根据 SQL 中的类型和日期计算不同的 ID

Question

我有 table 数据如下所示。我正在查找每个 TYPE 与 DATE 的总计数。

以上数据输出为：

我尝试了以下查询，但结果包含重复计数不止一次。

SELECT 
    DATE,
    SUM(CASE WHEN TYPE = 'Existing' THEN 1 ELSE 0 END) AS "EXISTING",
    SUM(CASE WHEN TYPE = 'New' THEN 1 ELSE 0 END) AS "NEW"
FROM 
    mytab
WHERE 
    DATE >= '2022-01-01'
GROUP BY    
    DATE ;

上述查询的输出：

Answer 1

一种方法是在子查询之上进行条件聚合，该子查询首先只获取与原始 table 不同的记录。

SELECT 
    DATE,
    COUNT(CASE WHEN TYPE = 'Existing' THEN 1 END) AS "EXISTING",
    COUNT(CASE WHEN TYPE = 'New' THEN 1 END) AS "NEW"
FROM (SELECT DISTINCT ID, TYPE, DATE FROM mytab) t
WHERE DATE >= '2022-01-01'
GROUP BY DATE;

Answer 2

未经测试，但也许是您想要的：

SELECT 
    DATE,
    COUNT(DISTINCT CASE WHEN TYPE = 'Existing' THEN ID END) AS "EXISTING",
    COUNT(DISTINCT CASE WHEN TYPE = 'New' THEN ID END) AS "NEW"
FROM 
    mytab
WHERE 
    DATE >= '2022-01-01'
GROUP BY    
    DATE ;

根据 SQL 中的类型和日期计算不同的 ID

Count Distinct IDs against type and Date in SQL

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