Php Lumen whereRaw with Prepared语句不返回任何内容

我正在尝试使用此查询从数据库中获取值

$vouchers = MerchantVoucher::where([
            'merchant_id' => $this->merchant->id,
            'code' => $voucherCode
        ])
        ->whereRaw("
            (
                (valid_from between '{$voucherValidFrom}' and '{$voucherValidTo}' ) || 
                (valid_to between '{$voucherValidFrom}' and '{$voucherValidTo}')
            )"
        )
        ->count();

通过这个查询，我得到了预期的结果，它返回的行数为1。 我尝试使用下面这个查询的Prepared语句优化查询

    $vouchers = MerchantVoucher::where([
        'merchant_id' => $this->merchant->id,
        'code' => $voucherCode
    ])
    ->whereRaw("
        ( 
            (valid_from between '?' and '?' ) || 
            (valid_to between '?' and '?' ) 
        )", 
        [
            $voucherValidFrom, 
            $voucherValidTo, 
            $voucherValidFrom, 
            $voucherValidTo
        ]
    )
    ->count();

使用此查询时，它返回0

有人对此有解释吗？提前谢谢你

非PHP用户的原始Sql比较

 select count(*) from `merchant_vouchers` where (`merchant_id` = ? and `code` = ?) and 
                (
                    (valid_from between '2019-12-03 11:36:35' and '2019-12-10 11:36:35' ) || 
                    (valid_to between '2019-12-03 11:36:35' and '2019-12-10 11:36:35')
                ) and `merchant_vouchers`.`deleted_at` is null

返回1

  select count(*) from `merchant_vouchers` where (`merchant_id` = ? and `code` = ?) and 
                ( 
                    (valid_from between '?' and '?' ) || 
                    (valid_to between '?' and '?' ) 
                ) and 
                `merchant_vouchers`.`deleted_at` is null

返回0

使用的框架：内腔

---

数据库：MySQL

您需要删除占位符周围的引号：

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是否介于两者之间？而且？

现在无法验证我的假设，但我认为您需要删除标识符周围的引号：

有效吗？还有？

@Fitzi你的假设解决了我的问题：），谢谢！那我就回答你