SQL 兩列不同值的計數

Question

我有以下查詢，允許我從 Flipside API 匯總每一天的唯一賣家/買家的數量：

SELECT
date_trunc('day', block_timestamp) AS date,
COUNT(DISTINCT(seller_address)) AS unique_sellers,
COUNT(DISTINCT(buyer_address)) AS unique_buyers
FROM  ethereum.core.ez_nft_sales
GROUP BY date

現在，我一直在嘗試很多不同的事情，但我終生無法弄清楚如何在給定的一天獲得唯一活動地址的數量，因為我需要以某種方式合並賣家和買家，然后計算唯一地址。 我將不勝感激任何幫助。 提前致謝！

Answer 1

嘗試通過以下方式在組中添加其他 2 列：unique_sellers 和 unique_buyers

Answer 2

這就是我通過對 unique_active 使用單獨的查詢並合並它們來設法解決問題的方法：

WITH
other_values AS (
    SELECT
    date_trunc('day', block_timestamp) AS date,
    COUNT(DISTINCT seller_address) AS unique_sellers,
    COUNT(DISTINCT buyer_address) AS unique_buyers
    FROM  ethereum.core.ez_nft_sales
    GROUP BY date
),
unique_addresses AS (
    SELECT
    date,
    COUNT(*) as unique_active
    FROM (
        SELECT
        date_trunc('day', block_timestamp) as date,
        seller_address as address
        FROM  ethereum.core.ez_nft_sales
        GROUP BY date, seller_address
        UNION
        SELECT
        date_trunc('day', block_timestamp) as date,
        buyer_address as address
        FROM  ethereum.core.ez_nft_sales
        GROUP BY date, buyer_address
        )
    GROUP BY date
)

SELECT * FROM other_values
LEFT JOIN unique_addresses
ON other_values.date = unique_addresses.date
ORDER BY other_values.date DESC