Question

我希望能够从SQL Server返回一个ids的JSON数组作为JSON对象的一部分。

我试图使它起作用并接近，但这不是我想要构造JSON的方式。

此SQL我已经输出了

add_action( 'woocommerce_cart_calculate_fees', 'add_discount_line' );
function add_discount_line( $cart ) {

    if($cart->get_applied_coupons() && $cart->subtotal==0):
    $getDetails = ( new WC_Coupon($cart->get_applied_coupons()));
    $discount  =  $getDetails->amount;

     $cart->add_fee( __( 'Special Discount', 'woocommerce' ) , -$discount );
    endif;
}

这会生成具有我想要的所有内容的JSON，但我想将rigIds构建为JSON数组，以生成类似的内容，所以我不会无缘无故地复制对象。

SELECT c.Pk_Company_Id AS [id],
       c.Name AS [name],
       r.Pk_Rig_Id AS [rigId]
FROM Company c
LEFT JOIN Rig r
ON c.Pk_Company_Id = r.Fk_Company_Id
FOR JSON PATH, ROOT('companies')

{"companies":[{"id":1,"name":"Company 1","rigId":100},{"id":1,"name":"Company 1","rigId":101},{"id":2,"name":"Company 2"}]}

Answer 1

带有REPLACE的变体：

SELECT
  c.Pk_Company_Id AS [id],
  c.Name AS [name],
  JSON_QUERY(REPLACE(REPLACE((SELECT r.Pk_Rig_Id id FROM Rig r WHERE c.Pk_Company_Id = r.Fk_Company_Id FOR JSON PATH),'{"id":',''),'}','')) [rigIds]
FROM Company c
FOR JSON PATH, ROOT('companies')

-- {"companies":[{"id":1,"name":"Company 1","rigIds":[100,101]},{"id":2,"name":"Company 2"}]}

带有STRING_AGG的变体：

SELECT
  c.Pk_Company_Id AS [id],
  c.Name AS [name],
  JSON_QUERY((SELECT '['+STRING_AGG(r.Pk_Rig_Id,',')+']' FROM Rig r WHERE c.Pk_Company_Id = r.Fk_Company_Id)) [rigIds]
FROM Company c
FOR JSON PATH, ROOT('companies')

SQL Server为匹配的行返回ID的JSON数组

1 个答案: