Question

我在where / whereOr查询中遇到了一些麻烦。我想检查fitting和dimmability是否相同。那么light_color_code可以是2700K或2800K。这是我当前的查询：

if ($lightColorCode === '2800K') {
    $correspondingLamps = Lamp::where(['fitting' => $fitting, 'dimmability' => $dimmable, 'light_color_code' => $lightColorCode])
        ->orWhere(['fitting' => $fitting, 'dimmability' => $dimmable, 'light_color_code' => '2700K'])
        ->get();
}

但是它返回所有Lamps，其中fitting或dimmability或light_color_code匹配，但它们都应该匹配。我看不出此查询有什么问题吗？

更新：如建议中的注释所示，我查看了https://laravel.com/docs/5.7/queries#parameter-grouping，并创建了以下查询：

        $lamps = DB::table('lamps')
            ->where('fitting', '=', $fitting)
            ->where('dimmability', '=', $dimmable)
            ->where(function ($query, $lightColorCode) {
                $query->where('light_color_code', '=', $lightColorCode)
                    ->orWhere('light_color_code', '=', '2700K - 827 Zeer warm wit');
            })
            ->get();

但这返回：

函数的参数太少 App \ Http \ Controllers \ CompareController :: App \ Http \ Controllers {closure}（）， 1个通过，恰好2个预期

我猜这是因为我将$lightColorCode作为参数传递，但是我需要在where中使用该参数。

Answer 1

您可以使用use关键字将必要的变量从父范围传递到闭包中。

$correspondingLamps = Lamp::where(['fitting' => $fitting, 'dimmability' => $dimmable])->where(function($query) use ($lightColorCode){
            $query->where(['light_color_code' => $lightColorCode])->orWhere('light_color_code' => '2700K');})->get();

检查此项以获取详细信息https://www.php.net/manual/en/functions.anonymous.php

Answer 2

尝试一下：

$correspondingLamps = Lamp::where(['fitting' => $fitting, 'dimmability' => $dimmable])
                    ->where(function($query) use ($lightColorCode){
                      $query->where(['light_color_code' => $lightColorCode]) // 2800K
                      ->orWhere('light_color_code' => '2700K');
                    })
                    ->get();

Laravel在哪里和哪里或

2 个答案: