Question

我有以下查询：

SELECT s."Description",
sp.*
FROM "Supplier" as s
OUTER APPLY (
    SELECT p."Id", p."Description", p."Price"
    FROM "Products" as p
    WHERE p."SupplierId" = s."Id"
    FOR JSON auto 
) as sp

我正在尝试根据OUTER APPLY的结果构建json数组，但由于存在错误No column name was specified for column 1 of 'sp'.，我在这里卡住了，我在stackoverflow上发现了类似的问题，但是没有适用于外部的示例。

您能解释一下该查询出了什么问题吗？

Answer 1

您只需要添加列别名：

SELECT 
    s."Description",
    sp.*
FROM "Supplier" as s
OUTER APPLY (
    SELECT p."Id", p."Description", p."Price"
    FROM "Products" as p
    WHERE p."SupplierId" = s."Id"
    FOR JSON auto 
) as sp(json)

没有为“ sp”的列1指定列名

1 个答案: