Question

假设我有一个类似的地址：

model<-ksvm(STT~.,data=prediction1_train,kernel="linear",cost=10,scale=FALSE)

我希望输出像

2001:04E8:0000:4001:0000:0000:0000:0000/64

所以基本上，它会删除前导零，同时用一个零替换四个零。

2001:4E8:0:4001:0:0:0:0/64

上面的正则表达式完成了这项工作，但输出为

replaceAll("(:(0)*)|(^0+)",":")

这四个零都被省略了。

注意：输入可能是任何有效的ipv6地址

Answer 1

您可以使用REPLACE两次，SUBSTR删除前导0。

SELECT ipv6,
       SUBSTR(REPLACE(REPLACE(':' || ipv6, ':00', ':'),':0', ':'),2) AS result
FROM tab;

的 SqlFiddleDemo

演示正在使用MySQL，因为Oracle没有响应。唯一的变化是字符串连接。

输出：

╔═════════════════════════════════════════════╦════════════════════════════╗ ║ ipv6 ║ result ║ ╠═════════════════════════════════════════════╬════════════════════════════╣ ║ 2001:04E8:0000:4001:0000:0000:0000:0000/64 ║ 2001:4E8:0:4001:0:0:0:0/64 ║ ║ 0000:0001:0012:0123:1234:0000:0000:0000/64 ║ 0:1:12:123:1234:0:0:0/64 ║ ║ 0001:04E8:0000:4001:0000:0000:0000:0000/64 ║ 1:4E8:0:4001:0:0:0:0/64 ║ ║ 0012:04E8:0000:4001:0000:0000:0000:0000/64 ║ 12:4E8:0:4001:0:0:0:0/64 ║ ║ 0123:04E8:0000:4001:0000:0000:0000:0000/64 ║ 123:4E8:0:4001:0:0:0:0/64 ║ ╚═════════════════════════════════════════════╩════════════════════════════╝

Answer 2

使用Oracle的Regexp_Replace：

的示例

   SQL> with t as (
  2  select '2001:04E8:0000:4001:0000:0000:0000:0000/64' str from dual union all
  3  select '0000:0001:0012:0123:1234:0000:0000:0000/64' from dual union all
  4  select '0001:04E8:0000:4001:0000:0000:0000:0000/64' from dual union all
  5  select '0000:0001:0012:0123:1234:0000:0000:0000/64' from dual
  6  ) -- end of sample data
  7  select regexp_replace(str, '(:|^)(0{0,3})([1-9]*)', '\1\3') str
  8    from t;
STR
--------------------------------------------------------------------------------
2001:4E8:0:4001:0:0:0:0/64
0:1:12:123:1234:0:0:0/64
1:4E8:0:4001:0:0:0:0/64
0:1:12:123:1234:0:0:0/64

Answer 3

在java中，类似这样的东西：

replaceAll("(:000)|(:0*)", ":");

Answer 4

试试这个：

WITH t AS (
  SELECT '2001:04E8:0000:4001:0000:0000:0000:0000/64' str FROM dual UNION ALL
  SELECT '0000:0001:0012:0123:1234:0000:0000:0000/64' FROM dual UNION ALL
  SELECT '0001:04E8:0000:4001:0000:0000:0000:0000/64' FROM dual UNION ALL
  SELECT '0000:0001:0012:0123:1234:0000:0000:0000/64' FROM dual UNION ALL
  SELECT '2001:04E8:0000:4001:0000:0:0000:0/64' FROM dual
  )
SELECT 
    REGEXP_REPLACE(str, '(:|^)0+([[:xdigit:]]+)', '\1\2')
from t;

2001:4E8:0:4001:0:0:0:0/64
0:1:12:123:1234:0:0:0/64
1:4E8:0:4001:0:0:0:0/64
0:1:12:123:1234:0:0:0/64
2001:4E8:0:4001:0:0:0:0/64

正则表达式格式化IPv6地址 - 删除前导零

4 个答案: