验证帐户号SQL表

我希望使用以下规则从表中选择数据，但在编写查询时遇到问题。我正在使用PostgreSQL，无法创建自定义项。该表如下所示：

id  | user_id  | account_number
-------------------------------
 1  |    1     | 12345671
 2  |    4     | 12356673
 3  |    7     | 12325678

SELECT * FROM table_name WHERE
  account_number SIMILAR TO '[0-9]{8}'
  AND (
    1 * CAST(SUBSTR(account_number, 1, 1) AS INTEGER) +
    2 * CAST(SUBSTR(account_number, 2, 1) AS INTEGER) +
    3 * CAST(SUBSTR(account_number, 3, 1) AS INTEGER) +
    1 * CAST(SUBSTR(account_number, 4, 1) AS INTEGER) +
    2 * CAST(SUBSTR(account_number, 5, 1) AS INTEGER) +
    3 * CAST(SUBSTR(account_number, 6, 1) AS INTEGER) +
    1 * CAST(SUBSTR(account_number, 7, 1) AS INTEGER)
  )%10 = CAST(SUBSTR(account_number, 8, 1) AS INTEGER)

id和用户id是整数，而帐号是字符串。我想选择符合以下条件的帐号：

• 帐号字符串正好包含8位数字
• 验证方案
• 取前7位数字
• 将第一个数字乘以1，第二个数字乘以2，第三个数字乘以3，第四个数字乘以1，第五个数字乘以2，第六个数字乘以3，第七个数字乘以1
• 将每个数字乘以相关数字的结果求和
• 如果第8位数字与mod（总和，10）相同，则选择此数字

在上面的这个表中，我应该只返回查询的前两行

重复一次，我无法创建UDF，所以我想知道在查询中使用普通SQL是否可行

---

谢谢

是的，你能做到。基本上，使用“类似于”检查精确的8位数字，然后使用子字符串和cast进行计算。大概是这样的：

id  | user_id  | account_number
-------------------------------
 1  |    1     | 12345671
 2  |    4     | 12356673
 3  |    7     | 12325678

SELECT * FROM table_name WHERE
  account_number SIMILAR TO '[0-9]{8}'
  AND (
    1 * CAST(SUBSTR(account_number, 1, 1) AS INTEGER) +
    2 * CAST(SUBSTR(account_number, 2, 1) AS INTEGER) +
    3 * CAST(SUBSTR(account_number, 3, 1) AS INTEGER) +
    1 * CAST(SUBSTR(account_number, 4, 1) AS INTEGER) +
    2 * CAST(SUBSTR(account_number, 5, 1) AS INTEGER) +
    3 * CAST(SUBSTR(account_number, 6, 1) AS INTEGER) +
    1 * CAST(SUBSTR(account_number, 7, 1) AS INTEGER)
  )%10 = CAST(SUBSTR(account_number, 8, 1) AS INTEGER)

当然，在您的示例中，这不会返回任何行，因为：

1×1 + 2×2 + 3x3 + 1×4 + 2×5 + 3×6 + 1×7 = 53
53 MOD 10 = 3
3 ≠ 1

---

PS：您确实意识到UDF可以用C以外的语言编写。例如，您可以用PL/pgSQL编写一个UDF。

只需创建一个表格，在其中拆分您的数字，然后进行算术运算（当然，您可以在

..

的位置填写其余内容）

当然，如果您不想创建一个单独的表，那么始终可以将用于创建

digits

表的

select

语句放入第二个查询的子查询中

select id, 
       user_id,
       digit_sum, 
       last_digit 
from (
  select id, 
         user_id,
         (substring(account_number,1,1)::int + 
          substring(account_number,2,1)::int * 2 + 
          substring(account_number,3,1)::int * 3 + 
          substring(account_number,4,1)::int + 
          substring(account_number,5,1)::int * 2 + 
          substring(account_number,6,1)::int * 3 +
          substring(account_number,7,1)::int ) as digit_sum, 
          substring(account_number,8,1)::int as last_digit
  from accounts
) t
where last_digit = digit_sum % 10

---

为了让生活更轻松，我将创建一个视图，对值进行拆分和求和。然后，您只需要从该视图中选择我用于派生表的where条件。

您可以尝试以下方法

select *,   (
    (substring(account_number from 1 for 1)::integer * 1) +
    (substring(account_number from 2 for 1)::integer * 2) +
    (substring(account_number from 3 for 1)::integer * 3) +
    (substring(account_number from 4 for 1)::integer * 1) +
    (substring(account_number from 5 for 1)::integer * 2) +
    (substring(account_number from 6 for 1)::integer * 3) +
    (substring(account_number from 7 for 1)::integer * 1)
   ) as sums,
  (
    (substring(account_number from 1 for 1)::integer * 1) +
    (substring(account_number from 2 for 1)::integer * 2) +
    (substring(account_number from 3 for 1)::integer * 3) +
    (substring(account_number from 4 for 1)::integer * 1) +
    (substring(account_number from 5 for 1)::integer * 2) +
    (substring(account_number from 6 for 1)::integer * 3) +
    (substring(account_number from 7 for 1)::integer * 1)
   ) % 10 as mod_10
from acct_no
where length(account_number) = 8 

---

我将计算结果写入SELECT子句而不是WHERE子句，因为要么我的算术错误，要么您的规格错误。

对于id=1的行，总和为'1+2*2+3*3+4+5*2+6*3+7=53，这将为mod（53，10）生成3。所以我不明白为什么该行应该返回，因为最后一位数字是1，而不是3UDF应该代表用户定义的函数或用户定义的字段？