行打印到反向顺序(从底部到第一行)并传递它的输出是 awk 程序的输入。在中,awk 程序设置 rs (记录分隔符)为null,并使用 sub 在其主要程序中替换&lt;/polygon&gt; < /代码>带有&lt;/Multigemetry&gt; 和本身(使用&amp; ),然后使用 1 awk SH打印当前行的方法。进一步发送 awk 代码的输出作为 tac 命令的输入,以获取实际的行顺序。

Here is another way of doing this, with tac + awk + tac combination.

tac Input_file |
awk -v RS="" '{sub(/<\/Polygon>/,"</MultiGeometry>\n&")} 1' | 
tac

Explanation: Simple explanation would be, using tac on Input_file first to print lines of file into reverse order(from bottom to first line) and passing its output as an input to awk program. In awk program setting RS(record separator) as NULL and in its main program using sub to substitute </Polygon> with </MultiGeometry> and itself(using &), then using 1 is an awksh way to print current line. Further sending awk code's output as an input to tac command to get actual order of lines.

最终模式匹配后,将字符串添加到新线路

脸赞 2025-02-08 03:05:15

我已经能够实现所需的行为。这是一个最小的可重现示例:

  1. 我创建了一个没有权限的用户“ foobar ”。我给了该用户一个标签环境带有值 test 。我给了该用户以下内联策略:
{
    "Version": "2012-10-17",
    "Statement": [
        {
            "Effect": "Allow",
            "Action": "s3:ListBucket",
            "Resource": "arn:aws:s3:::*${aws:PrincipalTag/environment}*"
        }
    ]
}
  1. 我创建了两个存储桶,一个称为 myDevbucket ,另一个称为 myTestBucket 。我将一个文件添加到 myTestBucket ,“ foo.txt ”中。
  2. 我以“ foobar ”的用户登录AWS帐户,并尝试导航到两个存储桶: https://s3.console.aws.amazon。 com/s3/buckets/mytestbucket?region = eu-west-2&amp; tab = objects for Test Bucket, https://s3.console.aws.aws.amazon.com/s3/buckets/mydevbucket ?

    正如预期的那样,我只能看到测试桶(因为环境标签具有a值 test ):

/Y5773.png“ rel =” nofollow noreferrer“>

I've been able to achieve the desired behavior. Here's a minimal reproducible example:

  1. I created a user "foobar" with no permissions. I gave this user a tag environment with the value test. I gave this user the following inline policy:
{
    "Version": "2012-10-17",
    "Statement": [
        {
            "Effect": "Allow",
            "Action": "s3:ListBucket",
            "Resource": "arn:aws:s3:::*${aws:PrincipalTag/environment}*"
        }
    ]
}
  1. I created two buckets, one called mydevbucket and another called mytestbucket. I added one file to the mytestbucket, "foo.txt".
  2. I logged in the AWS account as the "foobar" user, and tried to navigate to the two buckets: https://s3.console.aws.amazon.com/s3/buckets/mytestbucket?region=eu-west-2&tab=objects for the test bucket, https://s3.console.aws.amazon.com/s3/buckets/mydevbucket?region=eu-west-2&tab=objects for the dev bucket.

    As expected, I can only see the test bucket (because the environment tag has a the value test):

enter image description here

whereas in the dev bucket:

enter image description here

IAM策略以根据用户标签确定S3存储牌访问

脸赞 2025-02-07 09:07:34

这对我来说很好。尝试 grouped_list https://pub.dev/packages/grouped_list with with with with with with with with with with with with with with with with with with with with groupheaderbuilder

GroupedListView<StaticChat, String>(
      controller: listScrollController,
      elements: widget.privateChat.messages,
      groupBy: (element) => element.date,
     groupHeaderBuilder: (element) => ListTile(
        leading: CircleAvatar(
          backgroundImage: NetworkImage(element.sentBy.profilePicture!),
        ),
        title: Text(element.date),
      ),
      indexedItemBuilder: (context, StaticChat element, index) =>
          chatBubble(element, index),
      itemComparator: (item1, item2) => item1.time.compareTo(item2.time),
      useStickyGroupSeparators: true,
      floatingHeader: true,
      order: GroupedListOrder.ASC,
    );

This works fine for me. Try grouped_list https://pub.dev/packages/grouped_list package with groupHeaderBuilder

GroupedListView<StaticChat, String>(
      controller: listScrollController,
      elements: widget.privateChat.messages,
      groupBy: (element) => element.date,
     groupHeaderBuilder: (element) => ListTile(
        leading: CircleAvatar(
          backgroundImage: NetworkImage(element.sentBy.profilePicture!),
        ),
        title: Text(element.date),
      ),
      indexedItemBuilder: (context, StaticChat element, index) =>
          chatBubble(element, index),
      itemComparator: (item1, item2) => item1.time.compareTo(item2.time),
      useStickyGroupSeparators: true,
      floatingHeader: true,
      order: GroupedListOrder.ASC,
    );

如何使用粘性标头创建聊天页面,例如Telegram Group聊天?

脸赞 2025-02-07 04:00:29

可能的解决方法是使用 new request()构造函数,然后检查 request.bodyed boolean 属性

bodysed 属性的getter必须返回true,如果 distranded ,并且
否则为否。

确定流是否为分布式

实现 body mixin的对象被认为是 distranced 如果
body 是非零子,其 distranced

返回 fetch() promise 中。 readableStream request.bodyed.body 等于 true

请注意,该方法不会读取 request.body 的字节,因为将字节流传输到端点。同样,在将任何响应完全返回到浏览器之前,上传可以很好地完成。

const [input, progress, label] = [
  document.querySelector("input")
  , document.querySelector("progress")
  , document.querySelector("label")
];

const url = "/path/to/server/";

input.onmousedown = () => {
  label.innerHTML = "";
  progress.value = "0"
};

input.onchange = (event) => {

  const file = event.target.files[0];
  const filename = file.name;
  progress.max = file.size;

  const request = new Request(url, {
    method: "POST",
    body: file,
    cache: "no-store"
  });

  const upload = settings => fetch(settings);

  const uploadProgress = new ReadableStream({
    start(controller) {
        console.log("starting upload, request.bodyUsed:", request.bodyUsed);
        controller.enqueue(request.bodyUsed);
    },
    pull(controller) {
      if (request.bodyUsed) {
        controller.close();
      }
      controller.enqueue(request.bodyUsed);
      console.log("pull, request.bodyUsed:", request.bodyUsed);
    },
    cancel(reason) {
      console.log(reason);
    }
  });

  const [fileUpload, reader] = [
    upload(request)
    .catch(e => {
      reader.cancel();
      throw e
    })
    , uploadProgress.getReader()
  ];

  const processUploadRequest = ({value, done}) => {
    if (value || done) {
      console.log("upload complete, request.bodyUsed:", request.bodyUsed);
      // set `progress.value` to `progress.max` here 
      // if not awaiting server response
      // progress.value = progress.max;
      return reader.closed.then(() => fileUpload);
    }
    console.log("upload progress:", value);
    progress.value = +progress.value + 1;
    return reader.read().then(result => processUploadRequest(result));
  };

  reader.read().then(({value, done}) => processUploadRequest({value,done}))
  .then(response => response.text())
  .then(text => {
    console.log("response:", text);
    progress.value = progress.max;
    input.value = "";
  })
  .catch(err => console.log("upload error:", err));

}

A possible workaround would be to utilize new Request() constructor then check Request.bodyUsed Boolean attribute

The bodyUsed attribute’s getter must return true if disturbed, and
false otherwise.

to determine if stream is distributed

An object implementing the Body mixin is said to be disturbed if
body is non-null and its stream is disturbed.

Return the fetch() Promise from within .then() chained to recursive .read() call of a ReadableStream when Request.bodyUsed is equal to true.

Note, the approach does not read the bytes of the Request.body as the bytes are streamed to the endpoint. Also, the upload could complete well before any response is returned in full to the browser.

const [input, progress, label] = [
  document.querySelector("input")
  , document.querySelector("progress")
  , document.querySelector("label")
];

const url = "/path/to/server/";

input.onmousedown = () => {
  label.innerHTML = "";
  progress.value = "0"
};

input.onchange = (event) => {

  const file = event.target.files[0];
  const filename = file.name;
  progress.max = file.size;

  const request = new Request(url, {
    method: "POST",
    body: file,
    cache: "no-store"
  });

  const upload = settings => fetch(settings);

  const uploadProgress = new ReadableStream({
    start(controller) {
        console.log("starting upload, request.bodyUsed:", request.bodyUsed);
        controller.enqueue(request.bodyUsed);
    },
    pull(controller) {
      if (request.bodyUsed) {
        controller.close();
      }
      controller.enqueue(request.bodyUsed);
      console.log("pull, request.bodyUsed:", request.bodyUsed);
    },
    cancel(reason) {
      console.log(reason);
    }
  });

  const [fileUpload, reader] = [
    upload(request)
    .catch(e => {
      reader.cancel();
      throw e
    })
    , uploadProgress.getReader()
  ];

  const processUploadRequest = ({value, done}) => {
    if (value || done) {
      console.log("upload complete, request.bodyUsed:", request.bodyUsed);
      // set `progress.value` to `progress.max` here 
      // if not awaiting server response
      // progress.value = progress.max;
      return reader.closed.then(() => fileUpload);
    }
    console.log("upload progress:", value);
    progress.value = +progress.value + 1;
    return reader.read().then(result => processUploadRequest(result));
  };

  reader.read().then(({value, done}) => processUploadRequest({value,done}))
  .then(response => response.text())
  .then(text => {
    console.log("response:", text);
    progress.value = progress.max;
    input.value = "";
  })
  .catch(err => console.log("upload error:", err));

}

上传获取进度指标?

脸赞 2025-02-07 01:11:31

我不知道它是否在这样做,因为时钟的代码需要额外的时间才能执行

,您的计时器的节奏取决于代码的运行时。因此,制作此脚本需要更多的计算机资源也将减慢计时器的速度,反之亦然。

“不要重新发明轮子。” - 编程谚语。

使用时间模块在计时器中获得更准确的时间。更具体地说, time.time()和格式化以使其可读。

编辑:

这是一个显示您问题的示例。

import time

time_zone = 0 # UTC
while True:
    current_time = time.time()

    ms = (current_time * 1000) % 1000
    s = (current_time % 60)
    m = (current_time % (60*60))//60
    h = (current_time % (60*60*24))//(60*60)

    print(f"h:{int(h+time_zone)} - m:{int(m)} - s:{int(s)} - ms:{int(ms)}")
    time.sleep(1)

从理论上讲,此脚本应每秒打印时间。但是,由于代码运行时,添加了几毫秒。
删除 time.sleep(1)应该使您非常接近实时时钟。

因此,如果您想要接近计时器或秒表附近的东西,则可以在当前时期时间( time.time.time())获得时间戳,以每秒的tick tick tick的每一个x amount更新您的计时器计数MS的效果,当再次按下按钮时,您将获得一个新的时间戳,然后将其与第一个时间戳进行比较,并减去以获取每个按钮之间的确切时间。


我为您制作了一个模板,以解决:

import tkinter as tk
import time

time_zone = 0  # UTC

# Convert epoch time to readable format
def readable_epoch(time_epoch):
    ms = (time_epoch * 1000) % 1000
    s = (time_epoch % 60)
    m = (time_epoch % (60*60))//60
    h = (time_epoch % (60*60*24))//(60*60)
    return (h, m, s, ms)


class App(tk.Tk):
    def __init__(self):
        super().__init__()
        self.time1 = 0
        self.time2 = 0

        self.geometry("300x200")
        self.button1 = tk.Button(
            self, text="Click me first", command=self.button1, bg="teal")
        self.button1.pack(ipadx=10, ipady=10, expand=True, side=tk.LEFT)
        self.button2 = tk.Button(
            self, text="Click me second", command=self.button2, bg="pink")
        self.button2.pack(ipadx=10, ipady=10, expand=True, side=tk.LEFT)

    # Get current time
    def button1(self):
        self.current_time = time.time()
        h, m, s, ms = readable_epoch(self.current_time)
        print(f"h:{int(h+time_zone)}, m:{int(m)}, s:{int(s)}, ms:{int(ms)}")

    # Elapsed time since button 1 was pressed:
    def button2(self):
        new_time = time.time()
        new_time = new_time - self.current_time
        h, m, s, ms = readable_epoch(new_time)
        print(f"h:{int(h+time_zone)}, m:{int(m)}, s:{int(s)}, ms:{int(ms)}")


if __name__ == "__main__":
    app = App()
    app.mainloop()

I don't know if it's doing that because the code of the clock takes some extra time to execute

In this case you are right, the tempo of your timer is dependent on the runtime of your code. So making this script require more resources from your computer would also slow down the timer, and vice versa.

"Don't reinvent the wheel." - Programming proverb.

Use the time module to get a more accurate time in your timer. More specifically time.time() and format it to make it readable.

Edit:

Here's an example to show your problem.

import time

time_zone = 0 # UTC
while True:
    current_time = time.time()

    ms = (current_time * 1000) % 1000
    s = (current_time % 60)
    m = (current_time % (60*60))//60
    h = (current_time % (60*60*24))//(60*60)

    print(f"h:{int(h+time_zone)} - m:{int(m)} - s:{int(s)} - ms:{int(ms)}")
    time.sleep(1)

In theory this script should print the time exactly every second. However a couple of milliseconds are added due to the code runtime.
Removing time.sleep(1) should get you pretty close to a live clock.

So if you want something close to a timer or stopwatch, you get a timestamp for the current epoch time (time.time()) update your timer about every x-amount of ticks per second to get the effect of counting ms, when the button is pressed again you get a new timestamp, and then you compare it to the first timestamp and subtract to get the exact time between each button press.


I've made a template for you to tinker a bit around with:

import tkinter as tk
import time

time_zone = 0  # UTC

# Convert epoch time to readable format
def readable_epoch(time_epoch):
    ms = (time_epoch * 1000) % 1000
    s = (time_epoch % 60)
    m = (time_epoch % (60*60))//60
    h = (time_epoch % (60*60*24))//(60*60)
    return (h, m, s, ms)


class App(tk.Tk):
    def __init__(self):
        super().__init__()
        self.time1 = 0
        self.time2 = 0

        self.geometry("300x200")
        self.button1 = tk.Button(
            self, text="Click me first", command=self.button1, bg="teal")
        self.button1.pack(ipadx=10, ipady=10, expand=True, side=tk.LEFT)
        self.button2 = tk.Button(
            self, text="Click me second", command=self.button2, bg="pink")
        self.button2.pack(ipadx=10, ipady=10, expand=True, side=tk.LEFT)

    # Get current time
    def button1(self):
        self.current_time = time.time()
        h, m, s, ms = readable_epoch(self.current_time)
        print(f"h:{int(h+time_zone)}, m:{int(m)}, s:{int(s)}, ms:{int(ms)}")

    # Elapsed time since button 1 was pressed:
    def button2(self):
        new_time = time.time()
        new_time = new_time - self.current_time
        h, m, s, ms = readable_epoch(new_time)
        print(f"h:{int(h+time_zone)}, m:{int(m)}, s:{int(s)}, ms:{int(ms)}")


if __name__ == "__main__":
    app = App()
    app.mainloop()

Python tkinter:倒数计时器不准确

脸赞 2025-02-06 05:25:37

您可以添加自定义样式表并将其加载到站点的标题(load_static)中。在该样式表中,您可以覆盖Bootstrap的颜色。这些是Bootstrap 5.0.0-Beta2的颜色:

:root {
  --bs-blue: #0d6efd;
  --bs-indigo: #6610f2;
  --bs-purple: #6f42c1;
  --bs-pink: #d63384;
  --bs-red: #dc3545;
  --bs-orange: #fd7e14;
  --bs-yellow: #ffc107;
  --bs-green: #198754;
  --bs-teal: #20c997;
  --bs-cyan: #0dcaf0;
  --bs-white: #fff;
  --bs-gray: #6c757d;
  --bs-gray-dark: #343a40;
  --bs-primary: #0d6efd;
  --bs-secondary: #6c757d;
  --bs-success: #198754;
  --bs-info: #0dcaf0;
  --bs-warning: #ffc107;
  --bs-danger: #dc3545;
  --bs-light: #f8f9fa;
  --bs-dark: #212529;
  --bs-font-sans-serif: system-ui, -apple-system, "Segoe UI", Roboto, "Helvetica Neue", Arial, "Noto Sans", "Liberation Sans", sans-serif, "Apple Color Emoji", "Segoe UI Emoji", "Segoe UI Symbol", "Noto Color Emoji";
  --bs-font-monospace: SFMono-Regular, Menlo, Monaco, Consolas, "Liberation Mono", "Courier New", monospace;
  --bs-gradient: linear-gradient(180deg, rgba(255, 255, 255, 0.15), rgba(255, 255, 255, 0));
}

如果您的CSS在Django-bootstrap5中加载后,则应使用(未测试)

You can add a custom stylesheet and load it in the header of the site (load_static). In that stylesheet you can overwrite the colors of Bootstrap. These are the colors from Bootstrap 5.0.0-beta2:

:root {
  --bs-blue: #0d6efd;
  --bs-indigo: #6610f2;
  --bs-purple: #6f42c1;
  --bs-pink: #d63384;
  --bs-red: #dc3545;
  --bs-orange: #fd7e14;
  --bs-yellow: #ffc107;
  --bs-green: #198754;
  --bs-teal: #20c997;
  --bs-cyan: #0dcaf0;
  --bs-white: #fff;
  --bs-gray: #6c757d;
  --bs-gray-dark: #343a40;
  --bs-primary: #0d6efd;
  --bs-secondary: #6c757d;
  --bs-success: #198754;
  --bs-info: #0dcaf0;
  --bs-warning: #ffc107;
  --bs-danger: #dc3545;
  --bs-light: #f8f9fa;
  --bs-dark: #212529;
  --bs-font-sans-serif: system-ui, -apple-system, "Segoe UI", Roboto, "Helvetica Neue", Arial, "Noto Sans", "Liberation Sans", sans-serif, "Apple Color Emoji", "Segoe UI Emoji", "Segoe UI Symbol", "Noto Color Emoji";
  --bs-font-monospace: SFMono-Regular, Menlo, Monaco, Consolas, "Liberation Mono", "Courier New", monospace;
  --bs-gradient: linear-gradient(180deg, rgba(255, 255, 255, 0.15), rgba(255, 255, 255, 0));
}

If your css is loaded after the one from django-bootstrap5 it should work (not tested)

Django Bootstrap 5-自定义样式

脸赞 2025-02-06 01:51:08

您可能会缺少HandlerEdirectPromise ...
重定向完成后,承诺应抓住帐户...如果不尝试另一个aquiresilenttoken,以在下面的承诺中抓住它。

instance.handleRedirectPromise().then(resp => {
    if (resp && resp.account) instance.setActiveAccount(resp.account);
});

you are probably missing the handleRedirectPromise...
once the redirect is done the promise should catch the account... if not try another aquireSilentToken to catch it in the promise below.

instance.handleRedirectPromise().then(resp => {
    if (resp && resp.account) instance.setActiveAccount(resp.account);
});

用MSAL反应包装器的React-Apollo MSAL-BROWSER无法获得帐户

脸赞 2025-02-06 00:47:37

今天,当我的Mac上的Visual Studio代码1.89.1调试到运行.NET Core应用程序的Docker容器时,我们遇到了这个问题。我们还遇到了错误“致命错误:无法使用错误80131534初始化调度器”。

我们发现使用VSDBG版本17.0.10712.2为我们工作。我们通过更改将VSDBG下载的Dockerfile的线路来完成此操作:

curl -sSL https://aka.ms/getvsdbgsh | bash /dev/stdin -v 17.0.10712.2 -l /remote_debugger

如果它可以帮助其他任何人,我们使用.NET SDK 6.0.423,然后使用Microsoft.aspnetcore.App.App 6.0.31和Microsoft.netcore.netcore.app 6.0。 31。我们的容器正在运行Debian.11-ARM64。

We had this problem today when trying to debug from Visual Studio Code 1.89.1 on my Mac into a docker container running a .NET Core application. We were also getting the error "FATAL ERROR: Failed to initialize dispatcher with error 80131534".

We found that using version 17.0.10712.2 of vsdbg worked for us. We did this by altering the line of our Dockerfile that downloads vsdbg to this:

curl -sSL https://aka.ms/getvsdbgsh | bash /dev/stdin -v 17.0.10712.2 -l /remote_debugger

In case it helps anyone else, we are using .NET SDK 6.0.423, with Microsoft.AspNetCore.App 6.0.31 and Microsoft.NETCore.App 6.0.31. Our container is running debian.11-arm64.

使用&#x27; Docker .Net Core attect&#x27;不再工作了

脸赞 2025-02-05 23:21:47

使用 std :: array&lt; std :: variant&lt; type1,type2,type3,...&gt;,100000&gt;缓存;

Use std::array<std::variant<Type1, Type2, Type3, ...>, 100000> cache;

C&#x2B;&#x2B; 20预先分配的数组存储多种类型

脸赞 2025-02-05 18:00:50

您有一个列表的列表,并且想在嵌套列表中使用流中执行一些计算。 FlatMap()方法可以解决您的问题。

而且您想返回布尔值,如果流中有任何元素匹配给定条件,则AnyMatch()可能是一个不错的选择。

partn.getcontrolUnit().stream()
.flatMap(cu -> cu.getcontrolUnitVersion().stream())
.anyMatch(r -> r.getWorkflowState().getKey().equals("locked") || r.getWorkflowState().getKey().equals("released"));

You have a list of a list, and you want to perform some computation in the nested list with the stream. The flatMap() method could resolve your problem.

And you want to return a boolean, if there are any elements in stream that match a given condition, anyMatch() might be a good choice.

partn.getcontrolUnit().stream()
.flatMap(cu -> cu.getcontrolUnitVersion().stream())
.anyMatch(r -> r.getWorkflowState().getKey().equals("locked") || r.getWorkflowState().getKey().equals("released"));

如何在返回带有多种条件的布尔值的方法上使用流

脸赞 2025-02-05 16:29:12

关于:

我不想获得可用于从用户个人帐户中获取数据的访问令牌,只有与组织关联的帐户。

所有Esign数据访问都需要与具有访问数据的用户关联的访问令牌。

如果您想对帐户中的任何数据进行程序化访问,则需要与该帐户的系统管理员关联的访问令牌。 (帐户系统管理员可能无法访问一些数据,我不是这个问题的专家。)

Re:

I don't want to get an access token that can be used to fetch data from the user's personal account, just the account associated with the organisation.

All eSign data access requires an access token that is associated with a user that has access to the data.

If you want programmatic access to any data in an account then you need an access token associated with a system administrator for that account. (There may be data that is not accessible to an account system admin, I'm not expert on that issue.)

是否有任何方法可以使用任何验证流量获得对用户数据(特定于组织的特定)的有限访问?

脸赞 2025-02-05 12:40:20

您需要 compare.push(secondword [j])而不是 compare.push(secondword [i])

function mutation(arr) {
  var compare = [];
  var firstword = arr[0].toLowerCase();
  var secondword = arr[1].toLowerCase();
  var j = 0;
  for (let i = 0; i < firstword.length; i++) {
    if (firstword[i] === secondword[j]) {
      compare.push(secondword[j]); // Correction here
      i = -1;
      j++;
    }
  }
  let result = compare.join("");
  if (result.length === secondword.length) {
    return true;
  } else {
    return false;
  }
}

console.log(mutation(["Noel", "Ole"]));

另外,您可以考虑使用 array.prototype。每个

const mutation = ([first, sec]) => {
  const lowerCaseFirst = first.toLowerCase();
  const lowerCaseSec = sec.toLowerCase();
  return Array.from(lowerCaseSec).every((ch) => lowerCaseFirst.includes(ch));
};

console.log(mutation(["Noel", "Ole"]));

如果字符串很小,则工作正常,但是如果它们很大,则应考虑使用 set

const mutation = ([first, sec]) => {
  const firstSet = new Set(first.toLowerCase());
  const lowerCaseSec = sec.toLowerCase();
  return Array.from(lowerCaseSec).every((ch) => firstSet.has(ch));
};

console.log(mutation(["Noel", "Ole"]));

You need to compare.push(secondword[j]) instead of compare.push(secondword[i])

function mutation(arr) {
  var compare = [];
  var firstword = arr[0].toLowerCase();
  var secondword = arr[1].toLowerCase();
  var j = 0;
  for (let i = 0; i < firstword.length; i++) {
    if (firstword[i] === secondword[j]) {
      compare.push(secondword[j]); // Correction here
      i = -1;
      j++;
    }
  }
  let result = compare.join("");
  if (result.length === secondword.length) {
    return true;
  } else {
    return false;
  }
}

console.log(mutation(["Noel", "Ole"]));

Also, you can consider using Array.prototype.every.

const mutation = ([first, sec]) => {
  const lowerCaseFirst = first.toLowerCase();
  const lowerCaseSec = sec.toLowerCase();
  return Array.from(lowerCaseSec).every((ch) => lowerCaseFirst.includes(ch));
};

console.log(mutation(["Noel", "Ole"]));

If the strings are small then String.prototype.includes works fine but if they are large then you should consider using a Set.

const mutation = ([first, sec]) => {
  const firstSet = new Set(first.toLowerCase());
  const lowerCaseSec = sec.toLowerCase();
  return Array.from(lowerCaseSec).every((ch) => firstSet.has(ch));
};

console.log(mutation(["Noel", "Ole"]));

返回true如果数组的第一个元素中的字符串包含第二个元素中字符串的所有字母

脸赞 2025-02-05 08:50:31

您几乎在那里:

  • 不确定您的目标是
  • 在外循环内部定义一个新变量 时,您需要 calc [] ;
  • 在内部循环中,而不是 calc.push ,使用 row.push
  • 在内部循环之后,您有一个完整的行,您可以使用Array 来输出该行。
  • ​在我看来,没有必要。
const n = 3;
//const calc = []; //may not be necessary

for (let i = 1; i <= n; i++) {
    const row = [];
    for (let j = 1; j <= n; j++) {
        row.push(i + " x " + j + " = " + (i * j));
    }
    console.log( row.join(' | ') );
    //calc.push(row);//not sure if you still need this
}
/* OUTPUT
1 x 1 = 1 | 1 x 2 = 2 | 1 x 3 = 3
2 x 1 = 2 | 2 x 2 = 4 | 2 x 3 = 6
3 x 1 = 3 | 3 x 2 = 6 | 3 x 3 = 9
*/

You're almost there:

  • Not sure that you need the array calc if your aim is to print the table
  • Define a new variable row inside the outer loop as an empty array, [];
  • In the inner loop, instead of calc.push, use row.push
  • After the inner loop, you have a complete row, which you can output using the array .join() method
  • If you need to, then add the row to calc with calc.push(row); not necessary in my view.

const n = 3;
//const calc = []; //may not be necessary

for (let i = 1; i <= n; i++) {
    const row = [];
    for (let j = 1; j <= n; j++) {
        row.push(i + " x " + j + " = " + (i * j));
    }
    console.log( row.join(' | ') );
    //calc.push(row);//not sure if you still need this
}
/* OUTPUT
1 x 1 = 1 | 1 x 2 = 2 | 1 x 3 = 3
2 x 1 = 2 | 2 x 2 = 4 | 2 x 3 = 6
3 x 1 = 3 | 3 x 2 = 6 | 3 x 3 = 9
*/

2D数组中的JavaScript乘法表

脸赞 2025-02-05 06:58:27

该工具 npm-lockfile 使用NPM的内部并在Monoreorepos内部进行工作。

不幸的是,他们正在使用@npm/Arborist 的旧版本,因此它不能完全正常工作,您可以使用我的补丁

The tool npm-lockfile uses npm's internals and works inside of monorepos.

Unfortunately, they're using an old version of @npm/arborist, so it doesn't quite work properly, and you can use my patch for now

如何将NPM-ShrinkWrap与工作区使用?

脸赞 2025-02-04 19:36:33

我已经检查了一下,并且能够将所有三个或多个角色一起发送在一起,请使用操作方法像下面一样执行此操作。

Actions action = new Actions(driver);
ele = driver.findElement(By.cssSelector("yourselector"))

action.moveToElement(ele).click().build().perform();

action.moveToElement(ele).sendKeys('LAHORE').build().perform();

导入

import org.openqa.selenium.interactions.Actions

确保使用正确的定位器。此外,我与Java进行了测试,我相信您可以将其更改为Desire Language

I have checked and I am able to send all three or more character together, please use the action method to do it like below.

Actions action = new Actions(driver);
ele = driver.findElement(By.cssSelector("yourselector"))

action.moveToElement(ele).click().build().perform();

action.moveToElement(ele).sendKeys('LAHORE').build().perform();

Import

import org.openqa.selenium.interactions.Actions

Make sure to use the correct locator.Also, i tested with java, I am sure you can change it to desire language

我如何每次发送一个字符,而不是用C#中的硒中的一个单词发送

更多

推荐作者

櫻之舞

文章 0 评论 0

弥枳

文章 0 评论 0

m2429

文章 0 评论 0

野却迷人

文章 0 评论 0

我怀念的。

文章 0 评论 0

更多

友情链接

    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