我想把地址分成两列，如streetno和streetname，比如说ex 从客户
中选择地址1前身的地址
2719 STONE CREEK DR
并将其存储为街道号码2719和街道名称STONE CREEK DR。
使用split(address1 ,'[\ ]')[0]不仅仅拆分了街道。
它只是我们用来查看数据的select语句。
示例数据：
address1
100 HORACE GREELEY RD
55 School Street
2893 MASHIE CIR
1200 JEWEL DR
201 W RIVER RD
预期输出
+--------------+---------------------+
| streetnumber | streetname |
+--------------+---------------------+
| 100 | HORACE GREELEY RD |
| 55 | School Street |
| 2893 | MASHIE CIR |
| 1200 | JEWEL DR |
| 201 | W RIVER RD |
+--------------+---------------------+
答案 0 :(得分：0)
select regexp_extract(address1,'(.*?)\\s',1) as streetnumber
,regexp_extract(address1,'\\s(.*)' ,1) as streetname
from mytable
;
+----------------+--------------------+
| streetnumber | streetname |
+----------------+--------------------+
| 100 | HORACE GREELEY RD |
| 55 | School Street |
| 2893 | MASHIE CIR |
| 1200 | JEWEL DR |
| 201 | W RIVER RD |
+----------------+--------------------+
select regexp_extract(address1,'.*?(?=\\s)',0) as streetnumber
,regexp_extract(address1,'(?<=\\s).*',0) as streetname
from mytable
;
+----------------+--------------------+
| streetnumber | streetname |
+----------------+--------------------+
| 100 | HORACE GREELEY RD |
| 55 | School Street |
| 2893 | MASHIE CIR |
| 1200 | JEWEL DR |
| 201 | W RIVER RD |
+----------------+--------------------+
select split(address1,'\\s')[0] as streetnumber
,split(address1,'^.*?\\s')[1] as streetname
from mytable
;
+----------------+--------------------+
| streetnumber | streetname |
+----------------+--------------------+
| 100 | HORACE GREELEY RD |
| 55 | School Street |
| 2893 | MASHIE CIR |
| 1200 | JEWEL DR |
| 201 | W RIVER RD |
+----------------+--------------------+