Question

在postgres中，我做了这个函数，当它被调用时，它会打印count值，而不是我传递给变量的c.first_name值。我不知道这是否是正确的方法。我该如何纠正？

declare
    mostExperienced text;
BEGIN
    SELECT COUNT(*) AS count, c.first_name INTO mostExperienced 
    FROM crew_members c
    INNER JOIN aircrafts_members am ON am.crew_id = c.id 
    GROUP BY am.crew_id, c.first_name ORDER BY count DESC LIMIT 1;
    RETURN mostExperienced;
END;

Answer 1

该SELECT ... INTO中有两个变量，因此记录可能是一个更好的选择

declare
    mostExperienced RECORD;
BEGIN
    SELECT COUNT(*) AS count, c.first_name 
    INTO mostExperienced 
    FROM crew_members c
    INNER JOIN aircrafts_members am ON am.crew_id = c.id 
    GROUP BY am.crew_id, c.first_name ORDER BY count DESC 
    LIMIT 1;

    RETURN mostExperienced.first_name;
END;

如何使用INTO分配postgres函数内部查询的选择结果

1 个答案: