如何为每个 id 生成一个字符串列表？ [复制]

Question

我想列出每个客户购买的品牌。我有相同的“customer_id”字段反复出现，并带有他们在此代码中购买的不同品牌的名称（显示为列标题：“名称”）。我想按 customer_id 分组并显示每个 customer_id 的品牌列表。我收到错误消息：

“错误：函数 group_concat（字符变化，未知）不 存在 ^ 提示：没有函数匹配给定的名称和参数类型。你可能需要 添加显式类型转换。

    CREATE TEMP TABLE customer_brandids AS 
SELECT receipts.receipt_id, receipts.customer_id, receipt_item_details1.brand_id
FROM receipts
LEFT JOIN receipt_item_details1
ON receipts.receipt_id = receipt_item_details1.receipt_id;

SELECT customer_brandids.customer_id, customer_brandids.brand_id, brands.name, GROUP_CONCAT(brands.name,',')
FROM customer_brandids
INNER JOIN brands
ON customer_brandids.brand_id = brands.brand_id
GROUP by customer_id

Answer 1

CREATE TEMP TABLE customer_brandids AS 
SELECT receipts.receipt_id, receipts.customer_id, receipt_item_details1.brand_id
FROM receipts
LEFT JOIN receipt_item_details1
ON receipts.receipt_id = receipt_item_details1.receipt_id;

SELECT customer_brandids.customer_id, customer_brandids.brand_id, brands.name, string_agg(brands.name,',')
FROM customer_brandids
INNER JOIN brands
ON customer_brandids.brand_id = brands.brand_id
GROUP by customer_id

Answer 2

这仅聚合品牌名称：

SELECT cb.customer_id, ARRAY_AGG(b.name) as brand_names
FROM customer_brandids cb
INNER JOIN brands b
ON cb.brand_id = b.brand_id
GROUP by cb.customer_id

如果您还想要品牌 ID 列表：

SELECT 
    cb.customer_id, 
    ARRAY_AGG(b.brand_id) as brand_ids,
    ARRAY_AGG(b.name) as brand_names
FROM customer_brandids cb
INNER JOIN brands b
ON cb.brand_id = b.brand_id
GROUP by cb.customer_id

如果您需要将列表作为字符串列表，请使用 string_agg 而不是 array_agg

SELECT 
    cb.customer_id, 
    string_agg(b.brand_id, ',') as brand_ids, -- delete this line if you only need the names
    string_agg(b.name, ',') as brand_names
FROM customer_brandids cb
INNER JOIN brands b
ON cb.brand_id = b.brand_id
GROUP by cb.customer_id