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sql array 数组基本用法(四)

發布時間:2025/4/5 编程问答 24 豆豆
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查詢嵌套數組
SELECT "800M" AS race,[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] as splits),STRUCT("Makhloufi" as name, [24.5, 25.4, 26.6, 26.1] as splits),STRUCT("Murphy" as name, [23.9, 26.0, 27.0, 26.0] as splits),STRUCT("Bosse" as name, [23.6, 26.2, 26.5, 27.1] as splits),STRUCT("Rotich" as name, [24.7, 25.6, 26.9, 26.4] as splits),STRUCT("Lewandowski" as name, [25.0, 25.7, 26.3, 27.2] as splits),STRUCT("Kipketer" as name, [23.2, 26.1, 27.3, 29.4] as splits),STRUCT("Berian" as name, [23.7, 26.1, 27.0, 29.3] as splits)]AS participants
這是一行數據,查詢的時候需要把數據展開
WITH races AS (SELECT "800M" AS race,[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] as splits),STRUCT("Makhloufi" as name, [24.5, 25.4, 26.6, 26.1] as splits),STRUCT("Murphy" as name, [23.9, 26.0, 27.0, 26.0] as splits),STRUCT("Bosse" as name, [23.6, 26.2, 26.5, 27.1] as splits),STRUCT("Rotich" as name, [24.7, 25.6, 26.9, 26.4] as splits),STRUCT("Lewandowski" as name, [25.0, 25.7, 26.3, 27.2] as splits),STRUCT("Kipketer" as name, [23.2, 26.1, 27.3, 29.4] as splits),STRUCT("Berian" as name, [23.7, 26.1, 27.0, 29.3] as splits)]AS participants) SELECTrace,participant FROM races r CROSS JOIN UNNEST(r.participants) as participant; raceparticipant
800M{Rudisha, [23.4, 26.3, 26.4, 26.1]}
800M{Makhloufi, [24.5, 25.4, 26.6, 26.1]}
800M{Murphy, [23.9, 26, 27, 26]}
800M{Bosse, [23.6, 26.2, 26.5, 27.1]}
800M{Rotich, [24.7, 25.6, 26.9, 26.4]}
800M{Lewandowski, [25, 25.7, 26.3, 27.2]}
800M{Kipketer, [23.2, 26.1, 27.3, 29.4]}
800M{Berian, [23.7, 26.1, 27, 29.3]}

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