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脸赞 2025-02-20 23:47:23

查看您遇到的主要问题,似乎不使用观察到,而是应该使用 state> stateBoct

由于SwiftUi可以随时构建或破坏视图,因此在一个内部创建一个 @观察> 是危险的。为了确保视图重新绘制时的一致结果,请使用@stateObject包装器,除非您将 @ obsedObject 作为依赖关系。

指南是,无论如何创建您的对象的任何视图都必须使用 @ stateBoct 通知Swiftui它是数据的所有者,并且可以维护它。所有其他视图都必须使用 @观察到向Swiftui表示他们要注意对对象的更改,但不能直接拥有它。

@ stateObject 和 @ obsedobject 共享许多特征,而SwiftUI则以不同的方式管理其生活周期。要确保当当前视图生成观察到的对象时,请使用状态对象属性包装器。每当您将观察到的对象作为依赖项注入 @ @ obsedObject

Reviewing the main issue you are getting, it seems that instead of using ObservedObject you should use StateObject.

Since SwiftUI could at any time construct or destroy a view, it is dangerous to create a @ObservedObject inside of one. To guarantee consistent results upon a view redraw, utilize the @StateObject wrapper unless you inject the @ObservedObject as a dependency.

The guideline is that whatever view creates your object first must use @StateObject to inform SwiftUI that it is the owner of the data and it is in charge of maintaining it. All other views must use @ObservedObject to indicate to SwiftUI that they want to keep an eye out for changes to the object but do not directly own it.

While @StateObject and @ObservedObject share many traits, SwiftUI manages their life cycles differently. To guarantee consistent outcomes when the current view generates the observed object, use the state object property wrapper. You can use the @ObservedObject whenever you inject an observed object as a dependency.

firebase observableObject在应用程序入睡后不更新

脸赞 2025-02-20 17:47:27

您应该以更大的力量按。

从源代码中看到:

class getureDetector扩展了ntecleselesswidget {

onforcepresspresspeak pourcepresspressegenturerecognizer inforcepresspresspresspresspeak


 if (onForcePressStart != null ||
        onForcePressPeak != null ||
        onForcePressUpdate != null ||
        onForcePressEnd != null) {
      gestures[ForcePressGestureRecognizer] = GestureRecognizerFactoryWithHandlers<ForcePressGestureRecognizer>(
        () => ForcePressGestureRecognizer(debugOwner: this),
        (ForcePressGestureRecognizer instance) {
          instance
            ..onStart = onForcePressStart
            ..onPeak = onForcePressPeak
            ..onUpdate = onForcePressUpdate
            ..onEnd = onForcePressEnd;
        },
      );
    }


geduretector 被调用,

即,当 Onpeak FoursPressPressUrerEcognizer

class pourspresspressurerecognizer延长了onsequence gesturerEcognizer {,您可以看到使用条件。

 if (onPeak != null && pressure > peakPressure &&
         (_state == _ForceState.started)) {
        _state = _ForceState.peaked;
        if (onPeak != null) {
          invokeCallback<void>('onPeak', () => onPeak!(ForcePressDetails(
            pressure: pressure,
            globalPosition: event.position,
            localPosition: event.localPosition,
          )));
        }
      }

关键条件是压力&GT; peakPressure

也来自类LowerPressPressurerEcognizer


  ForcePressGestureRecognizer({
    this.startPressure = 0.4,
    this.peakPressure = 0.85,

...


  /// The pressure of the press required to peak a force press.
  ///
  /// A value of 0.0 is no pressure, and 1.0 is maximum pressure. This value
  /// must be greater than [startPressure].
  final double peakPressure;

只需要尝试一下 更难的 即可。

You should press with more strength.

seen from the source code:

from class GestureDetector extends StatelessWidget {

onForcePressPeak is function of ForcePressGestureRecognizer


 if (onForcePressStart != null ||
        onForcePressPeak != null ||
        onForcePressUpdate != null ||
        onForcePressEnd != null) {
      gestures[ForcePressGestureRecognizer] = GestureRecognizerFactoryWithHandlers<ForcePressGestureRecognizer>(
        () => ForcePressGestureRecognizer(debugOwner: this),
        (ForcePressGestureRecognizer instance) {
          instance
            ..onStart = onForcePressStart
            ..onPeak = onForcePressPeak
            ..onUpdate = onForcePressUpdate
            ..onEnd = onForcePressEnd;
        },
      );
    }


then, when onForcePressPeak of GestureDetector gets called,

namely, when onPeak of ForcePressGestureRecognizer gets called,

from class ForcePressGestureRecognizer extends OneSequenceGestureRecognizer {, you can see the usage condition.

 if (onPeak != null && pressure > peakPressure &&
         (_state == _ForceState.started)) {
        _state = _ForceState.peaked;
        if (onPeak != null) {
          invokeCallback<void>('onPeak', () => onPeak!(ForcePressDetails(
            pressure: pressure,
            globalPosition: event.position,
            localPosition: event.localPosition,
          )));
        }
      }

the key condition is pressure > peakPressure

also from class ForcePressGestureRecognizer,


  ForcePressGestureRecognizer({
    this.startPressure = 0.4,
    this.peakPressure = 0.85,

...

and


  /// The pressure of the press required to peak a force press.
  ///
  /// A value of 0.0 is no pressure, and 1.0 is maximum pressure. This value
  /// must be greater than [startPressure].
  final double peakPressure;

you just need to try it harder.

``ementectector()的Flutter onforcepress()不是回调

脸赞 2025-02-20 12:31:09

您需要升级到yarn 3.2.0 或以上。

yarn set version stable 
# or 
yarn set version 3.2.0

并运行纱线安装以更新 .pnp.cjs
请参阅更改日志他们已经处理ESM和concomjs模块。

You need to upgrade to yarn 3.2.0 or above.

yarn set version stable 
# or 
yarn set version 3.2.0

And run yarn install to update the .pnp.cjs.
See more at the Change log that they have handle the ESM and commonjs modules.

纱3.1,vite 2.9,找不到包装vite

脸赞 2025-02-19 10:51:41

这是因为对象的变量实际上是对象的指针。这就是为什么当它们传递到函数时,指针属性的任何更改都会改变实际对象。鉴于,如果您简单地将一个值分配给类型对象的变量,则指针本身会更改,而不是对象。

因此, obj1 = obj2 将OBJ1指针设置为OBJ2指针。现在,如果 obj1.value =“ c” ,则对象OBJ1(现在是OBJ2)的属性将会改变,从而影响原始对象。

This is because variables of objects are actually pointers to the object. That's why when they are passed to a function, any changes of the pointer properties would change the actual object. Whereas, if you simply assign a value to a variable of type object, then the pointer itself is changed, not the object.

So obj1 = obj2 sets the obj1 pointer to be obj2 pointer. Now, if obj1.value = "c" then a property of the object obj1 (which is now obj2) will change, affecting the original object.

在全局对象值上的函数中的对象分配的影响

脸赞 2025-02-19 08:38:47

升级后,我获得了拒绝的许可。
执行'mkb'邮政安装步骤没有任何更改,因为我的用户已经在'docker'组中;我重试了两次,但没有成功。

搜索时间后,以下解决方案最终工作:

sudo chmod 666 /var/run/docker.sock

解决方案来自 olshansk

看起来升级未经足够的“ Docker”组重新创建了插座。

问题

这个硬性 chmod 打开安全孔,每个重新启动后,此错误一次又一次地开始,您必须每次重新执行上述命令。我想要一劳永逸的解决方案。为此,您有两个问题:

  • 1) systemd :仅使用所有者'root'和group创建套接字'root'。

    您可以使用此命令检查第一个问题:

    如果一切都很好,您应该看到' root/docker '不是' root/root/root '。

  • 2) 图形登录 的问题: linux-account-lyly-belongs-to-One Group

    您可以使用此命令检查第二个问题:

      

    如果一切都正确,则应在列表中看到 docker 组。
    如果不尝试命令

      sudo su $ user -c groups
     

    如果您看到 docker 组是因为错误。

解决方案

如果您设法获得图形登录的解决方法,这应该可以完成工作:

sudo chgrp docker /lib/systemd/system/docker.socket
sudo chmod g+w /lib/systemd/system/docker.socket

但是,如果您无法管理此错误,那么一个不好的解决方案可能就是这样:

sudo chgrp $USER /lib/systemd/system/docker.socket
sudo chmod g+w /lib/systemd/system/docker.socket

这项工作是因为您处于图形环境中,并且可能计算机上唯一的用户。
在两种情况下

After an upgrade I got the permission denied.
Doing the steps of 'mkb' post install steps don't have change anything because my user was already in the 'docker' group; I retry-it twice any way without success.

After an search hour this following solution finaly worked :

sudo chmod 666 /var/run/docker.sock

Solution came from Olshansk.

Look like the upgrade have recreate the socket without enough permission for the 'docker' group.

Problems

This hard chmod open security hole and after each reboot, this error start again and again and you have to re-execute the above command each time. I want a solution once and for all. For that you have two problems :

  • 1 ) Problem with SystemD : The socket will be create only with owner 'root' and group 'root'.

    You can check this first problem with this command :

    ls -l /lib/systemd/system/docker.socket
    

    If every this is good, you should see 'root/docker' not 'root/root'.

  • 2 ) Problem with graphical Login : https://superuser.com/questions/1348196/why-my-linux-account-only-belongs-to-one-group

    You can check this second problem with this command :

    groups
    

    If everything is correct you should see the docker group in the list.
    If not try the command

    sudo su $USER -c groups
    

    if you see then the docker group it is because of the bug.

Solutions

If you manage to to get a workaround for the graphical login, this should do the job :

sudo chgrp docker /lib/systemd/system/docker.socket
sudo chmod g+w /lib/systemd/system/docker.socket

But If you can't manage this bug, a not so bad solution could be this :

sudo chgrp $USER /lib/systemd/system/docker.socket
sudo chmod g+w /lib/systemd/system/docker.socket

This work because you are in a graphical environnement and probably the only user on your computer.
In both case you need a reboot (or sudo chmod 666 /var/run/docker.sock)

如何修复Docker:获得许可的问题

脸赞 2025-02-19 02:44:21

对于未来的开发人员,可能有同样的问题!

建议的解决方案是解决方法,因为Kotlin和Cocoapods框架之间存在差异:

  • 因为Kotlin插件框架默认情况下是动态的。
  • 对于CocoApods插件框架默认情况下是静态的。

因此,通过JetBrains的支持,解决方案是将此代码添加到共享模块中的gradle文件中。

cocoapods {
    ...
    framework {
        isStatic = false
        ...
    }
}

For future developers that may have same issue!

The suggested solution is a workaround, because there is a difference between Kotlin and Cocoapods frameworks:

  • For Kotlin plugin frameworks are DYNAMIC by default.
  • For Cocoapods plugin frameworks are STATIC by default.

So the solution is to add this code into build.gradle file in shared module, thanks to JetBrains support:

cocoapods {
    ...
    framework {
        isStatic = false
        ...
    }
}

kmm&amp; KTOR- iOS项目带有框架库的iOS项目在两个目标中导入共享POD时崩溃

脸赞 2025-02-19 02:04:37

尝试使用Linq

string[] k = { a, b, c, d, e, f };
var se = k.Where((a) => int.TryParse(a, out int l))
          .Select(x => int.Parse(x))
          .Aggregate((a, b) => a + b);
System.Console.WriteLine(se);

tried using LINQ

string[] k = { a, b, c, d, e, f };
var se = k.Where((a) => int.TryParse(a, out int l))
          .Select(x => int.Parse(x))
          .Aggregate((a, b) => a + b);
System.Console.WriteLine(se);

简单的方法避免在C#中添加字符串

脸赞 2025-02-18 19:36:21

据我了解,此代码将保留重复名称的顺序(这是概念证明的原始版本)

 merged = []
 visited = []
  
  for (let i=0; i<names.length; i++) {
    for (let j=0; j<sections.length; j++) {
      if ((names[i].name === sections[j].name) && !visited.find(e => e === sections[j].id) ) {
        const obj = {...names[i]}
        obj['order'] = sections[j].order
        obj['id'] = sections[j].id
        merged.push(obj)
        visited.push(sections[j].id)
        //continue
        break
      }
    }
  }

console.log(merged)

输出:

[
  { name: 'B', order: 1, id: 3638 },
  { name: 'C', order: 2, id: 3658 },
  { name: 'D', order: 3, id: 3659 },
  { name: 'B', order: 5, id: 3661 },
  { name: 'A', order: 4, id: 3636 }
]

As far as I understood, this code will preserve the order for duplicate names (it's a raw version for a proof of concept)

 merged = []
 visited = []
  
  for (let i=0; i<names.length; i++) {
    for (let j=0; j<sections.length; j++) {
      if ((names[i].name === sections[j].name) && !visited.find(e => e === sections[j].id) ) {
        const obj = {...names[i]}
        obj['order'] = sections[j].order
        obj['id'] = sections[j].id
        merged.push(obj)
        visited.push(sections[j].id)
        //continue
        break
      }
    }
  }

console.log(merged)

output:

[
  { name: 'B', order: 1, id: 3638 },
  { name: 'C', order: 2, id: 3658 },
  { name: 'D', order: 3, id: 3659 },
  { name: 'B', order: 5, id: 3661 },
  { name: 'A', order: 4, id: 3636 }
]

JavaScript合并了2个阵列的一些属性

脸赞 2025-02-18 18:42:34

经过几次审判和错误,我制定了以下言论,以照顾上述所有所需标准。在此处添加答案,为将来偶然发现的任何人。

/^(?=.{1,254}$)(?=.{1,64}@)[a-zA-Z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-zA-Z0-9!#$%&'*+/=?^_`{|}~-]+)*@[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?(?:\.[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?)+$/

After a few trial and errors I formulated the following regexp which takes care of all the above required criteria. Adding the answer here for anyone that stumbles across this in the future.

/^(?=.{1,254}$)(?=.{1,64}@)[a-zA-Z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-zA-Z0-9!#$%&'*+/=?^_`{|}~-]+)*@[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?(?:\.[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?)+$/

电子邮件验证正则态度以在主机名中包含期限

脸赞 2025-02-18 13:18:10

我会稍微调整Tahmid的出色答案。为了允许空行(无子弹)在you functions.php 文件中使用此信息:

/**
 * create a ACF text field with bullets
 * Empty lines stay empty
 * @param string $field_name Name of the field to bullet
 */
function acf_bullets(string $field_name): void {
    $format_acf_bullet = function(
        string $value,
        int $post_id,
        array $form ):string {
        if( empty($value) ) {
            return '';
        }
        $lines = explode( "\n", $value );
        $result = "<ul class=\"theme-ul\">\n";
        foreach( $lines as $line) {
            if( strlen($line)<=1 ) { // empty line, start a new list
                $result .= "</ul><p/><ul class=\"theme-ul\">\n";
            } else {
                $result .= "<li>".$line."</li>\n";
            }
        }
        $result .= "</ul>\n";
        return $result;
    };
    add_filter("acf/format_value/name=$field_name", $format_acf_bullet, 10, 3);
}

使用 acf_bullets调用此('your-fieldname-here'here');

dixplisation

  • 为特定字段添加过滤器调用 $ format_acf_bulltets
  • 跳过完整的空截面
  • 行创建一个阵列,
  • 如果
  • 为所有
  • 行为空启动新&lt; ul&gt; 列表
  • 如果不包装&lt; li&gt;&lt;/li&gt;
  • 关闭&lt; ul&gt;

完整列表内容应用于字符串中变量 $ result 并在最后返回。练习使班级名称 theme-ul acf_bullets 的参数

I'd slightly adjust Tahmid's excellent answer. In order to allow for empty lines (without a bullet) use this in you functions.php file:

/**
 * create a ACF text field with bullets
 * Empty lines stay empty
 * @param string $field_name Name of the field to bullet
 */
function acf_bullets(string $field_name): void {
    $format_acf_bullet = function(
        string $value,
        int $post_id,
        array $form ):string {
        if( empty($value) ) {
            return '';
        }
        $lines = explode( "\n", $value );
        $result = "<ul class=\"theme-ul\">\n";
        foreach( $lines as $line) {
            if( strlen($line)<=1 ) { // empty line, start a new list
                $result .= "</ul><p/><ul class=\"theme-ul\">\n";
            } else {
                $result .= "<li>".$line."</li>\n";
            }
        }
        $result .= "</ul>\n";
        return $result;
    };
    add_filter("acf/format_value/name=$field_name", $format_acf_bullet, 10, 3);
}

Call this with acf_bullets('your-fieldname-here');

Explanation

  • Add a filter calling $format_acf_bulltets for the specific field that:
  • Skips complete empty sections
  • create an array for all the lines
  • open a new list
  • for all entries
  • if the line is empty start a new <ul> list
  • if not wrap the line in <li></li>
  • close the <ul>

The complete list content is appended in string variable $result and returned at the end. As exercise make the class name theme-ul a parameter for acf_bullets

高级自定义字段:显示为子弹点

脸赞 2025-02-18 10:59:06

有两种类型的全局负载平衡器:

  • 全局HTTP(S)负载平衡器-L7
  • 全局TCP/SSL代理负载均衡器-L4

两个负载平衡器都使用GFE代理将客户端的流量代理到后端应用程序。 为Google APIS流量共享,以及GCLB流量以及GCLB流量。 So in the port scan, it would establish and respond to TCP connection.但是,如果负载平衡器的VIP:端口号与目标IP:端口不匹配,则流量将不会转发到其后端,并且该应用程序不会面临任何安全风险。

There are two types of global Load balancers:

  • Global HTTP(S) Load balancer - L7
  • Global TCP/SSL proxy Load Balancer - L4

Both load balancers use GFE Proxies to proxy client's traffic to the backend application. GFE proxies are distributed around the world and are shared for Google APIs traffic as well as GCLB traffic. So in the port scan, it would establish and respond to TCP connection. However, if the Load balancer's VIP:Port number does not match with the Destination IP:Port, the traffic will not be forwarded to their backend and the application does not face any security risk.

GCP HTTP负载平衡器打开许多端口

脸赞 2025-02-18 09:44:28

如果有一个值重复数组中的最后一个值的值,则您的代码将无法正常工作,因为 $ value === end($ arrelement)对重复是正确的。这将使您在查询中添加额外的

不要使用循环,请使用 umpode()。那么,您无需特别治疗最后的迭代。

$columns = implode(', ', array_keys($arrElement);
$values = implode(', ', array_map(function($col) { return ':' . $col; }, array_keys($arrElement)));

$this->setQuery("$query ($columns) VALUES ($values)");

Your code won't work if there's a value that duplicates the last value in the array, because $value === end($arrElement) will be true for the duplicate. That will cause you to add an extra VALUES to the query.

Don't use loops, use implode(). Then you don't need to treat the last iteration specially.

$columns = implode(', ', array_keys($arrElement);
$values = implode(', ', array_map(function($col) { return ':' . $col; }, array_keys($arrElement)));

$this->setQuery("$query ($columns) VALUES ($values)");

PHP -SQL语句错误

脸赞 2025-02-18 07:55:43

首先,当前工作草案中的规则在 [lex.name] p3 << /a>:

此外,某些标识符以令牌预处理 - token-token 保留为C ++实现,否则不得使用;无需诊断。

  • 每个包含双重下划线的标识符 __ 或以下划线开头,然后将大写字母保留给实现。
  • 以下划线开头的每个标识符都保留在实现中,以用作全局名称空间中的名称。

此外,标准库保留在名称空间std 和一些僵尸名称中定义的所有名称;请参阅 [Reserved.names.general]

Posix呢?

正如公认的答案所指出的那样,实现的其他部分,例如POSIX标准,它限制了您可以使用的标识符。

标题部分中描述的每个标识符都保留在同一名称空间中的文件范围的标识符中,如果包括标题。

任何标头 [储备] 后缀 _t

v2_chap02.html“ rel =“ nofollow noreferrer”> posix 2008标准,2.2.2

在C ++中,几乎所有与POSIX相关的问题都可以通过命名空间来避免。
这也是为什么C ++标准可以添加诸如 std :: enable_if_t 之类的符号,而不会破坏POSIX的兼容性。

可视化

int x;      // OK
int x_;     // OK
int _x;     // RESERVED
int x__;    // RESERVED (OK in C)
int __x;    // RESERVED
int _X;     // RESERVED
int assert; // RESERVED (macro name)
int x_t;    // RESERVED (only by POSIX)

namespace {
int y;      // OK
int y_;     // OK
int _y;     // OK
int y__;    // RESERVED (OK in C, ignoring namespaces)
int __y;    // RESERVED
int _Y;     // RESERVED
int assert; // RESERVED (macro name)
int y_t;    // OK
}

y 的上述规则适用于命名和未命名的名称空间。
无论哪种方式,在以下名称空间中,全局名称空间的规则
不再应用(请参阅 [namespace.unamed] )。

y 的上述规则还适用于类,功能等中的标识符。除了全球范围外。

即使断言在此处不像功能风格的宏一样使用,但该名称已保留。这也是为什么到目前为止取得了一些成功。

推荐的练习

要安全,始终避免双重下降,并始终避免使用领先的下划线。
在某些情况下,后者还可以,但是很难记住这些规则,而且要安全比后悔要好。

_ 本身呢?

有些人使用 _ 表示未使用某些变量或功能参数。但是,您可以避免以下内容:

void foo(T _) { /* ... */ }
// replace with:
void foo(T) { /* ... */ }

std::scoped_lock _{mutex};
// replace with:
std::scoped_lock lock{mutex};

您也可以施放参数 p to void 喜欢(void)p ,如果这是关于沉默警告关于 p 未使用,您需要C兼容性。请参阅为什么为什么将未使用的返回值投入到void?

Firstly, the rules in current working draft are laid out in [lex.name] p3:

In addition, some identifiers appearing as a token or preprocessing-token are reserved for use by C++ implementations and shall not be used otherwise; no diagnostic is required.

  • Each identifier that contains a double underscore __ or begins with an underscore followed by an uppercase letter is reserved to the implementation for any use.
  • Each identifier that begins with an underscore is reserved to the implementation for use as a name in the global namespace.

Furthermore, the standard library reserves all names defined in namespace std and some zombie names; see [reserved.names.general].

What about POSIX?

As the accepted answer has pointed out, there may be other parts of the implementation, like the POSIX standard, which limit the identifiers you can use.

Each identifier with file scope described in the header section is reserved for use as an identifier with file scope in the same name space if the header is included.

ANY Header [reserves] Suffix _t

- POSIX 2008 Standard, 2.2.2

In C++, almost all problems associated with POSIX can be avoided through namespaces.
This is also why the C++ standard can add tons of symbols like std::enable_if_t without breaking POSIX compatibility.

Visualization

int x;      // OK
int x_;     // OK
int _x;     // RESERVED
int x__;    // RESERVED (OK in C)
int __x;    // RESERVED
int _X;     // RESERVED
int assert; // RESERVED (macro name)
int x_t;    // RESERVED (only by POSIX)

namespace {
int y;      // OK
int y_;     // OK
int _y;     // OK
int y__;    // RESERVED (OK in C, ignoring namespaces)
int __y;    // RESERVED
int _Y;     // RESERVED
int assert; // RESERVED (macro name)
int y_t;    // OK
}

The above rules for y apply to both named and unnamed namespaces.
Either way, in the following namespace, the rules of the global namespace
no longer apply (see [namespace.unnamed]).

The above rules for y also apply to identifiers in classes, functions, etc.; anything but global scope.

Even though assert isn't used like a function-style macro here, the name is reserved. This is also why proposal P2884 contemplates making it a keyword in C++26, with some success so far.

Recommended Practice

To be safe, always avoid double underscores, and always avoid nams with leading underscores.
The latter are okay in some cases, but it's difficult to memorize these rules, and it's better to be safe than sorry.

What about _ in itself?

Some people use _ to indicate that some variable or function parameter isn't used. However, you can avoid this with:

void foo(T _) { /* ... */ }
// replace with:
void foo(T) { /* ... */ }

std::scoped_lock _{mutex};
// replace with:
std::scoped_lock lock{mutex};

You can also cast a parameter p to void like (void)p, if this is about silencing warnings about p being unused, and you need C compatibility. See Why cast unused return values to void?.

在C&#x2b;&#x2B中使用下划线的规则是什么。标识符?

脸赞 2025-02-18 02:43:13

是的,可以代表用户同意。要实现相同的目标,您必须拥有全局管理员特权管理员角色。

转到Azure Portal -&gt; Azure Active Directory-&GT;企业应用程序 - &GT;同意和权限 - &GT;用户同意设置

输入映像”

“在此处

“在此处输入映像”

参考:

授予授予范围范围范围的管理员同意 - Microsoft Entra | Microsoft Docs

Azure Ad应用需要管理批准错误:应用程序需要许可以访问组织中只能授予的组织中的资源 - 堆栈溢出

Yes, it is possible to consent on behalf of the users. To achieve the same you must have Global Administrator or Privileged Administrator role.

Go to Azure Portal -> Azure Active Directory -> Enterprise applications -> Consent and permissions - > User consent settings

enter image description here

Check if the below option is set to No in Enterprise Application Users settings:

enter image description here

References:

Grant tenant-wide admin consent to an application - Microsoft Entra | Microsoft Docs

Azure AD app Need admin approval error: App needs permission to access resources in your organization that only an admin can grant - Stack Overflow

用户的管理员同意

脸赞 2025-02-18 01:57:56

如果您负责激活Virtualenv,请指定您的IDE+吗?

无论如何,这里是确保python模块预测安装在虚拟环境中的步骤:

  1. 中创建虚拟env
  2. python3 -m venv env_name :这将在您的当前路径 source> source incouse env/bin 在Windows上仅在CD中激活您的Env Name
  3. /activate :这将激活您的虚拟env(如果 选择解释器,然后转到您创建的Env
  4. ,现在我们拥有虚拟环境,只需点击 pip install antspyx ,然后使用 pip list 检查您的终端中,如果该软件包是在软件包中安装的,
  5. 如果 安装了该软件包,您仍然有错误重新启动您的IDE

Can you please specify your IDE+ if you took care of activating the virtualenv?

Anyways here are the steps to make sure that the python module is propoerly installed in your virtual environment :

  1. python3 -m venv env_name : this will create a virtual env in your current path
  2. source env/bin/activate : this will activate your virtual env (if on windows just cd into activate within your env name .\env\Scripts\activate
  3. If your using vscode hit ctrl+shift+p then select interpreter then go to your created env
  4. now we have our virtual environment just hit pip install antspyx then check with pip list and see in your terminal if the package is propoerly installed
  5. If you still have the error restart your IDE

模块&#x27; ants&#x27;没有属性&#x27; from_numpy&#x27;

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