Question

代码示例如下：

DB::table('orders')->select([
    DB::raw('SELECT COUNT(id) AS order_count WHERE type = 2'),
    DB::raw('SELECT SUM(id) AS order_sum WHERE type = 2')
])->paginate(20);

选中的字段是统计字段，子查询包含几个条件。我输入这样的代码，但代码无法工作并产生语法错误，那么我无法找出错误的位置。谁可以帮我？谢谢！

Answer 1

它应该是这样吗？

DB::table('orders')->select(
    DB::raw('COUNT(id) AS order_count'),
    DB::raw('SUM(id) AS order_sum')
)->where(type, 2)
->paginate(20);

Answer 2

我的原始SQL就是这样：

SELECT
    o1.store_id,
    (SELECT Count(id) FROM `order` WHERE type = 2 AND `order`.store_id = o1.store_id) AS package_order_count,
    (SELECT SUM(price) FROM `order` WHERE type = 2 AND `order`.store_id = o1.store_id) AS package_order_sum,
    (SELECT Count(id) FROM `order` WHERE type = 3 AND `order`.store_id = o1.store_id) AS member_order_count,
    (SELECT SUM(price) FROM `order` WHERE type = 3 AND `order`.store_id = o1.store_id) AS member_order_sum
FROM
    `order` as o1
WHERE
    `o1`.store_id > 0 AND `o1`.`from` = 1
GROUP BY `o1`.store_id

和我的PHP代码一样：

DB::table('orders')->select([
    DB::raw('SELECT COUNT(id) AS order_count WHERE type = 2'),
    DB::raw('SELECT SUM(id) AS order_sum WHERE type = 2')
    DB::raw('SELECT COUNT(id) AS member_order_count WHERE type = 3'),
    DB::raw('SELECT SUM(id) AS member_order_sum WHERE type = 3')
])->paginate(20);

所以，我的代码出错了。

如何在laravel 5的select方法中添加子查询

2 个答案: