请开发一个这样的实用的TOOLS，各位自认称职的SYS/NET ADMIN or Developer 请进

hillman(山大王)

目的是输入一个ip地址或地址段
然后在一个保存有许多IP地址和地址段的文本文件中查找
看此地址/地址段是否<b>包含</b>在其中

比如以下是保存有许多IP地址和地址段的文本文件 — ips.txt
182.3.58.0/23
192.168.172.0/22
127.0.0.0
188.54.0.0/15
...
...

输入
127.0.0.0
输出
Found: 127.0.0.0
输入
127.0.0.1
输出
Not Found.

输入
182.3.59.223
输出
Found:182.3.58.0/23
输入
182.3.59.8/27
输出
Found:182.3.58.0/23
.....
.....

编程的语言不限
(最好是 ShellScript, PERL or Java ). 跨平台嘛
结构越简单越好（算法不限）

挺有用也挺有趣吧。
做Sys Net Admin 的，别丢脸哦

(#875221@0)
Last Updated: 2002-11-25
This post has been archived. It cannot be replied.

俺不是什么啊命，但是你这个问题确实简单得不得了 -interview(intervieweree); 2002-11-25 {40} (#875284@0)

我现在不想马上告诉你，否则的话你印象不深

use NT/2000 script, it is not big deal. -bjduck(北京烤鸭); 2002-11-25 (#875362@0)

here is one implementation. It is really simple. Why don't you do it yourself?? -lusi(丑小鸭); 2002-11-25 (#875575@0)

I am not challenging anyone. Programming a simple tool is full of FUN and good exercise to your brain, right? If you Wrote that tool, could you share your source code here. Thanks. -hillman(山大王); 2002-11-25 (#875750@0)

You should really think about it yourself since this is so so so simple. -jeffrey815(Smartiecat); 2002-11-25 (#875753@0)
don't know why you don't develop it by yourself. As Smartiecat said, it is so so simple. Anyway, go to the same place for the source code in VC++ 6.0. -lusi(丑小鸭); 2002-11-25 (#875756@0)

Hahaaha, the code is 'so simple so naive :)' but wrong -hillman(山大王); 2002-11-25 {691} (#875777@0)
It is what in the VC++ program to do the search:

while ( m_ipFile.ReadString(lineBuffer) )
{
if ( lineBuffer.GetLength() < 1 )
continue;

nPos = lineBuffer.Find(m_ip, 0);
if ( nPos != -1 )
{
lineBuffer.Insert(0, "Found:");
m_resultList.AddString(lineBuffer);
break;
}
}
if ( nPos == -1 )
{
lineBuffer.Format("%s : Not Found", m_ip);
m_resultList.AddString(lineBuffer);
}

PLEASE read my descriptions of the question
Can it find
182.3.59.8/27
and show output of
182.3.58.0/23??

It is simple, but not as simple as your program. :)

I have no knowledge of system admin, and I don't understand the last example of your requirement, the input 182.3.59.8/27 and output 182.3.58.0/23. -lusi(丑小鸭); 2002-11-25 {102} (#875831@0)
To me, they are totally different two strings and the result should be: "Not Found". Please comment.

Sorry for confusing you. 182.3.59.96/27 means ip block 182.3.59.96 netmask bits length is 27 bits long ( 255.255.255.224 ). ALL Network vendors now use this way to descript IP blocks. -hillman(山大王); 2002-11-25 (#875847@0)

I see. So the real requirement is: if netmask bits is specified by "/##", find out the entry which has the same netmask. Otherwise, find the exact entry. Is my understanding correct? -lusi(丑小鸭); 2002-11-25 (#875916@0)

partially correct. but still not that Simple :). For example: IP/IP Blocks "182.3.59.96/27 , 182.3.59.214, 182.3.59.8/28 ....." are all in the IP block 182.3.58.0/23. Not as simple as you thought. right? :) -hillman(山大王); 2002-11-25 (#875923@0)

one question before I continue: /23 means the first 23 bits in the IP address, right? -lusi(丑小鸭); 2002-11-25 (#875987@0)

Right. TCP/IP fundamental -hillman(山大王); 2002-11-25 (#875997@0)

the first 23 bits starting from MSB. -lusi(丑小鸭); 2002-11-25 (#875990@0)
version 1.1 done. Please go to the same location for the exe. Check whether it meets the requirement. I read about the netMask long ago, and almost forgot it already. -lusi(丑小鸭); 2002-11-25 (#876133@0)

It is OK, could you show the source code. just want to see the algorithm. A minor flaw in your program, it should show ALL the matched entries. But that is OK. -hillman(山大王); 2002-11-25 (#876271@0)

here you go. It doesn't show all the matches because I return once a match is found. You want all the matches? -lusi(丑小鸭); 2002-11-25 {3267} (#876275@0)
本文发表在 rolia.net 枫下论坛void CIpSearchDlg::OnSearchBtn()
{
// to do a search, a ip address file must have been opened,
// and the ip address to be searched must have been keyed in
UpdateData(TRUE);

BYTE inField1, inField2, inField3, inField4, inMaskBits;
DWORD inIp, inMask;

BYTE field1, field2, field3, field4, maskBits;
DWORD Ip, Mask;

DWORD MaskWhenCompare;

int nRet;

if ( m_ip.IsEmpty() == TRUE )
{
MessageBox("Specify the IP address to be searched.",
NULL, MB_ICONEXCLAMATION | MB_OK );
return;
}
else
{
nRet = sscanf(m_ip, "%d.%d.%d.%d/%d",
&inField1, &inField2, &inField3, &inField4, &inMaskBits);
if ( nRet < 4 )
{
MessageBox("The specified IP address is not valid.",
NULL, MB_ICONEXCLAMATION | MB_OK);
return;
}

if ( nRet == 5 )
{
if ( inMaskBits > 32 )
{
MessageBox("The specified IP address mask is not valid.",
NULL, MB_ICONEXCLAMATION | MB_OK);
return;
}
// netmask is specified
inMask = MakeNetMask(inMaskBits);
}
else
inMask = 0;

inIp = MakeDWord(inField1, inField2, inField3, inField4);
}

if ( m_ipFile.m_hFile == CFile::hFileNull )
{
MessageBox("Specify the IP address file to search in.",
NULL, MB_ICONEXCLAMATION | MB_OK );
return;
}

// do the string search
BeginWaitCursor();

m_ipFile.SeekToBegin();
CString lineBuffer;
BOOL found = FALSE;
while ( m_ipFile.ReadString(lineBuffer) )
{
if ( lineBuffer.GetLength() < 1 )
continue;

nRet = sscanf(lineBuffer, "%d.%d.%d.%d/%d",
&field1, &field2, &field3, &field4, &maskBits);
if ( nRet < 4 )
continue;

if ( nRet == 5 )
{
if ( maskBits > 32 )
continue;

Mask = MakeNetMask(maskBits);
}
else
Mask = 0;

Ip = MakeDWord(field1, field2, field3, field4);

if ( inMask || Mask )
{
if ( inMask && Mask )
MaskWhenCompare = inMask > Mask? Mask : inMask;
else if ( inMask )
MaskWhenCompare = inMask;
else
MaskWhenCompare = Mask;
}
else
MaskWhenCompare = MakeNetMask(32); // this can be made as a constant global
// for a better speed performance

if ( (Ip & MaskWhenCompare) == (inIp & MaskWhenCompare) )
{
CString displayMsg;
displayMsg = m_ip + " ==> Found: " + lineBuffer;
m_resultList.AddString(displayMsg);
found = TRUE;
break;
}
}
if ( !found )
{
lineBuffer.Format("%s ==> Not Found", m_ip);
m_resultList.AddString(lineBuffer);
}

m_resultList.SetCurSel(m_resultList.GetCount() - 1);

EndWaitCursor();
}更多精彩文章及讨论，请光临枫下论坛 rolia.net

It is good. The algorithm is quite straight forward. Just another minor flaw: -hillman(山大王); 2002-11-25 {504} (#876369@0)
If I input 127.0.0.0/8
it will show: Found-> 127.0.0.1
It is not correct
because 127.0.0.0/8 is not INCLUDED in 127.0.0.1
(127.0.0.1 is included in 127.0.0.0/8)
That's ok. easy to fix

But any way, the algorithm is quite straight forward

Seems you used quite a lot of Subs like MakeDWord
MakeSubMask etc. life don't need to be so complicated. ;)

And I also learned the sscanf as C++'s compromise of Regular Expressions. Haha, Thanks.

I will wait for other more simpler solutions..

You are the ONLY person here want to do something instead of saying 'Oh, simple' then hide... :D -hillman(山大王); 2002-11-25 (#876019@0)

i read what u say and tested some. you are right, it is not easy. -bjduck(北京烤鸭); 2002-11-25 {161} (#876098@0)
i am trying use script, but not easy. it is seemed that it is not system admin job, more like programer. so script is not very easy for that. anyway, i will try.

scripts CAN do it, try try :) -hillman(山大王); 2002-11-25 (#876115@0)

by the way, for i am NT/2000 admin. so i am using VBScript not ShellScript. -bjduck(北京烤鸭); 2002-11-25 (#876147@0)

Kixtart -ssi(文文); 2002-11-25 (#875647@0)

this is too simple! -macsym(日进斗斤); 2002-11-25 (#875789@0)

Could anyone give a script less than 25 lines to solve this simple question?? Come On Admins. GaoShou should show up now la... -hillman(山大王); 2002-11-25 (#876502@0)

IT人是不是“贱－不值钱”了一下啊，这种话题也有这么多人回复！ -arthurxu(ArthurXu OCP8,9i); 2002-11-25 (#876564@0)

Take easy, take easy. Just for Fun, not for 贱. Heehee. If you are good, could you give a solution in SQL? Pls don't be so aggressive -hillman(山大王); 2002-11-25 (#876734@0)

Pay me $100, I will do a perfect application to you -arthurxu(ArthurXu OCP8,9i); 2002-11-27 (#879703@0)

My clumsy perl implementation ... all most no error checking ... with some effort you should be able to get it under 25 lines (currently ~40 lines). -dennis2(Dennis); 2002-11-26 {1073} (#877439@0)

本文发表在 rolia.net 枫下论坛<pre>
===============
#!/usr/bin/perl

while (<>) {
$low = limit(1, getipmask($_));
$high = limit(0, getipmask($_));
open(MYIPS,"./ips.txt") || die "open err\n";
while (<MYIPS>) {
if ($low >= limit(0, getipmask($_)) && $high <= limit(1, getipmask($_))) {
print "Found: ", $_, "\n";
}
}
close MYIPS;
}

sub limit {
($highlow, $mask, @octet) = @_;
my $ip = 0;
for ($i=0; $i<4; $i++) {
$ip += $octet[$i]*256**(3-$i);
}
$ip = $ip & (2 ** 32 - 2 ** (32-$mask));
if ($highlow == 1) {
return $ip + 2 ** (32-$mask) - 1;
} else {
return $ip;
}
}

sub getipmask {
chomp;
if ($_ =~ /\//) {
($myip, $mymask) = split (/\//, $_);
} else {
$myip = $_;
$mymask = 32;
}
return($mymask, split (/\./, $myip));
}
===============
</pre>更多精彩文章及讨论，请光临枫下论坛 rolia.net

It is also good. but not better than 丑小鸭's. his/her algorithm is simpler and more efficient. Heehee. Is it a good exercise? -hillman(山大王); 2002-11-26 (#877948@0)

i did it in VbScript , but about 80 lines.But i can not put into 20 ines. i gave up. -bjduck(北京烤鸭); 2002-11-26 (#877560@0)

Let me conclude the currently BEST algorithm -hillman(山大王); 2002-11-26 {731} (#878102@0)

1. Get ip and netmask from input
if no NetMask then NetMask = 32

2. ipnumber = ip1*16777216 + ip2 * 65536 + ip3 * 256 + ip4
(this is 丑小鸭's MakeDWord and Denis2's FOR Loop )

3. get InFileIPnumber and InFileMask from the file = 'ips.txt'
using the same why above
if InFileMask > NetMask then show "not found"
(both of you two forgot this la. actually you don't need to calculate
InFileMask > NetMask :)

4. ipnumber = ipnumber / 2 ** ( 32 - InFileMask )
inFileIPNumber = inFileIPNumber / 2 ** ( 32- InFileMask )
if (ipnumber == ipFileIPNumber) then show "FOUND!"

I think it is the simplest algorithm.
So language is not important. The way to do it really is.

Enjoy. :-)

不要把简单复杂化。 -pagecho(later); 2002-11-27 {38} (#879841@0)
你这个东东grep一下就完了，还要算法吗？

PLEASE read the question carefully before you say something. DON'T be TOO confident... hehe -hillman(山大王); 2002-11-27 (#880540@0)
if you can grep it. You must be Grep Expert. ;-) . But obvious you are not... -hillman(山大王); 2002-11-27 (#880541@0)

@Florida

请开发一个这样的实用的TOOLS，各位自认称职的SYS/NET ADMIN or Developer 请进

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