Laravel 合集日期比较

Note: This answer is specific to Laravel <= 5.2. Laravel 5.3 updated the where() method on the Collection to match the query builder, so it now accepts an operator.

As you state in your question, $dinnerOrders is a Collection, not a query. The where() method on the Collection works a little differently.

对于集合，where()方法不接受运算符.它只做平等比较.第一个参数是键，第二个是值，第三个是布尔值，用于标记松散比较(==)与严格比较(==).

What you're looking for is the filter() method. You pass in a Closure that does your date comparison. If the Closure returns true, the item stays in the Collection. If it returns false, the item is removed from the Collection. Something like this (just an example, the logic may need to be tweaked):

$dinnerOrders = $dinnerOrders->filter(function ($item) use ($date) {
    return (data_get($item, 'date') > $date) && (data_get($item, 'date') < $date->endOfDay());
});

发布问题编辑

Based on the code provided in your edit, I'm guessing that a restaurant hasMany dinners, and a dinner hasMany orders. If this is correct, you can setup a relationship stating that a restaurant hasMany orders through dinners:

Restaurant.php:

public function orders() {
    // restaurant has many orders through dinners
    $this->hasManyThrough('App\Order', 'App\Dinner');
}

设置此关系后，您可以使用查询构建器获取信息:

$dinnerOrders = $restaurant->orders()->where('date','>',$date)->where('date','<', $date->endOfDay())->get();