Question

我有一张看起来像这样的表

id        carrier
A1   |  66.87.151.2,sprint pcs,2015-05-21,2015-05-21,66.87.151.145,sprint pcs,2015-05-21,2015-05-21,66.87.150.131,sprint pcs,2015-05-13,2015-05-13
     |
B1   |  67.83.18.128,optimum online,2015-05-09,2015-05-09,b8bcdb64-72c2-4578-9db5-c011263b1180  69.204.80.158,time warner cabl,
     |
C1   |  76.180.4.64,time warner cable,2015-07-01,2015-07-29,66.87.137.65,sprint pcs

我想返回一个看起来像这样的表

id     carrier
A1 |  sprint pcs
A1 |  sprint pcs
A1 |  sprint pcs
B1 |  optimum online
B1 |  optimum online
C1 |  time warner cable
C1 |  sprint pcs

我唯一能想到的是应用正则表达式运算符

regex_replace（carrier，'[^ a-z]'，''）或类似的东西，

但我没有运气。

Answer 1

将文本列转换为数组，取消它并检查正则表达式的元素：

select id, elem
from (
    select id, unnest(string_to_array(carrier, ',')) elem
    from the_table
    ) sub
where elem ~ '^[a-z ]+$';

 id |       elem        
----+-------------------
 A1 | sprint pcs
 A1 | sprint pcs
 A1 | sprint pcs
 B1 | optimum online
 B1 | time warner cabl
 C1 | time warner cable
 C1 | sprint pcs
(7 rows)

删除所有数字和符号并创建新行

1 个答案: