group by

時間 2021-01-04

原文原文鏈接

select min(sage),sname from student group by sname; 1.查詢數據 2.按sname分組 3.分組時可能每一組有多個值，取sage的最小值錯誤情況解釋：select 後面的列名必須是group by後面的列名或者非group by 的列名，但是需要使用聚合函數修飾.

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1. order by ， group by
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