如何生成每个ID的字符串列表？ [重复]

Question

这个问题在这里已有答案：

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我想列出每个客户购买的品牌。我有相同的“customer_id”字段，他们在此代码中购买了不同品牌的名称（显示为列名称：“名称”）。我想按customer_id进行分组，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。您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

这只汇总了品牌名称：

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