在oracle中生成IP地址

我有一個數據庫，其中一個字段具有IP地址的值。默認值是12.123.60.165。從這些默認值我必須生成IP地址。最大值將是256.256.256.256。在oracle中生成IP地址

這是我的功能，仍然不能掩蓋我需要

/* Formatted on 9/26/2017 9:44:51 AM (QP5 v5.115.810.9015) */ 
SET SERVEROUTPUT ON 

DECLARE 
    IP INTEGER; 
    IP1 INTEGER; 
    IP2 INTEGER; 
    IP3 INTEGER; 
    IP4 INTEGER; 
    IPI INTEGER := 1; 
BEGIN 

    IP1 := FLOOR (IPI/POWER (2, 24)); 
    IPI := IPI - (IP1 * POWER (2, 24)); 
    IP2 := FLOOR (IPI/POWER (2, 16)); 
    IPI := IPI - (IP2 * POWER (2, 16)); 
    IP3 := FLOOR (IPI/POWER (2, 8)); 
    IPI := IPI - (IP3 * POWER (2, 8)); 
    IP4 := IPI; 

    DBMS_OUTPUT.PUT_LINE (IP1 || '.' || IP2 || '.' || IP3 || '.' || IP4); 
END; 
/

任何幫助將感激這麼多

來源

2017-09-26 Ras Rass

你怎麼從這些默認值我有可能產生IP地址是什麼意思？您是否只需要IP地址> 12.123.60.256的完整IP4地址範圍？不知道你在做什麼。 – xQbert

我的意思是我有IP號碼的默認值。它是12.123.60.165。然後我必須生成新的IP號碼，分別是12.123.60.167,12.123.60.168,12.123.60.169等等。直到以（256）12.123.60.256結尾。直到IP數量的最大IP數將是256.256.256.256。 –

生成4,213,638,225條記錄......這就是40億2.13億美元需要一些時間。但是有256個記錄的表的4個交叉連接似乎工作;而你的起始IP只會消除約14,612,400條記錄;或0.3456％不要指望它很快......但你可以用這些＃表創建一個表，我想...... – xQbert

要生成一組所有IPv4的人可以使用一個交叉連接和CTE與16條記錄和交叉連接。但是我還不確定你真的是什麼。

with cte (A) as (
SELECT 0 from dual UNION ALL 
SELECT 1 from dual UNION ALL 
SELECT 2 from dual UNION ALL 
SELECT 3 from dual UNION ALL 
SELECT 4 from dual UNION ALL 
SELECT 5 from dual UNION ALL 
SELECT 6 from dual UNION ALL 
SELECT 7 from dual UNION ALL 
SELECT 8 from dual UNION ALL 
SELECT 9 from dual UNION ALL 
SELECT 10 from dual UNION ALL 
SELECT 11 from dual UNION ALL 
SELECT 12 from dual UNION ALL 
SELECT 13 from dual UNION ALL 
SELECT 14 from dual UNION ALL 
SELECT 15 from dual), 
--Get data set for 0 to 255 numbers. 
Get256Range as (SELECT row_number() over (order by A.A)-1 seg 
    FROM CTE A 
    CROSS JOIN cte B) 
--Now cross join the 256 4 times for each class of Ip. 

SELECT A.Seg ||'.'|| B.Seg||'.'|| C.Seg||'.'||D.Seg as IPAddress 
from Get256Range A 
CROSS JOIN Get256Range B 
CROSS JOIN Get256Range C 
CROSS JOIN Get256Range D 
--Not sure what part of the "set" you need but one could change these values 
--Limited for sizing; but one could create a table of all IPv4 addresses... 
WHERE A.seg = 12 
    and B.seg = 123;

您可以根據需要包括前導零...

lpad(A.Seg,3,'0') ||'.'|| lpad(B.Seg,3,'0')||'.'|| lpad(C.Seg,3,'0')||'.'|| lpad(D.Seg,3,'0') IpAddressWithLeadZero

這讓人們可以進行排序或基於特定範圍過濾器

所以只能返回這些較大（不是最有效的）

WHERE lpad(A.Seg,3,'0') ||'.'|| lpad(B.Seg,3,'0')||'.'|| lpad(C.Seg,3,'0')||'.'|| lpad(D.Seg,3,'0') >='012.123.060.165'

可能更爲有效：

WHERE (A.Seg = 12 and B.Seg = 123 and C.Seg = 60 and D.Seg > 165) OR (A.Seg = 12 and B.Seg = 123 and C.Seg > 60) OR (A.Seg = 12 and B.Seg > 123) OR (A.Seg > 12)

來源

2017-09-26 13:47:31 xQbert

您可以使用DBMS_RANDOM包生成隨機值。您需要此軟件包的特權。例如：：

SQL> select trunc(dbms_random.value(12,256)) ip1, 
    2  trunc(dbms_random.value(123,256)) ip2, 
    3  trunc(dbms_random.value(60,256)) ip3, 
    4  trunc(dbms_random.value(165,256)) ip4 
    5* from dual 

    IP1  IP2  IP3  IP4 
---------- ---------- ---------- ---------- 
    156  163   78  231

或者在循環中生成。這樣的事情：

declare 
    ip1 int; 
    ip2 int; 
    cnt int := 0; 
begin 
    for i in 12..256 loop 
    ip1 := i; 
    for j in 123..256 loop 
     ip2 := j; 
    end loop; 
    end loop; 
end; 
/

來源

2017-09-26 03:52:44

dbms_random隨機產生IP。但我想要它隨機產生的。在結構化1,2,3,4,5 ...無論如何非常感謝你的回覆花花公子 –

然後我們可以使用循環。我在答案中添加了一個例子 –

在oracle中生成IP地址

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