Question

我有以下列（is_featured、sort、created_at）。按升序排列的最大特色项目是 8 个。如果特色商品少于 8 个，则从最新的非特色商品中获取。

| id | message | is_featured | sort | created_at |
|----|---------|-------------|------|------------|
| 1  | ...     | 1           | 1    | 2021-06-07 |
| 2  | ...     | 1           | 2    | 2021-06-08 |
| 3  | ...     | 1           | 3    | 2021-06-09 |
| 4  | ...     | 1           | 4    | 2021-06-10 |
| 5  | ...     | 1           | 5    | 2021-06-11 |
| 6  | ...     | 0           | 0    | 2021-06-12 |
| 7  | ...     | 0           | 0    | 2021-06-13 |
| 8  | ...     | 0           | 0    | 2021-06-14 |

这是我目前尝试过的。

$stories = Story::where(function($query) {
    $query->where('is_featured', 0)
        ->latest();
    })->orWhere(function($query) {
        $query->where('is_featured', 1)
            ->orderBy('sort', 'asc');
    })->limit(8)
    ->get();

我无法理解这一点。任何帮助，将不胜感激。谢谢。

Answer 1

我认为您应该将这作为两个查询并根据条件合并到结果集合中。

$featuredStories = Story::where('is_featured', 1)
    ->orderBy('sort', 'asc')
    ->limit(8)
    ->get();

if ($featuredStories->count < 8) {
    $nonFeaturedCount = 8 - $featuredStories->count;
    $nonFeaturedStories = Story::where('is_featured', 0)
        ->limit($nonFeaturedCount)
        ->get();
    $stories = $featuredStories->merge($nonFeaturedStories);
} else {
    $stories = $featuredStories;
}

如何执行以下查询？

1 个答案: