如何从一列中提取 postcode/digits 并将其放入 SQL Oracle 9i 中的另一列

Question

Table ADRESSES 包含列 [ ID, STREET, TOWN, POSTCODE,COUNTRY ].

Oracle 9i 数据库中的热门语句：

如果第 COUNTRY 列包含字符串 'UK'，并且 POSTCODE 为空，但第 TOWN 列以 post 代码开头（格式为 xxxxx 或 xx 的数字-xxx)，然后将 postal 代码移动到 POSTCODE 列，并从 TOWN.

列中删除 post 代码

Answer 1

下面有两条sql语句，一条到select，看看update语句的影响，update语句。在运行任何更新之前备份您的表，并始终通过运行将其作为 select 语句来测试更新是否按照您的预期进行。

注意：我假设您希望将邮政编码保留为最初输入的格式。例如，如果输入为 xxxxx，它将保存为 xxxxx，如果输入为 xx-xxx，它将保存为 xx-xxx。

检查更新语句将做什么：

select ID, STREET, TOWN, POSTCODE, COUNTRY, 
case when substr(TOWN, 3, 1) = '-' like '%-%' then trim(substr(TOWN,7, length(TOWN)-6)) else trim(substr(TOWN,6, length(TOWN)-5)) end as NEW_TOWN,
case when substr(TOWN, 3, 1) = '-' like '%-%' then substr(TOWN, 1, 6) else substr(TOWN, 1, 5) end as NEW_POSTCODE--Assumes you want to keep the dash if it exists
from ADDRESSES
where COUNTRY like'%UK%' --contains string UK
and trim(POSTCODE) is null -- postcode is empty
and (
    length(trim(translate(substr(TOWN, 1, 5), '0123456789', ' '))) is null -- town starts with xxxxx digits
    or
    (length(trim(translate(substr(TOWN, 1, 2)||substr(TOWN,4,3, '0123456789', ' '))))  and  substr(TOWN, 3, 1) = '-') -- town starts with xx-xxx digits
    )
;

如果满意，运行更新语句。

update ADDRESSES
set
TOWN = case when substr(TOWN, 3, 1) = '-' like '%-%' then trim(substr(TOWN,7, length(TOWN)-6)) else trim(substr(TOWN,6, length(TOWN)-5)) end,
POSTCODE = case when substr(TOWN, 3, 1) = '-' like '%-%' then substr(TOWN, 1, 6) else substr(TOWN, 1, 5) end --Assumes you want to keep the dash if it exists
from ADDRESSES
where COUNTRY like'%UK%' --contains string UK
and trim(POSTCODE) is null -- postcode is empty
and (
    length(trim(translate(substr(TOWN, 1, 5), '0123456789', ' '))) is null -- town starts with xxxxx digits
    or
    (length(trim(translate(substr(TOWN, 1, 2)||substr(TOWN,4,3, '0123456789', ' '))))  and  substr(TOWN, 3, 1) = '-') -- town starts with xx-xxx digits
    )
;

我希望这至少可以作为一个起点。

Answer 2

我也没有 O9i 来测试这个，所以我怀疑我的回答将毫无用处，但我可以尝试这样做只是为了好玩。我希望以下内容能给您或其他人一个起点。以下接缝在 postgresql 9.3 中工作:-)

create table addresses(
    id int,
    street varchar(255),
    town varchar(255),
    postcode varchar(6),
    country varchar(3)
);

insert into addresses values (1, '1', '11-111 dupa jasio1','', 'UK');
insert into addresses values (2, '2', '22222 dupa jasio2','', 'UK');
insert into addresses values (3, '3', '33-333 dupa jasio3','', 'UK');
insert into addresses values (4, '4', '44444 dupa jasio4','', 'UK');
insert into addresses values (5, '5', '55555 dupa jasio5','', 'PL');

dawid=# select * from addresses ;
 id | street |        town        | postcode | country 
----+--------+--------------------+----------+---------
  1 | 1      | 11-111 dupa jasio1 |          | UK
  2 | 2      | 22222 dupa jasio2  |          | UK
  3 | 3      | 33-333 dupa jasio3 |          | UK
  4 | 4      | 44444 dupa jasio4  |          | UK
  5 | 5      | 55555 dupa jasio5  |          | PL

查询以更新类似于 12345 的邮政编码

update addresses set 
postcode=substring(town, '^([0-9]{5})'),
town=substring(town, '^[0-9]{5}\s(.*)')
where country like '%UK%' and 
(postcode is null or postcode='') and 
town ~ E'^[0-9]{5} ';

查询以更新类似于 12-123 的邮政编码

update addresses set 
postcode=substring(town, '^([0-9]{2}\-[0-9]{3})'),
town=substring(town, '^[0-9]{2}\-[0-9]{3}\s([\w\d\-\s]+)')
where country like '%UK%' and 
(postcode is null or postcode='') and 
town ~ E'^[0-9]{2}\-[0-9]{3} ';

结果

dawid=# select * from addresses ;
 id | street |       town        | postcode | country 
----+--------+-------------------+----------+---------
  5 | 5      | 55555 dupa jasio5 |          | PL
  2 | 2      | dupa jasio2       | 22222    | UK
  4 | 4      | dupa jasio4       | 44444    | UK
  1 | 1      | dupa jasio1       | 11-111   | UK
  3 | 3      | dupa jasio3       | 33-333   | UK

如何从一列中提取 postcode/digits 并将其放入 SQL Oracle 9i 中的另一列

How to extract postcode/digits from one column and put it in to another column in SQL Oracle 9i

regex

sql

substring

trim

oracle9i