地址字符串到地址字段 VIEW 或 SELECT答案

【问题标题】：Address String to Address Fields VIEW or SELECT地址字符串到地址字段 VIEW 或 SELECT
【发布时间】：2015-05-28 00:45:25
【问题描述】：

我有一个单行的地址字段，如下所示：

Dr Robert Ruberry, West End Medical Practice, 38 Russell Street, South Brisbane 4101

我想编写一个视图，将该地址拆分为 Name、Addr1、Addr2、Suburb、Postcode 字段以用于报告目的。

我一直在尝试像这样使用 SUBSTRING 和 CHARINDEX，但它似乎没有正确拆分它。

SUBSTRING([address_Field],CHARINDEX(',',[address_Field]),CHARINDEX(',',[address_Field]))

有人可以帮忙吗？ TIA

【问题讨论】：

标签： sql sql-server tsql street-address

【解决方案1】：

这可能适合您的要求

IF OBJECT_ID('tempdb..#test') IS NOT NULL
    DROP TABLE #test
CREATE TABLE #test(id int, data varchar(100))

INSERT INTO #test VALUES (1,'Dr Robert Ruberry, West End Medical Practice, 38 Russell Street, South Brisbane 4101')


DECLARE @pivot varchar(8000)
DECLARE @select varchar(8000)

SELECT 
        @pivot=coalesce(@pivot+',','')+'[col'+cast(number+1 as varchar(10))+']'
FROM 
        master..spt_values where type='p' and 
        number<=(SELECT max(len(data)-len(replace(data,',',''))) FROM #test)

SELECT 
    @select='
        select p.col1 As Name,p.col2 as Addr1,p.col3 as Addr3,p.col4 as Postcode
        from (
        select 
            id,substring(data, start+2, endPos-Start-2) as token,
            ''col''+cast(row_number() over(partition by id order by start) as varchar(10)) as n
        from (
            select 
                id, data, n as start, charindex('','',data,n+2) endPos
                from (select number as n from master..spt_values where type=''p'') num
                cross join 
                (
                    select 
                        id, '','' + data +'','' as data 
                    from 
                        #test
                ) m
            where n < len(data)-1
            and substring(data,n+1,1) = '','') as data
        ) pvt
        Pivot ( max(token)for n in ('+@pivot+'))p'

EXEC(@select)

【讨论】：

【解决方案2】：

这里有几个选项供您选择。如果您只是在寻找快速答案，请查看已回答的类似问题：

T-SQL split string based on delimiter

如果您想更深入地了解各种选项，请查看以下内容：

http://sqlperformance.com/2012/07/t-sql-queries/split-strings

【讨论】：

【解决方案3】：

此答案不适用于 SQL，但确实适用于 街道地址。

如果您愿意依赖第三方，您可以将街道地址发送到SmartyStreets International Street API 服务。

为此，您提交一个HTTP GET 请求。

您的示例地址如下所示：

curl -v 'https://international-api.smartystreets.com/verify?
    auth-id=YOUR+AUTH-ID+HERE&auth-token=YOUR+AUTH-TOKEN+HERE&
    address1=Dr%20Robert%20Ruberry%2C%20West%20End%20
        Medical%20Practice%2C%2038%20Russell%20Street%2C%20
        South%20Brisbane%204101
    &country=aus'

（请注意，地址是 url 编码的。请求被包装以提高可读性。）

响应将是一个JSON 字符串，分成多个组件，然后您可以根据需要插入到数据库中：

[
    {
        "organization": "Dr Robert Ruberry, West End Med.",
        "address1": "Dr Robert Ruberry, West End Med.",
        "address2": "Russell Street",
        "address3": "38 Practice",
        "address4": "South Brisbane QLD 4101",
        "components": {
            "administrative_area": "QLD",
            "building": "Russell Street",
            "country_iso_3": "AUS",
            "locality": "South Brisbane",
            "postal_code": "4101",
            "postal_code_short": "4101",
            "thoroughfare": "Practice",
            "thoroughfare_name": "Practice",
            "sub_building_number": "38"
        },
        "metadata": {},
        "analysis": {
            "verification_status": "Partial",
            "address_precision": "Locality",
            "max_address_precision": "DeliveryPoint"
        }
    }
]

另一个好处是该服务为您提供有关地址有效性的额外信息。

（披露：我在 SmartyStreets 工作。）

【讨论】：