在 oracle sql 中使用 regexp_replace 标准化地址

Question

非常感谢大家的帮助。

我们正在对客户的地址进行标准化。

我有一个主要客户table，其中包含客户数据，要标准化的地址字段。

我有一个映射 table，其中包含 from_str 到 to_str 值的映射。

我需要将对应于from_str的to_str获取到地址数据中。

如果标准化后地址的长度超过35个字符，则从前面的space(' ')字符串的右边到末尾，被创建为一个单独的字段address2

你能帮忙 sql 或 pl/sql 吗？使用 Oracle 12c 数据库。

我在下面编写的代码不适用于所有 from_str 值...仅适用于前 2 行。

感谢任何帮助。

到目前为止代码：

with addresses as
(
    select cust_id,address addr from
    (
    select 10 cust_id,'9 Help Street, Level 4' address from dual union all
    select 11 cust_id,'22 Victoria Street' address from dual union all
    select 12 cust_id,'1495 Franklin Str' address from dual union all
    select 13 cust_id,'30 Hasivim St.,Petah-Tikva' address from dual union all
    select 14 cust_id,'2 Jakaranda St' address from dual union all
    select 15 cust_id,'61, Science Park Rd' address from dual union all
    select 16 cust_id,'61, Social park road' address from dual union all
    select 17 cust_id,'Av. Hermanos Escobar 5756' address from dual union all
    select 18 cust_id,'Ave. Hermanos Escobar 5756' address from dual union all
    select 19 cust_id,'8000 W FLORISSANT AVE' address from dual union all
    select 20 cust_id,'8600 MEMORIAL PKWY SW' address from dual union all
    select 21 cust_id,'8200 FLORISSANTMEMORIALWAYABOVE SW' address from dual union all
    select 22 cust_id,'8600 MEMORIALFLORISSANT PKWY SW' address from dual
    ) t1
),
  replacements as
    (
    select id,to_str,from_string from_str from
    (
    select 1 id,'St' to_str,'Street' from_string from dual union all
  select 2 id,'St' to_str,'St.' from_string from dual union all
  select 3 id,'St' to_str,'Str' from_string from dual union all
  select 4 id,'St' to_str,'St' from_string from dual union all
  select 5 id,'Rd' to_str,'Rd.' from_string from dual union all
  select 6 id,'Rd' to_str,'road' from_string from dual union all
  select 7 id,'Av' to_str,'Av.' from_string from dual union all
  select 8 id,'Av' to_str,'Ave.' from_string from dual union all
  select 9 id,'Av' to_str,'Avenue' from_string from dual union all
  select 10 id,'Av' to_str,'Aven.' from_string from dual union all
  select 11 id,'West' to_str,'W' from_string from dual union all
  select 12 id,'South West' to_str,'SW.' from_string from dual
  ) t2
),
  r(addr,test_addr,l) as 
  (
      select  addr,regexp_replace(addr,'(^|\W)' || from_str || '(\W|$)','' || to_str || '') test_addr,
          id - 1
         from  
         addresses,
         replacements
         where id = (select count(*) from replacements)
      union all
         select addr,regexp_replace(addr,'(^|\W)' || from_str || '(\W|$)','' || to_str || '') test_addr,
          l - 1
         from  r,
         replacements
         where id = l
   )
  select addr,test_addr,l
  from  r
  where l=0
  ;

预期输出：

cust_id address

10      9 Help St, Level 4
11      22 Victoria St
12      1495 Franklin St
13      30 Hasivim St ,Petah-Tikva
14      2 Jakaranda St
15      61, Science Park Rd
16      61, Social park Rd
17      Av Hermanos Escobar 5756
18      Av Hermanos Escobar 5756
19      8000 West FLORISSANT Ave
20      8600 MEMORIAL PKWY South West

如果地址长度超过 35 个字符，则预期输出为：

cust_id address                              address2
21      8200 FLORISSANTMEMORIALWAYABOVE      South West
22      8600 MEMORIALFLORISSANT PKWY         South West

Answer 1

这是一种应用多个替换操作的...有趣的方式。所以 - 正如您提到的，您有 2 个问题。对于第一个，CTE 的递归部分是在 addr 上执行 regexp_replace() 而不是 test_addr（上一个递归步骤的修改输出）。因此，只会应用列表中的最后一条规则。

r(addr,test_addr, l) as 
(
    select  addr,regexp_replace(addr,'(^|\W)' || from_str || '(\W|$)','' || to_str || '') test_addr, 
        id - 1
       from  
       addresses,
       replacements
       where id = (select count(*) from replacements)
    union all
       -- if you do regexp_replace on addr, it throws out the previous replace (which is in r.test_addr)
       select addr,regexp_replace(test_addr,'(^|\W)' || from_str || '(\W|$)','' || to_str || '') test_addr, 
        l - 1
       from  r,
       replacements
       where id = l
 )

对于“超过 35 个字符”的问题，我建议使用 substr/instr - 虽然难以阅读，但它们通常很快。

select addr,test_addr,l, 
  case when length(test_addr) > 35 then
      substr(test_addr, 1, instr(substr(test_addr,1,35), ' ', -1))
    else test_addr
    end as addr1,
  case when length(test_addr) > 35 then
      substr(test_addr, instr(substr(test_addr,1,35), ' ', -1))
    else null
    end as addr2
from  r
where l=0
;

可能有更优雅的方式来完成这部分，这只是我想到的第一件事。

在 oracle sql 中使用 regexp_replace 标准化地址

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