PHP错误，回调函数中有变量 [英] PHP error with variable in callback function

问题描述

我在php(laravel)中具有此功能:

I have this function in php (laravel):

    public static function isUserParticipatesInTournament($tourId, $userId)
    {
        var_dump($userId); //dumped
        $user = User::find($userId);

        if(!$user)
        {
            return null;
        }

        $obj = $user->whereHas('tournaments', function($query)
        {
            var_dump($tourId); //error
            $query->where('id', '=', $tourId); //error
        })->get();

        return $obj;
    }

问题在于，在闭包$obj = $user->whereHas('tournaments', function($query){...}中，未定义$tourId变量.我收到此错误: Undefined variable: userId.

The problem is that in the closure $obj = $user->whereHas('tournaments', function($query){...} the $tourId variable is undefined in it. I am getting this error: Undefined variable: userId.

为什么会这样?该变量在内部函数的范围内声明.我唯一的想法是，这是一个回调函数.

Why this is happening? The variable is declared in the scope of the inner function. My only thought is that, that it is a callback function.

当我尝试执行此功能时:$obj = $user->whereHas('tournaments', function($query, $tourId){...}，然后出现此异常:

When I tried to execute this function : $obj = $user->whereHas('tournaments', function($query, $tourId){...} then I am getting this exception:

Missing argument 2 for User::{closure}()

推荐答案

您的$tourId变量不在匿名函数的范围内.查看use关键字，以了解如何将其添加到范围.请参阅此页面上的示例3: http://www.php.net//manual/zh/functions .anonymous.php

Your $tourId variable is not in the scope of your anonymous function. Have a look at the use keyword to see how you can add it to the scope. See example 3 on this page: http://www.php.net//manual/en/functions.anonymous.php

它应该类似于以下内容:

It should look something like the following:

$obj = $user->whereHas('tournaments', function($query) use($tourId)
    {
        var_dump($tourId); // Dumps OK
    })->get();

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