Question

我想生成userID，逻辑是，UserID是FirstName和LastName的组合。如果userID已存在，则在FirstName和LastName之后添加1。

代表：

用户John Doe的用户ID将是John Doe。

如果另一个John Doe加入公司，那么UserID应该是JohnDoe1，依此类推。

我用来实现这个目的的案例陈述如下。

Select distinct case
when lower(a.First_Name) || lower(a.Last_Name) = b.Name then a.First_Name || a.Last_Name || 1
when lower(a.First_Name) || lower(a.Last_Name) || 1 = b.Name then a.First_Name || a.Last_Name || 2

else a.first_name || a.last_name
end as user_id from Users a, accounts b

where a.unique_id = '12345' (This is matching key from table B)

在我的表中，我已经有了userID JohnDoe和JohnDoe1。

和输出我得到的以上查询是，

JOHNDOE JohnDoe1 JohnDoe2

我如何仅将JohnDoe2作为输出。

Answer 1

这是一个可能适用于Oracle的解决方案：

WITH oldnames AS (
    SELECT 'JohnDoe' AS user_id, 'John' AS fname, 'Doe' AS lname
      FROM dual
     UNION
    SELECT 'JohnDoe1' AS user_id, 'John' AS fname, 'Doe' AS lname FROM dual
     UNION
    SELECT 'MaxHeadroom' AS user_id, 'Max' AS fname, 'Headroom' AS lname FROM dual
), newnames AS (
    SELECT 'John' AS fname, 'Doe' AS lname FROM dual
     UNION
    SELECT 'Max' AS fname, 'Headroom' AS lname FROM dual
     UNION
    SELECT 'Mark' AS fname, 'Twain' AS lname FROM dual
)
SELECT n1.fname, n1.lname, MAX(n2.user_id)
     , n1.fname || n1.lname ||
       CASE WHEN MAX(n2.user_id) IS NOT NULL THEN TO_CHAR(COALESCE(TO_NUMBER(MAX(REGEXP_SUBSTR(n2.user_id, '\d+$'))), 0) + 1) END
  FROM newnames n1, oldnames n2
 WHERE n1.fname = n2.fname(+)
   AND n1.lname = n2.lname(+)
 GROUP BY n1.fname, n1.lname

我假设您有一张表，您可以在其中存储用户ID以及名字和姓氏。我在上面的查询中称之为oldnames，newnames是您要为其生成ID的用户表。

如何通过SQL生成UserID

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