blocks|key|2326780|text|请查看以下查询，谢谢|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2326781|select+U.id,+U.name+AS+UserName,+R.name+AS+RoleName,P.id+As+PermissionID,+P.name+AS+PermissionName
FROM+users+U+
INNER+JOIN+user_has_role+ur+on+U.id+=+ur.user_id
INNER+JOIN+roles+R+on+ur.role_id+=+R.id
INNER+JOIN+role_has_permissions+rp+on+R.id+=+rp.role_id
INNER+JOIN+permissions+P+on+rp.Permission_id+=+P.id
WHERE+P.id+in+(2,3)|code-block|syntax|javascript|2326782|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Please check the following query, thanks
<pre><code>select U.id, U.name AS UserName, R.name AS RoleName,P.id As PermissionID, P.name AS PermissionName
FROM users U 
INNER JOIN user_has_role ur on U.id = ur.user_id
INNER JOIN roles R on ur.role_id = R.id
INNER JOIN role_has_permissions rp on R.id = rp.role_id
INNER JOIN permissions P on rp.Permission_id = P.id
WHERE P.id in (2,3)
</code></pre>

blocks|key|28816|text|如果您只需要用户列表，可以使用下面的查询：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|28817|select+distinct+u.id+,+u.name+username
from+users+u
join+user_has_role+ur
+++on+u.id+=+ur.user_id
join+role_has_permissions+rp
+++on+ur.role_id+=+rp.role_id
where++rp.permission_id+in+(+2,+3)+|code-block|syntax|javascript|28818|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you only need list of users you can use query below:
<pre><code>select distinct u.id , u.name username
from users u
join user_has_role ur
 on u.id = ur.user_id
join role_has_permissions rp
 on ur.role_id = rp.role_id
where rp.permission_id in ( 2, 3) 
</code></pre>

blocks|key|28879|text|我没有在文档中找到它，但是有一个许可范围应该给您直接或通过一个角色拥有权限的所有用户：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|28880|public+function+scopePermission(Builder+$query,+$permissions):+Builder|code-block|syntax|javascript|28881|你可以这样使用它：|28882|User::permission([2,+3])->get();|28883|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/spatie/laravel-permission/blob/master/src/Traits/HasPermissions.php#L62^0|G|4|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

I did not find it mentioned in the documentation but there is a <a href="https://github.com/spatie/laravel-permission/blob/master/src/Traits/HasPermissions.php#L62" rel="nofollow noreferrer">permission scope</a> that should give you all users that either have the permission directly or through a role:
<pre><code>public function scopePermission(Builder $query, $permissions): Builder
</code></pre>
You can use it like this:
<pre><code>User::permission([2, 3])-&gt;get();
</code></pre>

blocks|key|1901489|text|空间权限包使用Laravel本机授权功能，这意味着您可以访问授权文档中定义的许多特性。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1901490|这些特性之一是can()指令。您可以将权限的名称传递给can()，它将根据检查返回true或false。例如：|style|CODE|1901491|@can('view+posts')

if+(auth()->user()->can('view+posts')+{+}

if+($user->can('view+posts')+{+}|code-block|syntax|javascript|1901492|使用您的示例，如果您想找到所有具有users权限的的：|1901493|//+Get+the+name+of+your+permissions
$permissions+=+Permission::whereIn('id',+[2,3])->pluck('name')->toArray();

//+Filter+users+based+on+if+they+have+any+of+the+permissions
$users+=+User::all()
++++->filter(function+($user)+use+($permissions)+{
++++++++if+($user->can($permissions))+{
++++++++++++return+$user;
++++++++}
++++++++return+false;
++++});|1901494|或者，如果users必须拥有all，则权限：|BOLD|1901495|//+Filter+users+based+on+if+they+have+all+of+the+permissions
$users+=+User::all()
++++->filter(function+($user)+use+($permissions)+{
++++++++foreach+($permissions+as+$permission)+{
++++++++++++if+(!$user->can($permission))+{
++++++++++++++++return+false;
++++++++++++}
++++++++}

++++++++return+$user;
++++});|1901496|查看空间网站上的通过角色使用权限文档。|1901497|entityMap|0|LINK|mutability|MUTABLE|url|https://laravel.com/docs/8.x/authorization|1|https://spatie.be/docs/laravel-permission/v3/basic-usage/role-permissions^0|U|4|0|0|7|5|R|5|15|4|1A|5|0|0|H|5|0|0|5|5|E|3|0|0|8|8|1|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]|$1|D|3|E|5|6|7|1B|8|@$A|1C|B|1D|F|G]|$A|1E|B|1F|F|G]|$A|1G|B|1H|F|G]|$A|1I|B|1J|F|G]]|9|@]|C|$]]|$1|H|3|I|5|J|7|1K|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|1L|8|@$A|1M|B|1N|F|G]]|9|@]|C|$]]|$1|O|3|P|5|J|7|1O|8|@]|9|@]|C|$K|L]]|$1|Q|3|R|5|6|7|1P|8|@$A|1Q|B|1R|F|G]|$A|1S|B|1T|F|S]]|9|@]|C|$]]|$1|T|3|U|5|J|7|1U|8|@]|9|@]|C|$K|L]]|$1|V|3|W|5|6|7|1V|8|@]|9|@$A|1W|B|1X|1|1Y]]|C|$]]|$1|X|3|-4|5|6|7|1Z|8|@]|9|@]|C|$]]]|Y|$Z|$5|10|11|12|C|$13|14]]|15|$5|10|11|12|C|$13|16]]]]

The Spatie permissions package uses Laravels native authorization functionality meaning you have access to a lot of the features defined in the <a href="https://laravel.com/docs/8.x/authorization" rel="nofollow noreferrer">authorization docs</a>.
One of these features is the <code>can()</code> directive. You can pass the name of a permission to <code>can()</code> and it will return <code>true</code> or <code>false</code> based on the check. For example:
<pre><code>@can('view posts')

if (auth()-&gt;user()-&gt;can('view posts') { }

if ($user-&gt;can('view posts') { }
</code></pre>
Using your example, if you want to find all <code>users</code> that have any of your permissions:
<pre><code>// Get the name of your permissions
$permissions = Permission::whereIn('id', [2,3])-&gt;pluck('name')-&gt;toArray();

// Filter users based on if they have any of the permissions
$users = User::all()
 -&gt;filter(function ($user) use ($permissions) {
 if ($user-&gt;can($permissions)) {
 return $user;
 }
 return false;
 });
</code></pre>
Alternatively if the <code>users</code> must have all the permissions:
<pre><code>// Filter users based on if they have all of the permissions
$users = User::all()
 -&gt;filter(function ($user) use ($permissions) {
 foreach ($permissions as $permission) {
 if (!$user-&gt;can($permission)) {
 return false;
 }
 }

 return $user;
 });
</code></pre>
Check out the docs for <a href="https://spatie.be/docs/laravel-permission/v3/basic-usage/role-permissions" rel="nofollow noreferrer">using permissions via roles</a> on the Spatie website.

I'm trying to figure out if there is a simple way to get all the users that have role with given permissions.
So i used permissions through roles it means that i don't have direct relation to them so i need to check relation of user model &quot;roles&quot; and then check permissions that associated to each of user role.
SqlFiddle with set of data - <a href="http://sqlfiddle.com/#!9/2fe35d" rel="nofollow noreferrer">http://sqlfiddle.com/#!9/2fe35d</a>
Task is to get users that have permissions with id 2 and 3 for example.

Spatie Laravel Permissions get users with permissions

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我试图找出是否有一种简单的方法可以让所有具有给定权限的用户。所以我通过角色使用权限，这意味着我与它们没有直接关系，所以我需要检查用户模型“角色”的关系，然后检查与每个用户角色关联的权限。带数据集的SqlFiddle - 例如，任务是获取具有id 2和3权限的用户。

问Spatie Laravel权限获得具有权限的用户
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