SQL从字符串中提取值
问题描述:
如何从字符串中提取值?我试图分成3个新的列。城市,州和邮编的单独列。SQL从字符串中提取值
我已经试过
select address2,
left(address2, charindex('',address2)-1)
from table
和---当我尝试下面的代码,我得到“传递给左或子功能无效的长度参数”
,LTRIM(substring(a.Address2, CHARINDEX(' ', a.Address2)+1, CHARINDEX(' ', substring(a.address2, charindex(' ',
a.address2)+1, len(a.address2)))-1))
我可以打出来这个城市(West Warwick除外)使用下面的代码,但不知道如何使它适用于状态和zip。这也消除了错误。
SUBSTRING(Address2,1,CHARINDEX(' ', a.address2+ ' ')-1) as city
任何想法要尝试什么?
答
它看起来像你的邮政编码和您的国家都是一样的长度。如果这是真的,你应该能够使用这样的事情:
SELECT
LEFT(a.Address2,LEN(a.Address2) - 13) AS City,
RIGHT(LEFT(a.Address2,LEN(a.Address2) - 11),2) AS State,
RIGHT(a.Address2,10) AS Zip_Code
FROM
table;
演示代码
创建表和数据:
CREATE TABLE MyTable (Address2 VARCHAR(100));
INSERT INTO MyTable
VALUES
('SAN DIEGO CA 92128-1234'),
('WEST WARWICK RI 02893-1349'),
('RICHMOND IN 47374-9409');
查询:
SELECT
LEFT(Address2,LEN(Address2) - 13) AS City,
RIGHT(LEFT(Address2,LEN(Address2) - 11),2) AS State,
RIGHT(Address2,10) AS Zip_Code
FROM
MyTable;
输出:
答
既然你只有3件(城市/州/邮编),你可以采取在SQL Server 2008及更高版本称为parsename功能的优势。 (该功能的初衷是解析出对象名称。)
使用replace
和parsename
函数的组合可以使您能够将数据分成3部分,即使状态长度(不太可能)或Zip(更可能)更改。
实施例的数据:
create table #my_table
(
address2 varchar(75) not null
)
insert into #my_table values ('CONNERSVILLE IN 47331-3351')
insert into #my_table values ('WEST WARWICK RI 02893-1349')
insert into #my_table values ('RICHMOND IN 47374-9409')
insert into #my_table values ('WILLIAMSBURG IN 47393-9617')
insert into #my_table values ('FARMERSVILLE OH 45325-9226')
--this record is an example of a likely scenario for when the zip length would change.
insert into #my_table values ('WILLIAMSBURG IN 47393')
解决方案:
with len_vals as
(
select t.address2
, len(parsename(replace(t.address2,' ','.'), 1)) as zip_len
, len(parsename(replace(t.address2,' ','.'), 2)) as st_len
from #my_table as t
group by t.address2
)
select left(a.address2, len(a.address2) - b.zip_len - b.st_len - 2) as city
, substring(a.address2, len(a.address2) - b.zip_len - 2, b.st_len) as st
, right(a.address2, b.zip_len) as zip_code
from #my_table as a
inner join len_vals as b on a.address2 = b.address2
结果:
我必须有where子句吗?我仍然收到“传递给LEFT或SUBSTRING函数的无效长度参数”。 – BIReportGuy
我确实更新了'State'逻辑 - 我被关闭了一个。如果仍然出现错误,那就意味着有一些记录'state'或'zip'(或两者)的长度与您提供的示例数据中的长度不相同。 – Nicarus
这似乎消除了错误,但我只知道如何使它适用于一列。 (West Warwick除外)SUBSTRING(Address2,1,CHARINDEX('',address2 +'')-1)作为城市 – BIReportGuy