如何为每个 id 生成一个字符串列表？ [复制]

Posted 2023-02-24

【中文标题】如何为每个 id 生成一个字符串列表？ [复制]

【英文标题】：How do I generate a list of strings per id? [duplicate]

【发布时间】：2019-03-10 06:59:55

【问题描述】：

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

【问题讨论】：

试试 string_agg。 group_concat 是不是mysql？

您在Postgres manual 的哪个位置找到group_concat？

【参考方案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

【讨论】：

感谢我运行它并将“GROUP BY customer_id”替换为“GROUP BY customer_brandids.customer_id”我收到此错误消息：“列“customer_brandids.brand_id”必须出现在 GROUP BY 子句中或被使用在聚合函数 LINE 1: SELECT customer_brandids.customer_id, customer_brandids.bran... 我如何只按 customer_id 聚合？

@lschra01，你不能，除非你把它变成一个聚合体。我不知道你在那里有非聚合字段并且不包含在 group by 中。如果您不关心每个 customer_id 获得哪个brand_id，或者每个 customer_id 是唯一的（brand.name 相同），那么您可以执行以下操作：max(customer_brandids.brand_id) as brand_id。 - 或者简单地将它们添加到分组依据。

【参考方案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

【讨论】：

谢谢！只需列出每个客户 _id 购买的品牌，用逗号分隔每个品牌。

我添加了一个 string_agg 查询。更好？

我没有？我以为我做到了