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View Full Version : Sorting Ip address within array



sephlaire
01-25-2012, 06:41 PM
I am trying to make an addition to a .js plugin for a wiki. Currently it does not support sorting my IP address. So far I have created the below code which does more than the script initially did, but is still not sorting 100% correctly.

Any tips?


sort_ipaddr: function(a,b){
aa = a[0].split(".",4);
bb = b[0].split(".",4);
var counti = 0;

for (var i=0; i<4; i++) {
if (parseInt(parseFloat(aa[i])) == parseInt(parseFloat(bb[i]))){}
if (parseInt(parseFloat(aa[i])) < parseInt(parseFloat(bb[i]))){counti--}
if (parseInt(parseFloat(aa[i])) > parseInt(parseFloat(bb[i]))){counti++}
}
return counti;
},

EDIT

I've also tried this which is closer but still not there.

sort_ipaddr: function(a,b){
aa = a[0].split(".",4);
bb = b[0].split(".",4);

var resulta = (aa[3]+(aa[2]*256)+(aa[1]*256*256)+(aa[0]*256*256*256));
var resultb = (bb[3]+(bb[2]*256)+(bb[1]*256*256)+(bb[0]*256*256*256));

return resulta-resultb;
},
This results in a list like so:
10.1.15.22
10.1.16.22
10.1.15.23
10.1.15.24
10.1.16.24
10.1.15.25

Old Pedant
01-25-2012, 08:19 PM
Your second one is close, but the very first addition in each resultX calculation will be a string concatenation instead of arithmetic add. So that means that you end up comparing something like "24xxxxx" vs. "25xxxxx" and get the results you are seeing.

Keep it simple:


sort_ipaddr: function(a,b){
aa = a[0].split(".");
bb = b[0].split(".");

var resulta = aa[0]*0x1000000 + aa[1]*0x10000 + aa[2]*0x100 + aa[3]*1;
var resultb = ab[0]*0x1000000 + ab[1]*0x10000 + ab[2]*0x100 + ab[3]*1;

return resulta-resultb;
},

sephlaire
01-25-2012, 08:32 PM
--EDIT Removed--

I spoke too soon! Works perfectly thank you. My mistake was that in the split I used variables aa and bb. In the resultsx line aa and ab were used as variables.

Old Pedant
01-25-2012, 08:57 PM
Oops...I missed that, too. But glad it works now.

If you didn't know, multiplying by anything (including 1) is one way to force the conversion of the string you get from split to become a number.

Philip M
01-26-2012, 09:07 AM
If you didn't know, multiplying by anything (including 1) is one way to force the conversion of the string you get from split to become a number.

I thought you considered that to be a hack! :eek:

Old Pedant
01-26-2012, 08:37 PM
LOL! Depends on usage. Here, the whole point was to multiply each part of the IP address by the apprpriate number. I certainly wouldn't do parseInt(ip[0[) * 0x1000000 so why not use the same pattern throughout?

Beside, if it really is an IP address we *know* it will be a number. No detection of invalid values needed, one hopes.

wesleyren
11-23-2014, 09:49 PM
This thread helped me to come up with this to sort an array of IP addresses:
myIpArray.sort(function(a,b){ // sort IP address.
aa = getIP(a).split(".");
bb = getIP(b).split(".");

for (var i=0, n=Math.max(aa.length, bb.length); i<n; i++) {
if (aa[i] !== bb[i]) return aa[i] - bb[i];
}
return 0;
});

wesleyren
11-23-2014, 09:58 PM
Better format


myIpArray.sort(function(a,b){ // sort IP address.
aa = getIP(a).split(".");
bb = getIP(b).split(".");

for (var i=0, n=Math.max(aa.length, bb.length); i<n; i++) {
if (aa[i] !== bb[i]) return aa[i] - bb[i];
}

return 0;
});

Old Pedant
11-23-2014, 10:21 PM
Yes, that works, but However, I would *bet* that my code, to convert the strings into numbers, would be faster because it would avoid the overhead of the loop.

In other words:

myIpArray.sort(
function( a, b )
{
var aa = getIP(a).split(".");
var bb = getIP(b).split(".");

return ( aa[0]*0x1000000 + aa[1]*0x10000 + aa[2]*0x100 + aa[3]*1 )
- ( bb[0]*0x1000000 + bb[1]*0x10000 + bb[2]*0x100 + bb[3]*1 );
}
);

However, since you are supplying that getIP() function [I have no idea what it is doing...I assume it is somehow extracting the IP address from the given object variable], why not *also* supply a getIpNumber( ) function that indeed returns the ip address *as* an integer number?

So then the code could be as simple as

myIpArray.sort( function( a, b ) { return getIpNumber(a) - getIpNumber(b); } );

Old Pedant
11-23-2014, 10:24 PM
And as a minor point: If you are using a MySQL DB, you can easily ask for a sort by IP address in your SQL code:

SELECT * FROM sometable ORDER BY INET_ATON( ipaddress )

Probably not relevant to this thread, but something to keep in mind.



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