Question

我有以下查询：

WITH cteCountryLanguageMapping AS (
    SELECT * FROM (
        VALUES
            ('Spain', 'English'),
            ('Spain', 'Spanish'),
            ('Sweden', 'English'),
            ('Switzerland', 'English'),
            ('Switzerland', 'French'),
            ('Switzerland', 'German'),
            ('Switzerland', 'Italian')
    ) x ([Country], [Language])
)
SELECT
    [Country],
    CASE COUNT([Language])
        WHEN 1 THEN MAX([Language])
        WHEN 2 THEN STRING_AGG([Language], ' and ')
        ELSE STRING_AGG([Language], ', ')
    END AS [Languages],
    COUNT([Language]) AS [LanguageCount]
FROM cteCountryLanguageMapping
GROUP BY [Country]

我希望瑞士的“语言”列中的值以逗号分隔，即：

  | Country     | Languages                                 | LanguageCount
--+-------------+-------------------------------------------+--------------
1 | Spain       | Spanish and English                       | 2
2 | Sweden      | English                                   | 1
3 | Switzerland | French, German, Italian, English          | 4

相反，我得到以下输出（这4个值用and分隔）：

  | Country     | Languages                                 | LanguageCount
--+-------------+-------------------------------------------+--------------
1 | Spain       | Spanish and English                       | 2
2 | Sweden      | English                                   | 1
3 | Switzerland | French and German and Italian and English | 4

我想念什么？

这里是另一个示例：

SELECT y, STRING_AGG(z, '+') AS STRING_AGG_PLUS, STRING_AGG(z, '-') AS STRING_AGG_MINUS
FROM (
    VALUES
        (1, 'a'),
        (1, 'b')
) x (y, z)
GROUP by y

  | y | STRING_AGG_PLUS | STRING_AGG_MINUS
--+---+-----------------+-----------------
1 | 1 | a+b             | a+b

这是SQL Server中的错误吗？

Answer 1

是的，这是一个Bug（tm），存在于（截至编写时）SQL Server 2017 CU12之前的版本中（但根据@DanGuzman的说法，没有出现在Azure SQL数据库中，因此显然它已得到修复，并且该修复程序可能降落在下一个CU中）。具体而言，只要STRING_AGG(x, <separator>)匹配，优化器中执行公共子表达式消除（确保我们不会计算超出必要量的表达式）的部分会不正确地认为x形式的所有表达式都是相同的<separator>是什么，并将它们与查询中第一个计算出的表达式统一起来。

一种解决方法是通过对x进行某种（近）身份转换来确保不匹配。由于我们正在处理字符串，因此串联一个空字符串可以做到：

SELECT y, STRING_AGG(z, '+') AS STRING_AGG_PLUS, STRING_AGG('' + z, '-') AS STRING_AGG_MINUS
FROM (
    VALUES
        (1, 'a'),
        (1, 'b')
) x (y, z)
GROUP by y

Answer 2

不要重复自己*。您正在通过使用MAX(...)，LIST_AGG(...', ')和LIST_AGG(...' and ')来重复自己。您可以像这样简单地重写查询，并可能得到更好的计划：

WITH cteCountryLanguageMapping AS (
    SELECT * FROM (
        VALUES
            ('Spain', 'English'),
            ('Spain', 'Spanish'),
            ('Sweden', 'English'),
            ('Switzerland', 'English'),
            ('Switzerland', 'French'),
            ('Switzerland', 'German'),
            ('Switzerland', 'Italian')
    ) x (Country, Language)
), results AS (
    SELECT
        Country,
        COUNT(Language) AS LanguageCount,
        STRING_AGG(Language, ', ') AS Languages
    FROM cteCountryLanguageMapping
    GROUP BY Country
)
SELECT Country, LanguageCount, CASE LanguageCount
    WHEN 2 THEN REPLACE(Languages, ', ', ' and ')
    ELSE Languages
END AS Languages_Fixed
FROM results

结果：

| Country     | LanguageCount | Languages_Fixed                  |
|-------------|---------------|----------------------------------|
| Spain       | 2             | Spanish and English              |
| Sweden      | 1             | English                          |
| Switzerland | 4             | French, German, Italian, English |

DB Fiddle

^{*我也不想通过说这是一个错误来重复别人。}

STRING_AGG的行为异常

2 个答案: