Impala SQL left join on t1 vs t2

Question

我在此处连接时遇到问题，特别是在 x=y 上在 SQL Impala 中，有人可以帮助我了解它是如何工作的。 SUBSTR(cast(t1.employee_number as string), 3,10) as short_employee_number 工作正常，我测试过，但是当我构建这个左连接时，我不明白 x=y 上最终应该是什么我试过了

SUBSTR(cast(t1.employee_number as string), 3,10) =  t2.short_staff_number
t1.short_employee_number =  t2.short_staff_number
short_employee_number=  t2.short_staff_number

none 其中有效。

问题是什么？

 SELECT DISTINCT
    SUBSTR(cast(t1.employee_number as string), 3,10) as short_employee_number, 
    t1.begin_date_it0001,
    t1.end_date_it0001,
    t1.cost_center
    t1.position,
    t2.local_time_createddate,
    t2.area,
    t2. unit, 
    t2.short_staff_number, 
    t2.alias, 
    t2.email

    FROM dataone  as t1 
    LEFT JOIN datatwo as  t2 
    ON 
    short_employee_number  = t2.short_staff_number ;

Answer 1

试试这个

您不能在 FROM 处使用 SELECT FIELDS，只能在 ORDER BY

处使用

SELECT DISTINCT
    SUBSTR(cast(t1.employee_number as string), 3,10) as short_employee_number, 
    t1.begin_date_it0001,
    t1.end_date_it0001,
    t1.cost_center
    t1.position,
    t2.local_time_createddate,
    t2.area,
    t2. unit, 
    t2.short_staff_number, 
    t2.alias, 
    t2.email

    FROM dataone  as t1 
    LEFT JOIN datatwo as  t2 
    ON 
     SUBSTR(cast(t1.employee_number as string), 3,10)  = t2.short_staff_number ;

Impala SQL left join on t1 vs t2

Impala SQL left join on t1 vs t2

sql

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impala