发私信

生生不灭

文章 评论 浏览 31

生生不灭 2025-02-11 10:04:06

替换此代码段将完全解决问题。

public bool phase;
private void Update()
{
    if (!isMoving)
    {
        input.x = Input.GetAxisRaw("Horizontal");
        input.y = Input.GetAxisRaw("Vertical");
        
        if (input != Vector2.zero)
        {
            var targetPos = transform.position;

            if (phase) targetPos.x += input.x;
            else targetPos.y += input.y;

            StartCoroutine(Move(targetPos));

            phase = !phase;
        }
    }
}

Replacing this code snippet will solve the problem completely.

public bool phase;
private void Update()
{
    if (!isMoving)
    {
        input.x = Input.GetAxisRaw("Horizontal");
        input.y = Input.GetAxisRaw("Vertical");
        
        if (input != Vector2.zero)
        {
            var targetPos = transform.position;

            if (phase) targetPos.x += input.x;
            else targetPos.y += input.y;

            StartCoroutine(Move(targetPos));

            phase = !phase;
        }
    }
}

enter image description here

禁用Unity 2d的对角线运动（但两者之间波动）

老娘不死你永远是小三 2025-02-04 10:04:06 浏览 3 评论 0 收藏 0
回复 原文
生生不灭 2025-02-11 01:27:42

几分钟前，我遇到了类似的问题，但是我能够通过使用 - force 标志强制安装它来解决它，因此您可以通过：

npm install --force simple-react-lightbox

我相信问题是，您要安装的依赖关系/软件包是基于React V17.X构建的，并且您正在与React V18.X上的项目合作，

希望这能解决您的问题。

I faced similar problem some minutes ago, but I was able to solve it by forcefully installing it using the --force flag, so you'd install it by:

npm install --force simple-react-lightbox

What I believe the problem is, is that the dependency/package you want to install is built upon react v17.x and you're working with a project on react v18.x

I hope this solves your problem.

如何解决NPM错误！ ERESOLVE无法解析依赖树 - 简单的React Lightbox？

关于从前 2025-02-04 01:27:42 浏览 1 评论 0 收藏 0
回复 原文
生生不灭 2025-02-10 20:44:25

您可以尝试 errors ='coerce'

df["search_departure_date"] = pd.to_datetime(df['search_departure_date'], errors='coerce')

如果要过滤行，则可以使用 str.Match 

m = df['search_departure_date'].str.match('147(8|9)')

You can try errors='coerce'

df["search_departure_date"] = pd.to_datetime(df['search_departure_date'], errors='coerce')

If you want to filter out the rows, you can use str.match

m = df['search_departure_date'].str.match('147(8|9)')

固定OutofBoundSdateTime时过滤多行：超出范围纳米秒的时间戳多个值

睫毛上残留的泪 2025-02-03 20:44:25 浏览 2 评论 0 收藏 0
回复 原文
生生不灭 2025-02-10 18:53:04

通过遵循此

https://learn.microsoft.com/en-us/windows/win32/fileio/maximin-file-path-limitation?tabs = registry

摘要

  1. 集
    HKLM：\ System \ CurrentControlset \ Control \ Filesystem \ LongPathSeNabled
    到1
  2. 重新启动机器或无法正常工作

Disable Maximum Path Length Limitation by following this

https://learn.microsoft.com/en-us/windows/win32/fileio/maximum-file-path-limitation?tabs=registry

Summary

  1. Set
    HKLM:\SYSTEM\CurrentControlSet\Control\FileSystem\LongPathsEnabled
    to 1
  2. Restart the machine or it wont work

如何解决Xamarin.firebase.ios.Core -8.10.0.1 Nuget安装错误

夏花。依旧 2025-02-03 18:53:04 浏览 3 评论 0 收藏 0
回复 原文
生生不灭 2025-02-10 08:21:26

我将首先过滤它。例如：

x_list = [['422', '324', '733443'], ['342', '654', '674335'],
         ['953', '456', '828854'], ['345', '886', '446678'],
         ['3224', '533', '654333'], ['7545', '5567', '369990']]

x_list2 = [x for x in x_list if int(x[0]) > 400]
x_list2.sort(key=lambda x: int(x[2]))
x_list2 = x_list2[:3]

print(x_list2)
# [['7545', '5567', '369990'], ['3224', '533', '654333'], ['422', '324', '733443']]

或者您也可以单一顺序排序和过滤：

x_list = [['422', '324', '733443'], ['342', '654', '674335'],
         ['953', '456', '828854'], ['345', '886', '446678'],
         ['3224', '533', '654333'], ['7545', '5567', '369990']]

x_list2 = sorted((x for x in x_list if int(x[0]) > 400), key=lambda x: int(x[2]))[:3]

print(x_list2)
# [['7545', '5567', '369990'], ['3224', '533', '654333'], ['422', '324', '733443']]

I would first filter the list to only retain values where first element is greater than 400, and then sort the final result by ascending order based on last element, as you originally had it. For example:

x_list = [['422', '324', '733443'], ['342', '654', '674335'],
         ['953', '456', '828854'], ['345', '886', '446678'],
         ['3224', '533', '654333'], ['7545', '5567', '369990']]

x_list2 = [x for x in x_list if int(x[0]) > 400]
x_list2.sort(key=lambda x: int(x[2]))
x_list2 = x_list2[:3]

print(x_list2)
# [['7545', '5567', '369990'], ['3224', '533', '654333'], ['422', '324', '733443']]

Or you can also sort and filter in a single step as well:

x_list = [['422', '324', '733443'], ['342', '654', '674335'],
         ['953', '456', '828854'], ['345', '886', '446678'],
         ['3224', '533', '654333'], ['7545', '5567', '369990']]

x_list2 = sorted((x for x in x_list if int(x[0]) > 400), key=lambda x: int(x[2]))[:3]

print(x_list2)
# [['7545', '5567', '369990'], ['3224', '533', '654333'], ['422', '324', '733443']]

如何在Python中对列表进行排序和过滤？

总攻大人 2025-02-03 08:21:26 浏览 5 评论 0 收藏 0
回复 原文
生生不灭 2025-02-10 04:32:56

您需要动态SQL。您可以一起生成一个大的联合所有查询所有表格。

DECLARE @sql nvarchar(max) = (
    SELECT STRING_AGG(CAST('
SELECT
  ' + QUOTENAME(con.table_name, '''') + ' table_name,
  ' + QUOTENAME(con.column_name, '''') + ' column_name,
  ' + QUOTENAME(con.keyword, '''') + ' keyword,
  COUNT(*) cnt
FROM ' + QUOTENAME(s.name) + '.' + QUOTENAME(t.name) + ' t
WHERE t.' + QUOTENAME(c.name) + ' LIKE ' + QUOTENAME('%' + con.keyword + '%', '''')

       AS nvarchar(max)), '
UNION ALL')

    FROM YourConditions con
    JOIN sys.tables t ON t.table_name
    JOIN sys.schemas s ON s.schema_id = t.schema_id
    JOIN sys.columns c ON c.object_id = t.object_id
    WHERE t.name = con.table_name
      AND c.name = con.column_name  
);

PRINT @sql;    -- your friend

EXEC sp_executesql @sql;

如果每个表有许多列或关键字，则有更有效的方法可以做到这一点，但这应该使您开始。

You need dynamic SQL for this. You can generate a big UNION ALL query of all the tables together.

DECLARE @sql nvarchar(max) = (
    SELECT STRING_AGG(CAST('
SELECT
  ' + QUOTENAME(con.table_name, '''') + ' table_name,
  ' + QUOTENAME(con.column_name, '''') + ' column_name,
  ' + QUOTENAME(con.keyword, '''') + ' keyword,
  COUNT(*) cnt
FROM ' + QUOTENAME(s.name) + '.' + QUOTENAME(t.name) + ' t
WHERE t.' + QUOTENAME(c.name) + ' LIKE ' + QUOTENAME('%' + con.keyword + '%', '''')

       AS nvarchar(max)), '
UNION ALL')

    FROM YourConditions con
    JOIN sys.tables t ON t.table_name
    JOIN sys.schemas s ON s.schema_id = t.schema_id
    JOIN sys.columns c ON c.object_id = t.object_id
    WHERE t.name = con.table_name
      AND c.name = con.column_name  
);

PRINT @sql;    -- your friend

EXEC sp_executesql @sql;

There are more efficient ways to do this if there are many columns or keywords per table, but this should get you started.

如何使用循环进行类似或Ilike功能？

别低头，皇冠会掉 2025-02-03 04:32:56 浏览 5 评论 0 收藏 0
回复 原文
生生不灭 2025-02-10 02:30:39

使用背景计时器库。它解决了我的问题。

const timeoutId = BackgroundTimer.setTimeout(() => {
    // This will be executed once after 1 seconds
    // even when the application is the background
    console.log('tac');
}, 1000);

Use the Background Timer library. It solved my problem.

const timeoutId = BackgroundTimer.setTimeout(() => {
    // This will be executed once after 1 seconds
    // even when the application is the background
    console.log('tac');
}, 1000);

USESTATE SET方法不是立即反映更改

酒解孤独 2025-02-03 02:30:39 浏览 3 评论 0 收藏 0
回复 原文
生生不灭 2025-02-10 01:30:01

最后，我找到了解决方案：在 OpenAPI 3.x ，必须在参数的架构中指定默认值。换句话说，

    parameters:
      - name: page
        in: query
        default: 0
        description: Page number
        schema:
          type: integer

应该是

    parameters:
      - name: page
        in: query
        description: Page number
        schema:
          type: integer
          default: 0

Finally, I've found the solution: in OpenAPI 3.x, the default value must be specified within the parameter's schema. In other words, instead of

    parameters:
      - name: page
        in: query
        default: 0
        description: Page number
        schema:
          type: integer

it should be

    parameters:
      - name: page
        in: query
        description: Page number
        schema:
          type: integer
          default: 0

OpenAPI规格：查询参数的默认值被忽略

画尸师 2025-02-03 01:30:01 浏览 4 评论 0 收藏 0
回复 原文
生生不灭 2025-02-08 02:49:38

复制公式

快速修复

  • 只需在 m 之后删除 6 。
Selection.AutoFill Destination:=Range("H6:M" & LastRow), Type:=xlFillDefault

改进

  • 否选择
  • 否变量
  • no autofill 
With ActiveSheet ' improve!
    With .Range("H6:M" & .Cells(.Rows.Count, "A").End(xlUp).Row)
        .Formula = .Rows(1).Formula
    End With
End With

Copy Down Formulas

A Quick Fix

  • Just remove the 6 after the M.
Selection.AutoFill Destination:=Range("H6:M" & LastRow), Type:=xlFillDefault

An Improvement

  • No Select
  • No variable
  • No AutoFill
With ActiveSheet ' improve!
    With .Range("H6:M" & .Cells(.Rows.Count, "A").End(xlUp).Row)
        .Formula = .Rows(1).Formula
    End With
End With

将公式拖到最后一行（将第一个单元格作为A6）

爱她像谁 2025-02-01 02:49:38 浏览 3 评论 0 收藏 0
回复 原文
生生不灭 2025-02-07 17:41:04

首先：切勿解析 ls 命令的输出，而是使用 in 而不是。
由于查找基本上要通过所有子目录，因此您可能需要使用 -maxDepth switch switch，使用值1.

为了计算多个结果，您只需计算输出中的行数（如果您的输出显示为每行输出的一个输出，则是 find 命令的情况） 。使用 WC -L 命令来计算许多行。

因此，这取决于以下命令：

find ./ -maxdepth 1 -type f -name "k*" | wc -l

玩得开心！

To start with: never parse the output of the ls command, but use find instead.
As find basically goes through all subdirectories, you might need to limit that, using the -maxdepth switch, use value 1.
In order to count a number of results, you just count the number of lines in your output (in case your output is shown as one piece of output per line, which is the case of the find command). Counting a number of lines is done using the wc -l command.

So, this comes down to the following command:

find ./ -maxdepth 1 -type f -name "k*" | wc -l

Have fun!

bash脚本计算并打印出以特定字母开头的文件

追我者格杀勿论 2025-01-31 17:41:04 浏览 3 评论 0 收藏 0
回复 原文
生生不灭 2025-02-07 13:15:02

如果没有 break> break 语句在循环中，则执行 else 带有循环的块（for/while block之后）。

The else block with loops (after the for/while block) is executed if there was no exit with the break statement inside the loop.

很难理解Python中的代码

别挽留 2025-01-31 13:15:02 浏览 4 评论 0 收藏 0
回复 原文
生生不灭 2025-02-07 02:33:16 
<html>
    <head>
        <title>HTML Document</title>
        <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
    </head>

    <body>
        <div id="hover-id">
            Hello World
        </div>

        <script>
            jQuery(document).ready(function($){
                $(document).on('mouseover', '#hover-id', function(){
                    $(this).css('color','yellowgreen');
                });

                $(document).on('mouseout', '#hover-id', function(){
                    $(this).css('color','black');
                });
            });
        </script>
    </body>
</html>

<html>
    <head>
        <title>HTML Document</title>
        <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
    </head>

    <body>
        <div id="hover-id">
            Hello World
        </div>

        <script>
            jQuery(document).ready(function($){
                $(document).on('mouseover', '#hover-id', function(){
                    $(this).css('color','yellowgreen');
                });

                $(document).on('mouseout', '#hover-id', function(){
                    $(this).css('color','black');
                });
            });
        </script>
    </body>
</html>

事件绑定在动态创建的元素上？

℡Ms空城旧梦 2025-01-31 02:33:16 浏览 6 评论 0 收藏 0
回复 原文
生生不灭 2025-02-07 01:20:02

发生这种情况是因为您的消息是字符串，而不是JSX。

如果您想渲染Elments这样使用JSX：

setMessage(
  <>
    You don't have a membership for this training center.
    You can purchase 
    <Link href="/buy-membership"><a>here</a></Link>
    .
  <>
)

This happen because your message is a string and not a jsx.

If you want to render elments use jsx like this:

setMessage(
  <>
    You don't have a membership for this training center.
    You can purchase 
    <Link href="/buy-membership"><a>here</a></Link>
    .
  <>
)

如何在Usestate中呈现链接-NextJS

左岸枫 2025-01-31 01:20:02 浏览 1 评论 0 收藏 0
回复 原文
生生不灭 2025-02-05 18:35:58

您的尝试解决方案失败，因为列表 e 具有重复性，并且您使用的是 index 方法，

您可以通过两种方法来修复解决方案 -

列表，

for i in B:
    for j in [A, B, C, D, E]:
        G.append(j[B.index(i)])

选项1-使用无重复的 2-不要使用索引

H = []
for i in range(len(E)):
    for j in [A, B, C, D, E]:
        H.append(j[i])

输出

['asd', 1, 'ttt', 1, 0, 'qwe', 2, 'ttt', 2, 30, 'wer', 3, 'ttt', 5, 0]

Your attempted solution fails because the list E has repetitions and you are using the index method

There are two ways you can fix your solution -

Option 1 - use a list without repetitions

for i in B:
    for j in [A, B, C, D, E]:
        G.append(j[B.index(i)])

Option 2 - don't use index

H = []
for i in range(len(E)):
    for j in [A, B, C, D, E]:
        H.append(j[i])

output

['asd', 1, 'ttt', 1, 0, 'qwe', 2, 'ttt', 2, 30, 'wer', 3, 'ttt', 5, 0]

根据索引组合列表

仅一夜美梦 2025-01-29 18:35:58 浏览 4 评论 0 收藏 0
回复 原文
生生不灭 2025-02-05 08:39:08

从我可以收集到的问题上，这应该解决您的问题，

from time import sleep

# wait until seconds equal 30
while datetime.now().strftime("%H:%M:%S").split(":")[2] != "30":
    pass

for i in range(10):
    print(datetime.now().strftime("%H:%M:%S"))
    # sleep for 1 minute in every iteration
    sleep(60)

您需要修改它以适合您的事业，但这是一般的想法。

感谢@pranav在评论中的第一个时，循环可以优化如下

while datetime.now().second != 30:
    sleep(1)

From what I could gather, this should solve your problem

from time import sleep

# wait until seconds equal 30
while datetime.now().strftime("%H:%M:%S").split(":")[2] != "30":
    pass

for i in range(10):
    print(datetime.now().strftime("%H:%M:%S"))
    # sleep for 1 minute in every iteration
    sleep(60)

You will need to modify this to suit your cause, but this is the general idea.

Thanks to @Pranav in the comments the first while loop can be optimised as follows

while datetime.now().second != 30:
    sleep(1)

Python DateTime我需要特定的30秒执行循环

香草可樂 2025-01-29 08:39:08 浏览 3 评论 0 收藏 0
回复 原文
更多

热门标签

操作系统 程序设计 IT运维 Linux系统管理 JavaScript 服务器应用 solaris C/C++ PHP Shell BSD Vue.js aix Oracle Python HTML 系统管理 HTML5 CSS 前端
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

寻找一个思念的角度

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

文江博客
    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