Question

我有一种情况，我希望将地址拆分为Street Number和Street Name。

示例数据

1093 百线路
Flat 4442 Holly StreetAvondale
1401/142号公寓莎士比亚路
Unit K109 Northbridge 45 Akoranga Drive，Northcote

在这些示例中，粗体部分为Street Number，其余为Street Name。

到目前为止，我目前的努力并没有帮助我。有人能快速解决这个问题吗？

-Key是地址的Street Number部分，最后一次出现的数字结束。

Answer 1

如果字符串模式一致，您可以使用

select 
ltrim(rtrim(reverse(substring(reverse(address),patindex('%[0-9]%',reverse(address)),len(address))))) as streetNum,
ltrim(rtrim(reverse(substring(reverse(address),1,patindex('%[0-9]%',reverse(address))-1)))) as streetName
from tbl

使用patindex在反向字符串中查找第一个数字的出现位置，并使用substring和reverse将它们拆分为单独的字段。

Answer 2

示例

Declare @YourTable Table ([Id] varchar(50),[Addr] varchar(50)) Insert Into @YourTable Values (1,'1093 Hundred Line Road') ,(2,'Flat 4442 Holly StreetAvondale') ,(3,'Apartment 1401/142 Shakespeare Road') ,(4,'Unit K109 Northbridge 45 Akoranga Drive, Northcote') Select ID ,Addr1 = left(Addr,len(Addr)-patindex('%[0-9]%',reverse(Addr))+1) ,Addr2 = ltrim(right(Addr,patindex('%[0-9]%',reverse(Addr))-1)) From @YourTable

或只是为了好玩 - 使用交叉应用

Select ID ,Addr1 = substring(Addr,1,B.Pos) ,Addr2 = ltrim(substring(Addr,B.Pos+1,100)) From @YourTable Cross Apply (values (len(Addr)-patindex('%[0-9]%',reverse(Addr))+1)) B(Pos)

<强>返回

ID Addr1 Addr2 1 1093 Hundred Line Road 2 Flat 4442 Holly StreetAvondale 3 Apartment 1401/142 Shakespeare Road 4 Unit K109 Northbridge 45 Akoranga Drive, Northcote

我应该注意：

解析地址可能是一个滑坡。请考虑以下事项：Address standardization within a database

Answer 3

另一种选择......

IF OBJECT_ID('tempdb..#Address', 'U') IS NOT NULL 
DROP TABLE #Address;

CREATE TABLE #Address (
    StreetAddress VARCHAR(50) NOT NULL 
    );
INSERT #Address (StreetAddress) VALUES
    ('1093 Hundred Line Road'),
    ('Flat 4442 Holly StreetAvondale'),
    ('Apartment 1401/142 Shakespeare Road'),
    ('Unit K109 Northbridge 45 Akoranga Drive, Northcote');

--  SELECT * FROM #Address a;

--========================================================

SELECT 
    *,
    StreetNum = LEFT(a.StreetAddress, sl.SplitLocation),
    StreetName = SUBSTRING(a.StreetAddress, sl.SplitLocation + 1, 50)
FROM
    #Address a
    CROSS APPLY ( VALUES (PATINDEX('%[^0-9] [0-9]%', REVERSE(a.StreetAddress))) ) rs (ReverseSplit)
    CROSS APPLY ( VALUES (LEN(a.StreetAddress) - rs.ReverseSplit) ) sl (SplitLocation);

HTH，Jason

如何拆分字符串以取出SQL中的特定部分？

3 个答案: